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Karnataka 1st PUC Kannada Textbook Answers Sahitya Sanchalana Chapter 4 Halubidal Kalmaram Karaguvante

Halubidal Kalmaram Karaguvante Questions and Answers, Notes, Summary

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Karnataka 1st PUC Chemistry Question Bank Chapter 6 Thermodynamics

1st PUC Chemistry Thermodynamics One Mark Questions and Answers

Question 1.
What is thermodynamics ?
Answer:
Thermodynamics is a branch of chemistry that deals with the study of interconversion of with other forms energy during physical and chemical changes.

Question 2.
What you mean by enthalpy?
Answer:
The total heat content of a system is called enthalpy.

Question 3.
Write the relationship between ∆H and ∆E.
Answer:
∆H = ∆E + RT∆n, ∆H = Enthalpy change, ∆E = Internal energy change
∆n = Number of gaseous product number of gaseous reactant or number of moles.

Question 4.
Define (Heat of reaction) Enthalpy of reaction.
Answer:
Heat of a reaction is the change in enthalpy produced when the number of moles of the reactants as represented in the balanced chemical equation have completely reacted with each other.

Question 5.
Heat (enthalpy) of formation.
Answer:
Heat of formation of a compound is the change in enthalpy produced when one mole of the compound is formed from its elements denoted as ∆H_f.

Question 6.
What is Heat Capacity ?
Answer:
Het capacity of a system is defined as quantity of heat required to raise temperature by 1°.

Question 7.
What is specific heat capacity ?
Answer:
It is defined as the quantity of heat required to raise temperature of 1 gram of a substance by 1° C or 1K

Question 8.
What is the relation between C_P and C_v for an ideal gas.
Answer:
C_P– C_v = R

Question 9.
What is entropy ?
Answer:
A measure of degree of disorder of a system.

Question 10.
What is the unit of entropy ?
Answer:
J / K (Joule per Kelvin)

Question 11.
What is Gibbs-Helmholtz equation ?
Answer:
∆G° = ∆H° – T∆S°.

Question 12.
Give an expression for the work done in a reversible isothermal expansion of an ideal gas.
Answer:
w = -2.303nRT log \(\frac{V_{2}}{V_{1}}\)

Question 13.
Explain : Heat of formation a acetylene is + 54 kcal.
Answer:
When 1 mole of gaseous acetalylene is formed from its elements, carbon (s) and hydrogen (g), 54 kcal of heat is absorbed.

Question 14.
White phosphorous is less stable than red phosphorous. Mention whether the process of conversion of white phosphorous to red phosphorous is exothermic or endothermic reaction.
Answer:
Exothermic reaction.

Question 15.
What is the value of the standard enthalpy of formation of an element ?
Answer:
Zero

Question 16.
The standard enthalpies of formation of Hydrogen fluoride and Hydrogen chloride are -268.6 kJ and -92.3 kJ respectively. Which is between the two is stable?
Answer:
Hydrogen fluoride is more stable.

Question 17.
Give an example for a reaction in which AH = AE
Answer:
C(s) + O₂(g) → CO₂(g)

Question 18.
What is the meaning of the statement the enthalpy of formation of PCl₃ is -373 kJ ?
Answer:
When 1 mole of PCl₅ is formed from its elements, P(s) and Cl₂(g), 373 kJ of heat is liberated.

Question 19.
Define ‘standard enthalpy of formation’.
Answer::
Standard enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states under standard conditions i.e., at 298 K and 101.3 kPa pressure.

Question 20.
Define enthalpy of combustion.
Answer:
Enthalpy of combustion Of a substance is the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.

Question 21.
Write the thermochemical equation for the combustion of glucose.
Answer:
C₆H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H₂O_(l) ΔH = -xkJ

Question 22.
The enthalpy of combustion of graphite and diamond are -393.5 kJ and -395 kJ respectively. Which of them is more stable ?
Answer:
Graphite.

Question 23.
Explain : Heat of combustion of methane is -890 kJ.
Answer:
Heat of combustion of methane is -890 kJ means when one mole of methane (CH4)
i. e., 16 grams of methane is completely burnt in oxygen, 890 kJ of heat is liberated.

Question 24.
Write the thermochemical equation for the reaction involving the burning of hydrogen gas in excess of air forming water and liberating heat.
Answer:
2H₂(g) + O₂(g) → 2H₂O(l); ΔH = -xkJ

Question 25.
The enthalpy of combustion of ethyl alcohol is – 1360 kJ / mol. Write the thermochemical equation of the reaction.
Ans:
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = -1360 kJ / mole.

Question 26.
Define heat of enthalpy of a soultion.
Answer:
Heat of a solution is the change in enthalpy produced when one mole of the solute is dissolved in excess of a solvent so that further dilution does not produce any heat exchange.

Question 27.
Give one difference between isolated system and closed system.
Answer:
Isolated system can neigher exchange matter nor energy with the surroundings. Closed system can exchange energy but not matter with the surroundings.

Question 28.
Which of the following is an intensive property : Surface tension, mass, volume, enthalpy, density ?
Answer:
Surface tension and density are intensive properties.

Question 29.
What happens to the internal energy of the system if:
(a) Work is done on the system ? (b) Work is done by the system ?
Answer:
(a) If work is done on the system, internal energy will increase, (b) If work is done by the system, internal energy will decrease.

Question 30.
For the reaction, N₂(g) + 3H₂(g) → 2NH₃(g) predict whether the work is done on the system or by the system.
Answer:
Volume is decreasing therefore, work is done on the system.

Question 31.
What is the limitation of first law of thermodynamics ?
Answer:
It cannot tell us the direction of the process.

Question 32.
Which of the following is an extensive property ?
(a) Volume, (b) Surface tension, (c) Viscosity, (d) Density
Answer:
(a) Volume is an extensive property.

Question 33.
Write the relation between standard free energy change and equilibrium constant Kp for a reversible reaction.
Answer:
Relationship between standard free energy change ∆G° and the equilibrium constant ∆G° = -2.303 RT log K_p, where R is the gas constant = 8.314 JK^-1. T is the temperature in Kelvin.

1st PUC Chemistry Thermodynamics Two Marks Questions and Answers

Question 1.
What is exothermic reaction ? Give example.
Answer:
The reaction in which heat is evolved is called exothermic reaction.
Example : C_(s) + O_2(g) → CO_2(g) ∆H = -393.5 kJ

Question 2.
What is endothermic reaction ? Give example.
Answer:
The reaction in which heat is absorbed is called endothermic reaction.
Example : N_2(g) + O_2(g) → 2NO_(g) ∆H = +180.8kJ

Question 3.
Mention different types of process.
Answer:

1. Isothermal process
2. Adiabatic process
3. Isoehoric process
4. Isobasic process
5. Reversible process
6. Irriversible process
7. Cyclic process

Question 4.
What is open system ? Give example.
Answer:
An open system is one in which there is an exchange of both matter and energy with its surroundings.
Example : Water in an open beaker.

Question 5.
What is Extensive propery ? Give example.
Answer:
It is a property of which depends on the amount of the substance present in the system.
Example : Mass, Volume, Energy

Question 6.
Write any two difference between Isothermal and Adiabatic process.
Answer:

Question 7.
Mention the factors affecting enthalpy of reaction.
Answer:

1. Physical state of reactant and product.
2. Amount of reactant and product.
3. Temperature of the reaction.,
4. Allotropic ferrous.,
5. Condition of constant volume or constant pressure.,
6. Reaction stoichiometry.

Question 8.
State and illustrate Hess’s law.
Answer:
Hess’s law : Whether a chemical reaction takes place in a single step or in several steps, the total change in enthalpy remains the same.
Consider the formation of CO₂ from C.

Question 9.
Explain the meaning of thermochemical equation for the reaction.
2H₂(g) + O₂(g) → 2H₂O(l) +136 kcal.
Answer:
When two moles i.e., 4 grams of gaseous H₂ reacts with 1 mole (32g) of gaseous O₂ to give 2 moles of liquid water i.e., 36 gram of liquid water, 136 kcal of heat is liberated.

Question 10.
Explain heat of neutralisation with an example.
Answer:
Heat of neutralisation is the change in enthalpy produced when one gram equivalent weight of an acid is neutralised by one gram equivalent weight of a base in dilute solution.
Example : HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O;= ΔH – 57.3 kJ

Question 11.
With example explain the term ‘Heat of transition’.
Answer:
Heat of transition is the charge in enthalpy produced when one mole of one allotropic form of an element is converted into another allotropic form.
Example : C(diamond) > C(graphite); ΔH = -X₁ kJ When 1 mole of diamond is coverted into graphite x kJ of heat is liberated, -x kJ is the heat of transition in this case.

Question 12.
Define heat (enthalpy) of transition.
Answer:
It is enthalpy change when one mole of an element is changed from one equilibrium to another.

Question 13.
What do you understand by the statements :
(i) enthalpy of formation of nitric oxide is +90.7 kJ.
(ii) enthalpy of combustion of carbon disulphide is =1065.2 kJ
Answer:
(i) When 1 mole of NO(g) is formed from its elements N₂(g). and O₂(g), 90.7 kJ of heat is absorbed.
(ii) When 1 mole of CS₂ (l) is completely burnt in O₂,1065.2 kJ of heat is liberated.

Question 14.
Enthalpy of neutralisation of NH₄OH with HC is 51.45 kJ. Calculate the enthalpy of ionization of ammonium hydroxide.
Answer:
Enthalpy of ionzators of NH₄OH = -51-41 – (57.3) = 5.8500 J

Question 15.
Explain: “The enthalpy of formation of nitric oxide is +90.4 kJ”.
Answer:
\(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g) → ANO_(g)ΔH = +90.4kJ When 1 mole of nitric oxide is formed from its elements 90.4 kJ of heat is absorbed. (Actually ΔH\+90.4kJ Error in the question)

Question 16.
Distinguish between exothermic and endothermic reactions.
Answer:

Question 17.
State and explain the first law of thermochemistry.
Answer:
If enthalpy of formation of a compound is -x kJ, enthalpy of its decomposition intoits elements is + xkJ.
Example : C(s) + O₂(g) → CO₂; ΔH = -xkJ According to the law (Lavoisier & Laplae law) CO₂ (g) → C(s) + O₂ (g); ΔH = +x kJ

Question 18.
Which of the following has highest heat of combustion out of the following and why ?
(a) C₂H₆ (b) C₂H₄ (c) C₂H₂ (d) CH₄
Answer:
C₂H_6(g) will have highest heat of combustion because it has highest molecular weight
and second highest calorific value.
ΔH_c = calorific value of kJ/g × mol. wt. = 52 × 30 = 1560 kJ/mol.

Question 19.
What do you understand by state functions ? Neither q nor w is a state function but q + w is a state function ? Explain.
Answer:
State function is a property whose value depends only uopon the state of system an is independent of the path, q and w are not state functions because they depend upon path. But q + w = AE which is a state function. AE does not depend upon path.

Question 20.
Acetic acid and hydrochloric acid react With KOH solution. The enthalpy of neutralization of acetic acid is -55.8 kJ mol^-1 while that of hydrochloric acid is-57.3 kJ mol^-1. Why?
Answer:
It is because HCl is strong acid, it ionises completely in aqueous solution whereas acetic acid is weak acid which does not ionise completely and some energy is used in its complete ionisation.

Question 21.
Taking a specific example show that ΔS_total is a criterion for spontaneity of a change.
Answer:
ΔG = -TΔS_total
If ΔS_total is +ve, ΔG will be -ve, reaction will be spontaneous.
If ΔS_total is -ve, ΔG will be +ve, reaction will be non-spontaneous,
ΔS_total = ΔS_syst + ΔS_surroundings
For example in case of freezing of water at 273 K, TAStotai is +0.08 J/K/mol whereas at 274 K it is -0.080 J/K/mol. Since the ΔS total is +ve at 273 K, freezing of water will take place. On the other hand at 274 K, the total entropy change is negative, the freezing will not occur.

Question 22.
What is meant by entropy driven reaction ? How can a reaction with positive changes of enthalpy and entropy be made entropy driven ?
Answer:
Those reactions which are endothermic and entropy is increasing such that TΔS > ΔH are entropy driven reactions. If ΔS total is +ve, the reaction will be spontaneous. If ΔS is +ve and is small, but T is large, the reaction will be spontaneous.
A reaction which is endothermic, i.e., ΔH is +ve and ΔS is +ve, T must be large.
ΔG = ΔH – TΔS ; ΔG will be -ve if TΔS > ΔH, i.e., the reaction will be spontaneous at higher temperature because entropy will increase with increase in temperature which will make this reaction spontaneous.

Question 23.
Justify the following statements:
(a) An exothermic reaction is always thermodynamically spontaneous.
(b) The entropy of a substance increases on going from liquid to vapour state at any temperature.
Answer:
(a) It is false. Exothermic process are not always spontaneous. If ΔS = -ve and TΔS > ΔH, the process will be non spontaneous even if it is exothermic.
(b) The entropy of vapour is more than that of liquid therefore entropy increases during vaposisation.

Question 24.
Evaporation of water is an endothermic process but spontaneous. Explain.
Answer:
It is because entropy increases -during this process because water vapour have more entropy than liquid water. ΔG become -ve because TΔS > ΔH.

Question 25.
What is meant by toral entropy change of a process ? Assuming the thermodynamic relationship ΔG = ΔH – TΔS, derive the relationship ΔG = TΔS_total for a system.
Answer:
Total entropy change of a process if entropy change taking place in universe, i.e.
ΔS_system + ΔS_surrounding = ΔS_Total
ΔS_Total = ΔS_sys – \(\frac{q}{T}\)= ΔS_Sys – \(\frac{\Delta \mathrm{H}}{\mathrm{T}}\)
TΔS_Total = TAS_Sys – ΔH [∵ ΔG = ΔH – TΔS]
TΔS_Total = -ΔG
ΔG = -TΔS_Total

Question 26.
At a certain temperature ‘T’ endothermic reaction A → B proceeds virtually to end. Determine the sign of ∆S for the reaction A → B and ∆G for the reverse reaction B →A.
Answer:
∆S – +ve ∆H = +ve, ∆G = -ve the reaction proceeds virtually to end in reverse reaction B → A, ∆S = -ve, ∆G = +ve

Question 27.
Starting with the thermodynamics relationship G = H – TS, derive the following relationship ∆G = -T∆S_Total
Answer:
At initial state of the system G_i = H_i – T S_i
At final state of the system G_f = H_f – TS_f
∴ Change is free energy ∆G = G_f – G_i = (H_f – H_i) – T (S_t – S_i)
∆G = ∆H – T∆S

Question 28.
Explain the help of example, the difference between bond dissociation energy and bond energy.
Answer:
Bond disoociation energy is energy required to break 1 mole of bonds e.g.
H – H(g)) → 2H(g) ∆H = 436 kJ mol^-1
Bond energy is energy released when 1 mole of bonds are formed example
2H →H₂(g) ∆H = -436kJ mol^-1
In diatomic molecule both bond dissociation energy and bond energy are equal in magnitude but opposite in sign.

Question 29.
What is meant by the free energy of a system ? What will be the direction of chemical reaction when (i) ∆G = 0 (ii) ∆G > 0 (iii) ∆G < 0 Answer: Free energy is defined as energy which can be converted into useful work (i) ∆G = 0, the reaction will be in equilibrium (ii) ∆G > 0, the reaction will not take place
(iii) ∆G < 0, the reaction will be spontaneous

Question 30.
Why most of the exothermic process (reactions) spontaneous ?
Answer:
∆G = ∆H – ∆H – T∆S; For exothermic reactions ∆H is -ve. For a spontanec s process AG is to be -ve.
Thus decrease in enthalpy (-∆H) contributes significantly to the driving force (to make AG negative).

Question 31.
The enthalpy of combustion of sulphur is 297 kJ. Write the thermochemical equation for combustion of sulphur. What is the value of Δ_fH of SO₂ ?
Ans wer:
S(s) + O₂(g) → SO₂(g); ΔH = -297kJ and Δ_fH of SO₂ = -297 kJ mol^-1.

Question 32.
What is the most important condition for a process to be reversible in . thermodynmics ?
Answer:
The process should be carried out infinitesimally slowly of the driving force should be infinitesimally greater than the opposing force.

Question 33.
Why absolute value of enthalpy cannot be determined ?
Answer:
As H = E = +PV
Absolute value of E – the interned energy cannot be determined as it depends upon various factors whose value cannot be determined.
∴ Absolute value of H cannot be determined.

Question 34.
What are the applications of Hess’s Law of constant heat summation ?
Answer:

• It helps to calculate the enthalpies of formation of those compounds which cannot be determined experiementally.
• It helps in determine the enthalpy of allotropic transformation like C (graphite) → C (diamond)
• It helps to calculate the enthalpy of hydration.

Question 35.
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vaccum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer:
We have q = – w : p_ext (10 – 2) = 0 × 8 = 0
No. work done; No heat, is absorbed.

Question 36.
Define (i) Molar enthalpy of fusion (ii) Molar enthalpy of vaporization ?
Answer:
(i) The enthalpy change that accompanies melting of one mole of a solid substance at its melting point is called molar enthalpy of fusion.
(ii) Amount of heat required to vaporize one mole of a liquid at constant temperature and understand pressure (1 bar) is called molar enthalpy of vapourisaton.

Question 37.
Define : (a) Enthalpy of atomization (b) Lattice enthalpy.
Answer:
Enthalpy of atomization : It is defined as the enthalpy change accompyning the breaking of one mole of a substance completely into its atoms in the gas phase.
H₂(g) → 2H(g); ∆_aH° = 435.0kJ mol^-1
Lattice enthalpy ; Lattice of an comp; the enthalpy change which occure when one mole of an ionic compound dissociates into its gaseous ionic state.
NaCl(s) → Na⁺ (g) + Cl^– (g); ∆H = +788 ks / mol

Question 38.
Explain Laplace- Lavoisier law.
Answer:
The quantity of heat that must be supplied to decompose a compound into its elements is equal and opposite to the heat evolved when the same compound is formed from its elements.
C(s) + O_2(g) → CO_2(g) ∆H = -393.5 kJml^-1
CO_2(g) → C_(s) + O_2(g) ∆H = +393.5 kJml^-1

Question 39.
If the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ moh1. What is the enthalpy change for the C (graphite) → C (diamond) ?
Answer:
C(diamond) + O₂(g) → CO₂(g) ∆H =-395.4 kJmol^-1 …(1)
C(graphite) + O₂(g) → CO₂(g) ∆H = -393.6kJmol^-1 … (2)
C(graphite) → C(diamond ) substracting (1) from (2), we get
C(graphite) → C(diamond) ∆H = -393.6 kJ – (-395.4 kJ), ∆H = +1.8 kJ mol^-1

Question 40.
H₂(g) + \(\frac { 1 }{ 2 }\) O₂(g) → H₂O(g) Δ_fH = -242kJ mol^-1. Bond energy of H₂ and O₂ are 436 mol^-1 and 500 mol^-1 respectively. What is bond energy of O – H bond ?
Answer:
H₂(g) + \(\frac { 1 }{ 2 }\) O₂(g) → H₂O(g); Δ_fH° = -242kJ mol^-1
H = Bond energy of reactions – Bond energy of products
ΔH = B_H-H + \(\frac { 1 }{ 2 } \)B_{O – O} + B_{O – H}  ; 242kJ = 436 + \(\frac { 1 }{ 2 }\) (500) – 2B_{O – H} ⇔ -928kJ = -2B_{O – H}; B_{O – H} = \(\frac { 928 }{ 2 }\) = 464kJ mol^-1

Question 41.
What is spontaneous process ? Give example.
Answer:
Process which takes place itself, without any external aid under the given condition is called spontaneous process.
Example : Flow of heat from higher to lower temperature.

Question 42.
Define non-spontaneous process.
Answer:
Process which does not takes place itself or on its own but with the help of external aid under the given condition is called non-spontaneous process.
Example : Flow of heat energy from lower to higher temperatures.

1st PUC Chemistry Thermodynamics Three Marks Questions and Answers

Question 1.
Classify the following processes as reversible or irreversible :
(i) Dissolution of sodium chloride
(ii) Evaporation of water at 373 K and 1 atm. pressure
(iii) Mixing of two gases by diffusion
(iv) Melting of ice without rise in temperature
(b) When an ideal gas expands is vaccum, there is neither absorption nor revalutions of what ? why ?
Answer:
(a)(i) Irreversible, (ii) Irreversible, (iii) Irreversible, (iv) Reversible
(b) It is because no work is done,
i.e., w = 0 w = -p_ext × ∆V = 0 × ∆V = 0
In ∆U = q + w
q = 0 because gas chamber is insulated ∆U = 0 + 0 = 0

Question 2.
Justify the following statements :
(a) Reaction with ∆G° < 0 always have an equilibrium constant greater thanl.
(b) Many thermodynamically feasible reaction do not occur under ordinary conditions.
(c) At low temperatures enthalpy change dominates the AG expression and at high temperature it is the entropy which dominates the value of AG.
Answer:
(a) ∆G° = -2.303RT logK If K > 1, ∆G° will be less than zero because products formed are more than that of reactants, i.e., process is spontaneous in forward direction.
(b) It is because heat energy is required to overcome activation energy.
(c) ∆G = ∆H – T∆S, At low temperatures ∆H > T∆S whereas at high temperature T∆S >∆H
∴ ∆G decreases, i.e., becomes negative.

Question 3.
If for the reaction PbO₂ → PbO, ΔG° < 0
And for the reaction SnO₂ → SnO, ΔG° > 0
What are the most probable oxidation states of Pb and Sn ?
Answer:
Oxidation states of the elements in the following compounds are

Since PbO₂ → PbO change is accompanied by decrease in free energy it is a spontaneous change hence Pb in +2 state is more stable than in +4 state.

Question 4.
(i) What is the value Δ_rH° of the following reaction :
H⁺(aq) + OH^–(aq) →H₂O(l).
(ii) Give an example in which enthalpy change is equal to internal energy ?
Answer:
(i) Δ_rH° = -57.1kJmol^-1
(ii) H₂(g) + I₂(g) → 2HI(g) here ΔH = ΔU

Question 5.
Consider the reaction. A + B → C + D
(i) If the reaction is endothermic and spontaneous in the direction indicated, comment on the sign of ΔG and ΔS.
(ii) If the reaction is exothermic and spontaneous in the direction indicated, can you comment on the sign of G and S?
(iii) If the reactin is exothermic and spontaneous only in the direction opposite to the indicated, comment on the sign of AG and AS for the direction indicated in the equation.
Answer:
(i) ΔG = -ve, ΔS = +ve
(ii) ΔG = -ve, ΔS = -ve (if ΔH > TΔS)
(iii) ΔG = ΔS = -ve

Question 6.
Predict in which of the following entropy increases / decreases.
i) Temperature of a crystalline solid is raised from 0 K to 115 K.
ii) 2NaHCO₃ → Na₂CO₃(s) +CO₂ (g) + H₂O(g)
iii) H₂(g) → 2H(g
Answer:
(i) When temperature is raised, disorder in molecules increases and therefore entropy increases.
(iii) Reactant is a solid and hence has low entropy. Among the products there are two gases and one solid so products represent a condition of higher entropy.
(iv) Here 2 moles of H atoms have higher entropy than one mole of hydrogen molecule.

Question 7.
Which of the following is / are exothermic and which are endothermic ?
(i) Ca(g) → Ca²⁺(g) + 2e^–
(ii) O(g) + e^– → O^– (g)
(iii) N^2-(g) + e^– → N^3-(g)
Answer:
(i) Endothermic (ionisation energy is required)
(ii) Exothermic (first electron affinity – energy is released)
(iii) Endothermic (higher electron affinities energies are required)

1st PUC Chemistry Thermodynamics Numerical Problems and Answers

Question 1.
Calculate the heat of reaction of the following reaction :

C₆H₁₂O₆(s) + 6O_2(g) → 6CO_2(g)+ 6H₂O(g); ΔH = ?
C(graphite) + O₂(g) → CO₂(g); DH = -395.0 kJ … (1)
H₂ (g) + \(\frac { 1 }{ 2 }\)O₂(g) → H₂O(l); DH = -269.5 kJ … (2)
6C(graphit.e) + 6H₂(g) + O₂(g) → C₆H₁₂O₆(s); DH = -1169.8 kJ … (3)
Answer:
Multiplying equation (1) and (2) each by 6 reversing (3), we get,
6C_(graphite) + 6O_2(g) → 6CO_2(g); ΔH = -2370 kJ … (4)
6H_2(g) + 3O_2(g) → 6H₂O_(l); ΔH = -1616.4 kJ … (5)
C₆H₁₂O_6(s) → 6C_(graphite) + 6H_2(g) + 3O_2(g) ; ΔH = +1169-8 kJ …. (6)
Adding (4), (5) and (6), C₆H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H₂O_(g) ;
AH(C₆H₁₂O₆) = -2370.0 – 1616.4 + 1169,8 = -2816.6 kJ

Question 2.
Calculate the heat of reaction of the following reaction :
CO₂(g) + H₂(g) → CO(g) + H₂O(g)
Given that the Δ_fH°CO(g) =-110.5kJ, Δ_fH°CO₂(g) =-393.8kJ,
Δ_fH°H₂O(g) = -241.8kJ respectively.
Answer:
The required equation is CO₂(g) + H₂(g) → CO(g) + H₂O(g)
ΔH – Σ ΔH_f(products) — Σ ΔH_f(reactants)
= ΔH_f²CO(g) + ΔH_f²H₂O(g) – ΔH_f²CO_2(g) – ΔH_f²H_2(g)
= -110.5 kJ – 241.8 kJ – (-398.3 kJ) – 0 = -352.3 kJ + 393.8 kJ = 41.5 kJ.

Question 3.
1 m³ of C₂H₄ at STP is burnt in oxygen, according to the thermochemical
reaction: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l); ΔH = -1410 kJ mol^-1
Assuming 70% efficiency, determine how much of useful heat is evolved in the reaction.
Answer:
22.4 L of C₂H₄ at STP produces 1410 kJ of energy.

Question 4.
With the help of thermochemical equation, calculate Δ_fH° at 298 K for the following reactions:
C(graphite) + O₂(g) → CO₂(g) ; Δ_fH° =-393.5kJ/mol
H₂ (g) + \(\frac { 1 }{ 2 }\)O₂(g) → H₂O(l) ; Δ_fH° = -285.8kJ/mol
CO₂ (g) + 2H₂O(l) → CH₄ (g) + 2O₂(g) ; Δ_fH° = +890.3kJ/mol
Answer:
C(graphite) + O₂(g) → CO₂(g) ; Δ_fH° = -393.5kJ/mol …(1)
H₂(g) + \(\frac { 1 }{ 2 }\)O₂(g) → H₂O(l) ; Δ_fH° = -285.8kJ/mol …(2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) Δ_fH° = +890.3kJ/mol …(3)
Here we want one mole of C(graphite) as reactant, so we write down equation (1) as such, we want two moles of H₂(g) as reactant, so we multiply equation (2) by 2, we want

C(graphite) + O₂ (g) → CO₂(g) ; Δ_fH° = -393.5kJ / mol …(1)
2H₂(g) + O₂(g) → 2H₂O(1) ; Δ_fH° = 2(-285.8kJ/mol) …(2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ; Δ_fH° =+890.3kJ/mol . … (3)
Adding we obtain: .
C(graphite) + 2H₂(g) → CH₄(g); Δ_fH° =-74.8kJ/mol

Question 5.
The heat of combustion of C₂H₆ is -368.4 kcal. Calculate heat of combustion of C₂H₄, heat of combustion of H₂ is 68.32 kcal mol^-1.
C₂H₄(g) + H₂(g) → C₂H₆(g) ; ΔH = -37.1 kcl
Answer:
C₂H₄(g) + H₂(g) → C₂H₆(g); ΔH = -37.1 k cal.
ΔH_cC₂H₆ = -368.4 k cal, ΔH_cC₂H₂ = ?; ΔH_cH_2(g) = -68.32 k cal.

ΔH  = ΣΔH_c(reactanta) – ΣΔH_c(products)
ΔH = ΔH_cC₂H₄+ΔH_cH_2(g) – ΔH_cC₂H_6(c)
-37.1k cal = ΔH_cC₂H₄-68.32-(-368.4)
ΔH_cC₂H₆ =-337.18 kcal.

Question 6.
The following thermochemical equations represent combustion of ammonia and hydrogen:

4NH₃(g) + 3O₂(g) → 6H₂O(l) + 2N₂(g); ΔH = -1516 kJ
2H₂(g) + O₂(g) → 2H₂O(l); ΔH = -572kJ
Calculate enthalpy of formation of ammonia.
Answer:
Required equation is  \(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 3 }{ 2 }\)H_2(g) → NH_3(g) ΔH = ?
4NH_3(g) + 3O_2(g) → 2N_2(g) + 6H₂O_2(l) ΔH = -1516 kJ …(l)
2H_2(g) + O_2(g) → 2H₂O_(l) ΔH = -572 kJ …(2)
Reserving equation (1) and dividing by 4 we get:
\(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 3 }{ 2 }\)2H₂O_2(l) → NH_3(g)+ \(\frac { 3 }{ 4 }\)O_2(g) ΔH = +379 kJ …(3)
Multiplying equation (2), by 3/4 we get,
\(\frac { 3 }{ 2 }\)H_2(g) + \(\frac { 3 }{ 4 }\)O₂ → \(\frac { 3 }{ 2 }\)H₂O_(l) ΔH = -572 × \(\frac { 3 }{ 4 }\) = -429kJ
Adding (3) and (4) we get ,
\(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 3 }{ 2 }\)H_2(g) → NH_3(g) ;ΔH = +379 – 429 = -50 kJ/mol

Question 7.
The equilibrium constant at 25° C for the process.
CO³⁺(aq) + 6NH₃(aq) ⇌ [CO(NH₃)₆]³(aq) is. 2.5 × 10⁶. Calculate the value of ΔG° at 25° C. (R = 8.314 JK1 mol^-1). In which direction is the reaction spontaneous under standard conditions ?
Answer:
ΔG° =-2.303 RT log K .
= -2.303 × 3.134 × 298 log (2.5 x 10⁶)
= -5705.8 [0.3980 + 6.0000] = -5705.8 × 6.3980 = -36.505 k/mol.
The reaction is spontaneous in forward direction under standard conditions.

Question 8.
What is the value of equilibrium constant for the following reaction at 400 E?
2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
ΔH⁰ = 77.5kJmol^-1, R = 8.3124 J Mol^-1K^-1, AS = 135 J K^-1mol^-1
Answer:
ΔG = ΔH – TΔS
ΔG = 77.5 × 1000 J – 400 K × 135 J K^-1 = 77500 J – 54000 J = 23500 J
ΔG = -2.303 RT log K
23500 J = -2.303 × 8.314 × 400 K log K

Log k =- 3.068 k = antilog(-3.008)
= Antilog (0.932-4) = Antilog (0.93) × 10^-4 = 8055 × 10^-4

Question 9.
The standard free energy change for a reaction is -212.3 kJ mol^-1. If the enthalpy of the reaction is -216.7 kJ mol^-1, Calculate the entropy change for the reaction.
Answer:
ΔG° = -212.3 kJ mol^-1, ΔH⁰ = -216.7 kJ mol^-1, ΔS° = ?
T = 298 K (Because standard free energy is measured at 298 K)
ΔG° = -212.3 kJ mol^-1, ΔH° = 216.7 kJ mol^-1, ΔS° = ?
-212.3 kJ = 216.7 kJ – 298 K × ΔS°

Question 10.
Calculate the standard free energy change AG° for the reaction.
2HgO(s) → 2Hg(l) + O₂(g)
ΔH° = OlkJmol^-1 at 298 K, S°_(Hg0) = 72.0JK^-1 mol^-1
Answer:
ΔS = 2S°_Hg + S°_(O2) – 25°_Hgo
= (2 × 77.4 + 205 – 2 × 72.0) JK ^-1mol^-1 = 215.8 jK^-1 mol^-1
ΔG° = Δ_rH° – TΔ_rS°

Question 11.
From the data given below at 298 K for the reaction :
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Calculate the enthalpy of formation of CH₄(g) at 298 K.
Enthalpy of reaction is = -89.5 kJ
Enthalpy of formation of CO₂(g) – -393kJ mol^-1
Enthalpy of formation of H₂O(l ) = -286.0 kJ mol^-1
Answer:
ΔH = AH_fCO₂(g) + 2ΔH_fH₂O(l) – ΔH_fCH₄(g) – ΔH_f;O₂(g)
-890.5 kJ = -393.5 kJ +2 × -286 kJ – Δ H_fCH₄(g) – 0
ΔH_fCH₄ – 75.0kJ

Question 12.
Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given : Lattice energy of NaCl = – 777.8 kJ mol^-1), Hydration energy
= -774.1 kJ mol^-1 and ΔS = 0.043 kJK^-1 mol^-1 at 298 K.)
Answer:
ΔH = Hydration energy – Lattice energy
ΔH = -774.1 kJ mol^-1 (-777.8 kJ mol^-1) = 3.7 kJ mol^-1
ΔG = ΔH – TΔS = +3.7 kJ- 298 × 0.043 kJ = +3.7 kJ- 12.81 kJ
ΔG = -9.11 kJ moH

Question 13.
For the equilibrium, PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g) at 298 K, K = 1.8 × 10 ^-7.
Calculate ΔG° for the reaction (R = 8.314JK 1mol 1)
Answer:
PCl₅(g) ⇌ PCl₂(g) + Cl₂(g), K = 1.8 × 10^-7
ΔG° = -2.303RTlogK = -2.303 × 8.314JK^-1mol^-1 × 298K × log(1.8 × 10^-7)
= -2.303 × 8.314 × 298[log 1.8 + logl0^-7 ] = -19.147 × 298 [0.2553 – 7.000]

Question 14.
Calculate the entropy change involved in conversation of 1 mole of water at 373 K to vapours at the same temperature. Latent heat of vaporation of water = 2.257 kJ g1.
Answer:
k = 1.8 × 10^-7
ΔH = 2.257 × 18kJ= 40.626kJ mol^-1

Question 15.
Calculate ΔH_f of HCl if bond energy of H – H bond is 437 kJ Cl – Cl bond is 244, and H – C is 433 kJ mol1.
Answer:
\(\frac { 1 }{ 2 }\)H2(g) + \(\frac { 1 }{ 2 }\) cl₂(g) > HCl(g)
ΔH = \(\frac { 1 }{ 2 }\)B_{H – H} + \(\frac { 1 }{ 2 }\)B_{O – O} = \(\frac { 1 }{ 2 }\) × 437 + \(\frac { 1 }{ 2 }\) × 244 -433
= 218.5 kJ + 122 kJ – 433 kJ = -92.5 kJ mol^-1

Question 16.
Calculate bond energy of C – H bond if ΔHc of CH₄ is -891 kJ, ΔHc of C (s) is – 394 kJ, ΔHc of H₂(g) is -286 kJ, heat of sublimation of C(s) is 717 kJ, heat of dissociation of H₂ is 436 kJ.
Answer:
CH₄(g) + 2O₂(g) → C0₂(g) + 2H₂O(l) ; ΔH = -891 kJ
C(s) + O₂(g) → CO₂(g) ;ΔH = -394 kJ
H₂(g) + \(\frac { 1 }{ 2 }\) O₂(g) → H₂O(l) ; ΔH = -286 kJ
C(s) → C(g) ;ΔH = +717 kJ
H₂(g) → 2H(g);Δ H = +436 kJ
Target equation is CH₄ +4H(g)
Reversing (ii),Multiply (iii) by 2 and reverse, Multiply eqn. (v) 2 and then adding all together we get:
CH₄ → C(g) + 4H(g) ΔH = -891kJ + 394kJ + 575kJ + 717kJ + 872kJ
ΔH = 1664 kJ
Energy required to break 4(C – H) bond = 1664 kJ
Energy required to break one (C – H) bond =\frac{1664}{4} = 416 kJ mol^-1

Question 17.
What would be heat released when :
(i) 0.25 mole of HCl in solution is neutralized 0.25 mole of NaOH solution ?
(ii) 0.5 mole of HNO₃ in solution is mixed with 0.2 mole of KOH solution ?
(iii) 200 cm³ of 0.2 M HCl is mixed with 300 cm3 of 0.1 M NaOH solution ?
(iv) 400 cm³ of 0.2 M H₂SO₄ is mixed with 300 cm3 of 0.1 M NaOH solution ?
Answer:
(i) H⁺(0.25mole) + OH^–(0.25mole) → H₂O(0.25mole)
Heat released = 0.25 × 57.1 = 14.3 kJ
(ii) 0.5 mole HNO₃ (aq) + 0.2 mole KOH(aq)
The net reaction is (0.2 mole HNO₃ reacts with only 0.2 mole KOH)
0.2 mole is limiting reagent.
0.2 mole of H⁺ + 0.2 mole of OH^– → 0.2 mole of H₂O
0.3 mole of H⁺ will remain unreacted.
Heat evolved = 0.2 × 57.1 = 11.4 kJ

(iii)

0.03 mole of OH^– is limiting reactant
Heat evolved = 0.03 × 57.1 = 1.71 kJ

(iv)

0.03 is limiting reagent Heat evolved = 0.03 × 57.1 = 1.7 kJ

(v) Mass of solution = 200 + 300 = 500 g in (iii) part (Assuming d = 1 g cm 3)
q = m × c × T 1.71 × 1000 J = 500 g × 4.18 × T
T = 0.82 K For (iv), T = \(\frac{1.7 \times 1000}{700 \times 4.18}\) = 0.58K

You can Download Chapter 10 The S-Block Elements Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 10 The S-Block Elements

1st PUC Chemistry The S-Block Elements One Mark Questions and Answers

Question 1.
How does the density of alkali metal change from Li to Cs?
Answer:
Density increases down the group from Li to Cs.

Question 2.
Elements of which group in the periodic table belong to s-block?
Answer:
I and II groups.

Question 3.
In what way the electronic configuration of hydrogen is similar to that of the electronic configuration of alkali metals?
Answer:
Hydrogen and alkali metals both have one electron in the outermost orbital.

Question 4.
What similarity is found in the electronic configurations of hydrogen and halogen?
Answer:
Both hydrogen and halogen are in short of one electron for the completion of the outermost orbital containing electrons.

Question 5.
Name any four alkali metals.
Answer:
Lithium, Sodium, Potassium and Rubidium.

Question 6.
LiCI and MgCk dissolve in alcohol. How do you explain this?
Answer:
Both LiCI and MgCl₂ are covalent compounds and dissolve in alcohol. This is due to high polarizing power of Li⁺ and Mg²⁺ ions.

Question 7.
The alkali metals have no tendency to show variable oxidation states. Give reason.
Answer:
Alkali metals show oxidation state of +1. With the loss of valence electron it attains the stable configuration of nearest inert gas. Its second ionization potential is high. Hence an alkali metal does not show variable oxidation states.

Question 8.
Write the alkali metals in the increasing order of hydration energy.
Answer:
Li ⁺ > Ma⁺ > K⁺> > Rb⁺ > Cs⁺

Question 9.
Why are group 1 elements called alkali metals ?
Answer:
It is because their hydroxides are soluble bases called alkalies. Secondly their ashes are alkaline in nature.

Question 10.
Why do alkali metals have low ionisation energy ?
Answer:
It is due to largest atomic size, they can lose electrons easily.

Question 11.
Alkali and alkaline earth metals cannot be obtained by chemical reduction, why?
Answer:
Alkali and alkaline metals are good reducing agents, they cannot be obtained by chemical reduction.

Question 12.
Why does ionisation energy of alkali metals decrease with the increase in atomic number ?
Answer:
Atomic size increases with increase in atomic number, therefore, nuclear force of attraction between valence electrons and nucleus decreases, hence ionisation energy decreases down the group.

Question 13.
Why group 2 elements (Mg and Ca) are harder and denser than group 1 elements ?
Answer:
They have strong metallic bonds due to smaller size and have more number of valence electrons.

Question 14.
Why is potassium more reactive than sodium ?
Answer:
K has lower ionisation energy than sodium due to bigger atomic size, therefore, it is more reactive.

Question 15.
Why are alkali metals strong reducing agents?
Answer:
It is because of low ionisation energy. They can lose electrons easily, that is why they are strong reducing agents.

Question 16.
Why are alkali metals used in photoelectric cells ?
Answer:
They have low ionisation energy and can lose electrons when light falls on them, that is why they are used in photoelectric cells.

Question 17.
Write electronic configuration of Na (11) and K (19).
Answer:
Na(11): 1s²2s²2p⁶3s¹ K(19): ls²2s²2p⁶3s¹3p⁶4s¹

Question 18.
Why do alkali metals have low melting and boiling points ?
Answer:
It is due to weak metallic bonds which is due to bigger atomic size that is why they how low melting and boiling points.

Question 19.
How will you prepare sodium hydrogen carbonate from sodium chloride ?
Answer:

(NH₃)HCO₃ + NaCl → NaHCO₃ + NH₄Cl

Question 20.
Why do alkali metals not occur in free state ?
Answer:
They are highly reactive, therefore, they occur in combined state and do not occur in free state.

Question 21.
Why is second ionisation energy of alkali metals higher than alkaline earth metals ?
Answer:
Alkali metals acquire, noble gas configuration after losing 1 electron, therefore their second ionization energy is higher than alkaline earth metals.

Question 22.
Which out of K, Mg, Ca and Al from amphoteric oxide ?
Answer:
Al forms amphoteric oxide, i.e., acidic as well as basic in nature.

Question 23.
Which out of Na, K, Al, Mg occur as oxide in nature ?
Answer:
Al occurs as oxide in nature as bauxite Al₂O₃. 2H₂O

Question 24.
Why do alkali metals give characteristic flame colouration?
Answer:
They have low ionization energy and absorb energy from visible region of spectrum and radiate complementary colour.

Question 25.
What happens when K burns in air ? Give chemical equation.
Answer:
K + O₂ → KO₂, pottassium superoxide

Question 26.
What is quick lime ? How is it prepared ?
Answer:
Quick lime is calcium oxide. It is prepared by heating limestone.

Question 27.
Give two uses of plaster of paris. Also give its formula.
Answer:

1. It is used in plastering fracture bones,
2. It is used in preparations of chalks. Its formula is (CaSO₄)₂.H₂O

Question 28.
Arrange the following in order of their increasing covalent character : MCI, MBr, MF, Ml (Where M is alkali metal)
Answer:
MF < MCI < Ml, lesser the difference in electronegativity, more will be covalent character.

Question 29.
One reason on being heated in excess supply of air K, Rb and Cs from superoxide in preference to oxides and peroxides ?
Answer:
K, Rb and Cs are more reactive therefore, they form superoxide in preference to oxides and peroxides K⁺,Rb⁺ and Cs⁺ ions are large cations and superoxide ion \(\mathrm{O}_{2}^{-}\), is also large. Larger cations stabilize larger anions, therefore, they form superoxide.

Question 30.
What happens when KO₂ reacts with water ? Give balanced chemical equation.
Answer:
2KO₂ (Pottasium sup eroxide) + 2H₂O → 2KOH + O₂ + H₂O₂

Question 31.
Complete the reaction : Lil + KF →
Answer:
Lil + KF → LiF + KI; larger cation stabilizes larger anion and smaller cation stabilizes smaller anion.

Question 32.
Name the reagent or one process to distinguish between :

• BeSO₄ and BaSO₄
• Be(OH₂) and Ba(OH)₂

Answer:

• BeSO₄ is soluble in water while BaSO₄ is not.
• Be(OH)₂ dissolves in NaOH while Ba(OH)₂ is insoluble.

Question 33.
Why does Be resemble Al?
Answer:
Be resembles Al because charge over radius ratio is similar, i.e., they have similar polarizing power.

Question 34.
The second ionization enthalpy of Ca is higher than first and yet calcium forms CaCl₂ and not CalCl Why?
Answer:
The hydration energy of Ca²⁺ over comes the second ionization energy of Ca, that is why Ca forms CaCl₂ and not CaCl. Ca⁺ is not stable

Question 35.
Name the alkali metals which form superoxide when heated in air?
Answer:
K, Rb, Cs are alkali metals which form superoxide when heated in air.

Question 36.
Name the metal which floats on water without apparent reaction.
Answer:
Berylium.

Question 37.
Why is BeCl₂ soluble in organic solvents ?
Answer:
BeCl₂ is covalent, therefore, soluble in organic solvents.

Question 38.
Starting from quick lime how slaked lime is prepared ? Is this reaction exothermic or endothermic ?
Answer:
CaO + H₂O → Ca(OH)₂ +heat
When CaO is put in water, it forms calcium hydroxide. It is an exothermic reaction.

Question 39.
Carbon dioxide is passed through a suspension of limestone in water. Write balanced chemical equation for the above reaction.
Answer:
CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂

Question 40.
What do we get when crystals of washing soda exposed to air?
Answer:
We get amorphous sodium carbonate because it loses water molecules.

Question 41.
What happens when sodium dissolve in liquid ammonia?
Answer:
It results in the formation of intense blue colour solution which possess conducting power
\(\mathrm{Na}+(\mathrm{x}+\mathrm{y}) \mathrm{NH}_{3} \longrightarrow \mathrm{Na}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}}^{+}+\left[\mathrm{e}\left(\mathrm{NH}_{3}\right)_{3}\right]^{-}\)

Question 42.
Name the elements (alkali metals) which form superoxide when heated in excess of air.
Answer:
Potassium, rubidium and caesium

Question 43.
Why is oxidation state of Na and K always + 1?
Answer:
It is due to their high second ionization enthalpy and stability of their ions [Na⁺ K⁺]

Question 44.
What is meant by dead burnt plaster?
Answer:
It is anhydrous calcium suplhate (CasO₄)

Question 45.
What is the reason that sodium reacts with water more vigorously than lithium ?
Answer:
Because sodium is more electro-positive than Li.

Question 46.
Why is sodium thiosulphate used in photography ?
Answer:
Because of its complex forming behaviour.

Question 47.
Why does lithium show anomalous behaviour ?
Answer:
Due to its small size and high charge/size ratio.

1st PUC Chemistry The S-Block Elements Two Marks Questions and Answers

Question 1.
Give reason for diagonal relationship of lithium with magnesium.
Answer:
Both Lithium and magnesium have small size and high charge density. The electronegativities of Li is 1.0 and Mg is 1.2. They are low and almost same. Their ionic radii are similar. Hence they show similarities which is known as diagonal relationship between first element of a group with the second element in the next higher group.

Question 2.
What is photoelectric effect?
Answer:
Alkali metals have the lowest ionization energy in each period of the periodic table. Hence they emit electrons even when exposed to light. This phenomenon is called photoelectric effect. Rubidium and caesium are used in photoelectric cells.

Question 3.
Give two important ores each of Na and K.
Answer:
Rock salt (NaCl), Na₂CO_3, NaHCO₃, 2H₂O (trona) are important ores of Na. Sylvine (kCl); kCl.mgcl₂.6H₂O (carnallite) are of k.

Question 4.
Give one important use of following compounds.

1. NaHCO₃
2. Slaked lime
3. NaOH

Answer:

1. Sodium bicarbonate is used as antacid
2. Slaked lime is used for white washing,
3. NaOH is used in manufacture of soap.

Question 5.
What is effect of heat on the following compounds ? (Give equations for the reactions)
(i) CaCO₃
(ii) CaSO₄2H₂O
Answer:

Question 6.
Name the metals which are found in each of the following minerals :
(a) Chile salt petre
(b) Marble
(c) Epsomite
(d) Bauxite
Answer:
(a) Na
(b) Ca
(c) Mg
(d) Al.

Question 7.
What are the raw materials used in manufacture of Portland cement ? How is it manufactured ?
Answer:
Limestone and clay are raw materials used in manufacture of cement. It is prepared by heating powdery mixture of limestone and clay in dry process. In wet process, fine – powdered mixture is converted into slurry by adding water and then it is heated at a temperature 1500° C to 1600° C, the product formed is called clinkers. It is cooled down and mixed with gypsum (CaSO₄, 2H₂O) and then it is powdered.

Question 8.
What is composition of Portland Cement ? What is average composition of good quality cement ?
Answer:
CaO = 50% ; 60% ; SiO₂ = 20% to 25% ; Al₂O₃ = 5 to 10% ; MgO = 2% to 3% ; Fe₂O = 1 to .2% ; SO₂ = 1 to 2% is composition of Portland cement.
The ratio of SiO₂ (silica) to alumina (Al₂O₃) should be between 2.5 and 4.0 and the ratio of lime (CaO) to total oxides of silicon, aluminium and iron (SiO₂, Al₂O₃ and Fe₂O₃) should be as close to 2 as possible.

Question 9.
Describe in brief the manufacture of caustic soda using the Castner-Kellner cell.
Answer:
The Castner-Kellner cell consists of large rectangular trough divided into three compartments with partition short of reaching the bottom of the tank. Thus mercury in one compartment can flow into another but solution cannot mix. Graphite anodes are used in outer compartments filled with NaCl solution. The middle compartment contains very dilute solution of caustic soda and filled with iron rods as cathode.

On passing electric current CI₂ is liberated in outer compartments and sodium liberated at cathode. Mercury forms amalgam which is passed into middle compartment in which mercury acts as anode (having induced +ve potential).
At anode Na⁺ + e^– → Na
At cathode Na + Hg → Na – Hg
2(Na – Hg) + 2H₂O → 2NaOH + Hg + H₂
The concentration of NaOH goes on increasing in the middle compartment. When the concentration of NaOH reaches 20% the solution is replaced by dilute solution.

Question 10.
Compare four properties of alkali metals and alkaline earth metals.
Answer:

Alkali Metals	Alkaline earth metals
1. The show + 1 oxidation state.	1. They show + 2 oxidation state.
2. They are soft metals	2. They are harder than alkali metals.
3. They do not form complexes except Li.	3. They can form complex compounds
4. Their carbonates are soluble in water except Li2CO3	4. Their carbonates are insoluble water.

Question 11.
What happens when exhaling is made through a tube passing in lime water ? What will happen if continued exhaling is made through it ? If the solution thus obtained is heated, what do we observe ? Explain giving chemical reactions.
Answer:
Lime water turns milky due to formation of CaCO₃
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
Milkiness will disappear if continuously exhaling i.e., CO₂ is passed due to formation of calcium bicarbonate.
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂
Milkiness will reappear if heating is done due to formation. CaCO₃

Question 12.
Complete the following equations :
(i) Ca + N₂ →
(ii) Ca + SO₂ →
(iii) Ca(OH)₂+NH₄Cl →
(iv) Ca + CO₂ →
Answer:
(i) 3 Ca + N₂ → Ca₃N₂
(ii) 2 Ca + SO₂ → 2CaO + S
(iii)

(iv) 2 Ca + CO₂ → CaO + C

Question 13.
What is dead burnt plaster ? How is it obtained from gypsum?
Answer:
CaSO₄ is called dead burnt plaster. It is obtained by heating gypsum at high , temperature.

Question 14.
What is used for drying alcohol and non-acidic gases and why ?
Answer:
Calcium is used for drying alcohol and non-acidic gases because Ca does not react with alcohol.

Question 15.
What is the mixture of CaCN₂ and carbon called ? How is it prepared ? Give its uses.
Answer:
It is called Nitrolim. It is prepared by heating CaC₂ with N₂ at high temperature. It is used as fertilizer.

Question 16.
Convert limestone to calcium carbide.
Answer:

Question 17.
What are isomorphous salts ? Give two examples.
Answer:
Isomorphous salts are those which have same crystalline structure, e.g., MgSO₄, 7H₂O and ZnSO₄,7H₂O are isomorphous.

Question 18.
Which metal is present in chlrophyll ? How does this metal react with N₂ ?
Answer:
Mg is present in chlorophyll. N₂ reacts with Mg to form magnesium nitride.
3Mg + N₂ → Mg,N₂ (magnesium nitride)

Question 19.
Name an alkali metal carbonate which is thermally unstable and why ? Give its decomposition reaction.
Answer:
Li₂CO₃ is thermally unstable because it is covalent. It decomposes to form Li₂O and
CO₂ ; Li₂CO₃ → Li₂O + CO₂

Question 20.
Why are ionic hydrides of only alkali metals and alkaline earth metals are known ? Give two examples.
Answer:
Alkali metals and alkaline earth metals aye most electropositive due to low ionization energy or enthalpy. Therefore, they can form ionic hydrides, e.g., NaH, KH and CaH₂.

Question 21.
Which one of the alkaline earth metal carbonate is thermally most and last stable. Why ?
Answer:
BaCO₃ is thermally most satable due to greater ionic character and high lattice energy whereas BaCO₃ is thermally least stable because it is covalent and has less lattice energy.

Question 22.
Which out of Li, Na, K, Be, Mg, Ca has lowest ionization enthalpy and why ?
Answer:
K has lowest ionisation energy due to larger atomic size among these elements. The force of attraction between valence electron and nucleus is less, therefore, it can lose electron easily.

Question 23.
Which alkali metal ion forms largest hydrate ion in aqueous solution and why?
Answer:
Li⁺ forms largest hydrated cations because it has highest hydration energy. It has smallest size therefore, it is most hydrated.

Question 24.
What is responsible for the blue colour of the solution of alkali metal in liquid ammonia ? Give chemical equation also. [MSE (Chandigarh) 2003]
Answer:
The solvated electron, [e(NH3)]^– or ammoniated electron is responsible for blue colour of alkali metal solution in NH₃. It absorbs light from visible region and radiates complementary colour, (in the equation am = ammoniated)
Na⁺(am) + e^–(am) + NH₂(I) →NaNH₂(am) + \(\frac { 1 }{ 2 }\)H₂(g)

Question 25.
Heat of Hydration of Na+ (size 102 pm) = -397 kJ mol-1 whereas Caz 100 pm) = -1650 kJ mol-1. Explain the difference.
Answer:
Ca²⁺ is smaller in size than Na⁺ and also it has higher charge, therefore, its hydration energy is more than that of Na⁺.

Question 26.
Discuss the diagonal relationship of Be and A1 with regard to
1. action of alkali and
2. the structure of their chloride.
Answer:
1. Be and A1 both react with NaOH to form sodium beryllate and sodium meta aluminate respetively. Be dissolves in excess of NaOH to form [Be(OH)4]² where as
Al forms [ A1 (OH)6 ]^3- in excess of NaOH.
2H₂O + Be + 2NaOH → Na₂ [Be (OH)4 ] (saliun beryllali) + H₂
2A1 + 6NaOH + 6H₂O → 2Na₃ [ A1 (OH)₆ ] (Sodiam meta ilumivali) + 3H₂

2. BeCl₂ is electron deficient, therefore, has polymeric chain structure in solid state.

AlCl₃ is also electron deficient, It exists as dimmer, i.e., Al₂Cl₆

Question 27.
Complete the following:
(i) Li + N₂ →
(ii) \(\text { LiNO }_{3} \stackrel{\text { heat }}{\longrightarrow}\)
(iii) \(\mathrm{NaNO}_{3} \stackrel{\text { heat }}{\longrightarrow}\)
(iv) B₂H₆ →
Answer:

Question 28.
Arrange the (i) hydroxide and (ii) sulphates of alkaline earth metals in order of decreasing solubilities giving a suitable reason for each.
Answer:
Ba(OH)₂ > Sr(OH)₂ > Ca(OH)₂ > Mg(OH)₂ > Be(OH)₂
Solubility of hydroxides goes on increasing down the group because hydration energy dominates over lattice energy.
BaSO₄ < SrSO₄ < CaSO₄ < MgSO₄ < BeSO₄

Question 29.
What makes lithium show properties uncommon to the rest of alkali metals ? Write two points of similarly in properties between lithium and magnesium.
Answer:
Solubility of sulphates goes on decreasing down the group because lattice energy dominates over hydration energy. Lithium has smallest atomic size and highest ionization energy, highest polarizing power that is why it shows uncommon properties to the rest of alkali metals.

• Both Lithium and Magnesium react with N₂ to form nitrides.
• Li and Mg react with O₂ to form monoxides.

Question 30.
Write the chemical equations of the reactions involved in solvay process of preparation of sodium carbonate.
Answer:

Question 31.
Arrange the following in order of the increasing covalent character : MCI, MBr, MF, MI (where M = alkali metal)
Answer:
As the size the anion increases, covalent character increases and hence the order is ‘MF < MCI < MBr < MI.

Question 32.
What is the formula of gypsum? What happens when it is heated?
Answer:
CaSO₄. 2H₂O. When heated to 393 K, it gives plaster of paris (CaSO₄.1/2H₂O ) but at 473 K it gives dead burnt plaster (CaSO₄).

Question 33.
The E° for C1^–/Cl₂ is 1.36, for I^–/I₂ is +0.53, for Ag⁺/Ag is + 0.79, Na⁺ is -2.71 and for Li⁺ / Li is -3.04 V Arrange the following species in decreasing order of reducing strength. I^–, Ag, Cl^– Li, Na
Answer:
The more negative or less positive is the electrode potential, more is the reducing strength of the species. Since the electrode potentials inceases in the order : Li(-3.04V) < Na(-2.7V) < I^–(0.53 V) < Ag(+0.79V) < Cl^– (+1.36V), therefore, the reducing strength decreases in the order Li > Na > I^– > Ag > Cl^–

Question 34.
How do you prepare KO₃ ? Mention Magnetic Behaviour of \(\mathrm{O}_{3}^{-}\) .
Answer:
Potassium ozonide (KO₃) is formed when ozone is passed through KOH.
2KOH + 5O₃ → 2KO₃ + 5O₂ + H₂O
It is an orange coloured solid and contains the paramagnetic O₃ ion.

Question 35.
What are ionic polyhide compounds ?
Answer:
The alkali metals react with halogens and interhalogen compounds forming ionic polyhide compounds.
KI + I₂ → K[I₃] ; KBr + ICI → K[BrICI] ; KF + BrF₃ → K[BrF₄]

1st PUC Chemistry The S-Block Elements Three / Four Marks Questions and Answers

Question 1.
What are alkali metals? Describe their general properties.
Answer:
1st group elements of periodic table i.e., lithium, sodium, potassium, rubidium and caesium are called alkali metals.
General properties:

• Alkali metals have general electronic configuration ns¹E
• Alkali metals exhibit on oxidation state of+1.
• Atomic radius increases down the group from lithium to caesium.
• The metallic property of alkali metals increases from lithium to caesium.
• Alkali metals have low ionization potential.
• Due to low ionization power they are highly electropositive.
• Alkali metals are light metals.

Question 2.
Write the balanced equations from the reaction between
(a) Na₂ O₂ and water
(b) KO₂ and water
(c ) Na₂O and CO₂
Answer:
(a) 2Na₂O₂ + 2H₂O → 4NaOH + O₂
(b) 2KO₂ + 2H2_O → 2KOH + H₂O₂ + O₂
(c) Na₂O + CO₂ → Na₂CO₃

Question 3.
What is the action of heat on the following compound ?
(i) Na₂CO₃ and CaCO₃ ,
(ii) MgCl₂.6H₂O and CaCl₂.6H₂O
(iii) Ca(NO₃)₂ and NaNO₂
Answer:
(i) Na₂CO₃ and CaCO₃
Na₂CO₃ does not decompose on heating while CaCO₃ evolves CO₂
\(\begin{array}{l}{\mathrm{Na}_{2} \mathrm{CO}_{3} \stackrel{\text { heat }}{\longrightarrow} \text { Noaction }} \\ {\mathrm{CaCO}_{3} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}}\end{array}\)

(ii) MgCl₂.6H₂O and CaCl₂.6H₂O
On heating hydrated CaCl₂ is dehydrated while hydrated MgCl₂ changes into MgO

(iii) On heating the two nitrate form different products

Question 4.
Complete the following

1. Ca + H₂0 →
2. Ca(OH)₂ + Cl₂ →
3. BeO + NaOH →

Answer:

1. Ca + H₂O → CaO + H₂O
2. Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
3. BeO + NaOH → Be(OH)₂ + Na₂O

Question 5.
Explain what happens when
(i) Sodium hydrogen carbonate is heated
(ii) Sodium with mercury reacts with water
(iii) Fused sodium metal reacts with ammonia.
Answer:

Question 6.
Why is it that the s-block elements never occur in free state/nature? What are their usual modes of occurrence and how are they generally prepared?
Answer:
The elements belonging to s-block in periodic table (i.e. alkali and alkaline earth metals) are highly reactive because of their low ionization energy. They are highly electropositive forming positive ions. So they are never found in free state.
They are widely distributed in nature in the combined state. They occur in earth’s crust in the form of oxides, chlorides, silicates and carbonates.
Generally group I metals are prepared by the electrolysis of fused solution.
As for example:

These metals are highly reactive and therefore cannot be extracted by the usual methods, because they are strong reducing agents.

Question 7.
What is the effect of heat on the following compounds ?
(a) Magnesium chloride hexahydrate
(b) Gypsum
(c) Magnesium sulphate heptahydrate
Answer:

Question 8.
What is the approximate composition of Portland cement ? What raw materials are used in the manufacture of this cement ? Describe method.
Answer:
Raw materials : The raw materials required for the manufacture of cement are lime stone, stone and clay. Lime stone in calcium carbonate, (CaCO₃) and it provides calcium oxide. (CaO) Clay is hydrated aluminium silicate, (Al2O₃.2SiO₂.2H₂O) and it provides alumina as well as silica. A small amount of gypsum, CaSO₄.2H₂O is also required. It is added in calculated quantity in order to adjust the rate of setting of cement.

Manufacture : Cement is made by strongly heating a mixture of lime stone and clay in a rotatory kiln. Lime stone and clay are finely powdered and a little water is added to get a thick paste called slurry. The slurry is fed into a rotatory kiln from the top through the hopper.

The hot gases produce a temperature of about 1770-1870 K in the kiln. At this high temperature at the lime stone and clay present in slimy combine to form cement in the form of small pieces called clinker. This clinker is mixed with 2-3% by weight of gypsum (CaSO₄.2 H₂O) to regulate the setting time and is then ground to an exceedingly fine powder.
\(\text { Limestone }+\text { Clay } \frac{170-1870 \mathrm{K}}{\text { (Clinker) }} \text { , Cement }+\mathrm{CO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O} \uparrow\)
When mixed with water the cement reacts to form gelatinous mass which sets to a hard mass when three .dimensional cross lines are formed between silica oxygen silica and silica oxygen aluminium as .
…….. Si – O – Si …….. and Si – O – Al ……….. chains
Composition of cement:
CaO = 50 -60%
SiO₂ = 20 – 25%
Al₂O₃ = 5 – 10%
MgO = 2 – 3%
Fe₂G₃ = 1-2%
SO₃ = 1 – 2%
For a good quality cement the ratio of’silica (SiO₂) and alumina (AI₂O₃) should be between 2.5 to 4.0. Similarly the ratio of lime (CaO) to the total oxide mixtures consisting of SiO₂, Al₂O₃ and Fe₂O₃ should be roughly 2:1:1, If lime is in excess, the cement cracks during setting. On the other hand, if lime is less than the required, the cement is weak in strength. Therefore, a proper composition of cement must be maintained to get cement of good quality.

Question 9.
Identify A and B in the following reaction


Answer:
(i) ‘A’ is BeCh and ‘B’ is AlCl₃
(ii) A is CaCO₃, B is CO₂

Question 10.
A white crystalline solid ‘A’ on heating loses the water of crystallization to form a monohydrate ‘B’ above 373K, the monohydrate also becomes completely anhydrous and changes to white powder called soda ash. Identify ‘A’ and ‘B’. Also give two uses of ‘A’
Answer:
‘A’ is Na2CO3.10H₂O (Washing soda)
‘B’ is Na₂CO₃ (Anhydrous sodium carbonate)

Uses of‘A’
(a) Used for softening hard water.
(b) Used in glass and soap industries.

Question 11.
Write the uses, and any two reaction of KO₂.
Answer:
Potassium superoxide (KO₂) is used as a source of oxygen in submarines, space shuttles and in emergency breathing apparatus such as oxygen masks. Such masks are used in rescue work in mines and in other areas where the air is so deficient in oxygen that an artificial atmosphere must be generated.

The moisture of the breath reacts with superoxide to liberate oxygen, and at that same time the potassium hydroxide formed removes carbon dioxides as it is exhaled thereby allowing the atmosphere in the mask to be continuously regenerated.
4KO₂(s) + 2H₂O(g) → 4KOH(aq) + 3₂(g) → KHCO₃ (s)
KO₂ also combines directly with CO₂ forming K₂CO₃ and with CO₂ & moisture forming KHCO₃
4KO₂ + 2CO₂ → 2K₂CO₃ + 3O₂; 4KO₂ + 4CO₂ + 2H₂O → 4KHCO₃ + 3O₂

Question 12.
(a) Write any four uses of Calcium Hydroxide.
(b) Give chemical equation of the reaction of caustic soda with
1. ammonium chloride, and
2. carbon dioxide
Answer:
(a)

• It is used in the building material in the form of mortar.
• It is used in the manufacture of bleaching powder.
• It is used in glass making and the purification of sugar.
• It is used to absorbed acidic gases.

(b)
1. NH₄Cl + NaOH → NaCl + NH₄OH
2. 2NaOH + CO₂ → Na₂CO₃ + H₂O

Question 13.
What is plaster of paris ? How is it prepared ? Give its any two important uses.
Answer:
Plaster of paris is CaSO₄ \(\frac { 1 }{ 2 }\) H₂O. It is prepared by heating gypsum at 373K

Uses:

• It is used for manufacture of statues
• It is used for filling gaps before white washing

Question 14.
Discuss the trends of:
1. Thermal stability of alkaline earth metal carbonates
2. Solubility of sulphates of group 2 elements
3. Basic strength of alkaline earth metal hydroxides
Answer
1. Thermal stability of alkaline earth metal carbonates increases down the group due to increase in ionic character and therefore, increase in lattice energy

2. Solubility of sulphates of group 2 elements decreases down the group because lattice energy dominates over hydration energy.

3. Basic strength of alkaline earth metal hydroxides increases down the group because ionization energy of metal decreases and electropositive character increases down the group.

Question 15.
Explain the different oxides of metals or classify different metal oxides.
Answer:
Classification of oxides on the basis of oxygen content.
On the basis of oxygen content, oxides can be classified into the following types
1. Normal oxides: Those oxides inwhich the oxidation number of the element (M) can be deducted from the empirical formula MxOy by taking the oxidation number of oxygen as – 2 are called normal oxides. For example, H₂O, Na₂O, MgO AI₂O₃, CO₂ etc. All these oxides contain M – O bonds.

2. Polyoxides: These oxides contain more oxygen than would be expected from the oxidation number of the element (M). These have have been further classified into peroxides, and superoxides.

a : Peroxides : Metallic axides which on treatment with dilute acids produce hydrogen peroxide are called peroxides. For example, Na₂CO₂ and BaO₂. In these peroxides, the two oxygen atoms are linked by a single hand and each oxygen atom has an oxidation state -1. In other words, all peroxides contain a peroxide ion \(\left(0_{2}^{2-}\right)\) having the structure. In this structure,  all the electrons are paired and hence all peroxides are diamagnetic.

There are certain other oxides like PbCO₂ and MnO₂ which may be mistaken as peroxides. These compounds, however, do not give H₂O₂ on treatment with dilute acids. As such these compounds do not contain a peroxide ion \(\left(0_{2}^{2-}\right)\) and hence they cannot be called as peroxides. Actually in these compounds the two oxygen atoms are linked to the metal atom by a double bond and hence called dioxide i.e, 0 = Pb = 0 (Lead dioxide) and O = Mn = O (mangenses dioxide). In dioxides, the oxidation state of each oxygen atom is -2.

b. Superoxides: Besides peroxides, alkali metals also form higher oxides called superoxides. For example, potassium superoxide (KO₂), rubidium superoxide (RbO₂) cesium superoxide (CSO₂) etc. All these superoxides contain a superoxide ion, i.e., \(\mathrm{O}_{2}^{-}\) having the structure,  Thus all superoxides contain an odd number of electrons (i.e. 13) and hence are paramagnetic.

1st PUC Chemistry The S-Block Elements Give Reasons

Question 1.
The ionic compounds of alkali metals are colourless, why?
Answer:
Alkali metals form unipositive ions which have stable configuration of the nearest inert gas. Alkali metal salts are diamagnetic and colorless because they do not have impaired electrons.

Question 2.
Alkali metals are good conductors of electricity why ?
Answer:
Alkali metals have low ionization energy, Hence they show metallic character. They are good conductors of electricity due to the presence of mobile valance electrons