MPBSE Class 11 Chemistry Solutions For Chemical Thermodynamics

Chemical Thermodynamics Class 11 Notes

Question 1. In which of the following two processes, the change in entropy of the system will be negative?

• Fusion of Ice.
• Condensation of water vapor

Answer: The change in entropy of the system is negative during the condensation of water vapor. In this process \(\Delta S_{\text {sys }}=S_{\text {water }}-S_{\text {water vapour }}<0\)

As the molecules in water vapor have more freedom of motion than they have in the water, the molecular randomness is higher in water vapor than in water. Thus, \(S_{\text {water }}<S_{\text {water vapour }}\) Consequently, the value of AS becomes negative.

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Question 2. The change in internal energy in different steps of the process A→ B → C→ D are given: A→Bx.kJ-mol-1; B→ C, -y kj-mol-1; C-D,z kj-mol-1. What will the value of AH be for the change A → D?
Answer:

⇒ \(A →{\Delta U_1} B→{\Delta U_2} C →{\Delta U_3} D\)

Δ Ux = UR- UA = x kj.mol-1 ;

ΔU2 = Hc.— UB = -ykjmol-1;

ΔU3 = UD-uc = zkj.mol-1

∴ For the change A D

A U = UD-UA =(UD-UC) + (UC-UB) + (UB-UA)

=(z-y + x) kj-mol-1.

Chemical Thermodynamics Class 11 Notes

Question 3. Change in enthalpy in different steps of the process A→B→C→A are given: A→B, x kj-mol-1; C→A, y kj-mol-1. Find the value of AH for step B→C.
Answer:

The initial and final states (A) are the same in the given process. So it is a cyclic process. Since H is a state function, AH = 0 for the process. Thus in this process \(\begin{aligned}
& \Delta H=\left(H_B-H_A\right)+\left(H_C-H_B\right) \Psi\left(H_A-H_C\right)=0 \\
& \left.x+\left(H_C-H_B\right)+y=0 \text { or }\left(H_C\right)_B H_B\right)=-(x+y) \mathrm{kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

So change in enthalpy for BC =- (x + y) kj-mol-1.

Question 4. Why are the standard reaction enthalpies of the following two reactions different?
Answer:

⇒ \(\begin{aligned}
& \mathrm{C}(\text { graphite, } s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-393.5 \mathrm{~kJ} \\
& \mathrm{C}(\text { diamond }, s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-395.4 \mathrm{~kJ}
\end{aligned}\)
Anszwer: Graphite and diamond are two allotropes ofsolid carbon. Different allotropic forms have different enthalpies. As the given U reactions involve different allotropes, the standard reaction enthalpies ofthese reactions are different.

Question 5. At 25°C, the standard reaction enthalpy for the reaction is -221.0 kj. Does this enthalpy change indicate the standard enthalpy of the formation of CO(g)? If not, then what would be the value of the enthalpy of j formation of CO(g) at 25°C?
Answer:

In the given reaction 2 mol CO(g) is formed from its stable constituent elements. Sd,’ definition, AH0 of this reaction does not represent standard enthalpy of formation of CO(g). 2C(graphite,s) + O₂(g) 2CO(g) ; AH0 = -221.0 kj

Dividing both sides by 2, we get

⇒ \(\mathrm{C}(\text { graphite, } s)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g) ; \Delta H^0=-110.5 \mathrm{~kJ} \cdots[1]\)

In reaction [1], 1 mol of CO(g) is formed from its stable constituent elements. Thus, in this reaction AH0 = standard enthalpy of formation of CO(g).

So, at 25°C the standard enthalpy of formation of CO(g) =-110.5 kj-mol-1.

Chemical Thermodynamics Class 11 Notes

Question 6. Determine the standard reaction enthalpy for the reaction: \(\mathrm{A}_2 \mathrm{~B}_3(s)+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(s)+3 \mathrm{CB}_2(g)\) Given: \(2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow \mathrm{A}_2 \mathrm{~B}_3(s); \Delta H^0=-x \mathrm{~kJ}\)
Answer:  Reversing equation 1 we get \(\mathrm{A}_2 \mathrm{~B}_3(s) \rightarrow 2 \mathrm{~A}(s)+\frac{3}{2} \mathrm{~B}_2(\mathrm{~g}) ; \quad \Delta H^0=+x \mathrm{~kJ}\)

Multiplying the equation by 3, we get

⇒  \(3 \mathrm{CB}(g)+\frac{3}{2} \mathrm{~B}_2(g) \rightarrow 3 \mathrm{CB}_2(g) ; \Delta H^0=-3 y \mathrm{~kJ}\)

Adding equations [3] and [4], we get

⇒ \(\mathrm{A}_2 \mathrm{~B}_3(\mathrm{~s})+3 \mathrm{CB}(g) \rightarrow 2 \mathrm{~A}(\mathrm{~s})+3 \mathrm{CB}_2(g) ; \Delta H^0=(x-3 y) \mathrm{kJ}\)

Therefore, the standard reaction enthalpy for the given reaction =(x- 3y)kJ.

Question 7. In which ofthe following reactions does AH0 at 25 °C indicate the standard enthalpy of formation \(\left(\Delta \boldsymbol{H}_f^0\right)\) of the compound formed in each of the reactions?
Answer:

Here, 2 mol of NH₃(g) is produced from the stable constituent elements, H2(g) and N2(g). Hence according to the definition, at 25°C the standard reaction enthalpy of this reaction is not equal to the standard enthalpy of formation of NH3(g).

The standard state of oxygen at 25°C is O₂(g). So, the given equation does not represent the formation reaction of NO(g). Consequently, at 25°C the standard reaction enthalpy of this reaction is not equal to die standard enthalpy offormation of NO(g).

Here, 1 mol of solid NaCl is formed from its stable constituent elements. Hence at 25°C, the die standard reaction enthalpy of this reaction is equal to the die standard enthalpy of formation of NaCl(s).

Question 8. Discuss the change in die degree of randomness for the following cases— Combustion of kerosene, Sublimation of dry ice, Extraction of salt from seawater, Condensation of water vapour, Crystallisation of a solid from its aqueous solution,
Answer: Combustion of kerosene: Kerosene is a mixture of liquid hydrocarbons. The combustion of kerosene produces C02 and water vapour. As in the reaction, a liquid converts into a gaseous mixture, the molecular randomness ofthe system increases.

Sublimation of dry ice: In dry ice, (solid carbon dioxide), the molecules of C02 exist in an orderly state. When this solid sublimes, the gaseous molecules formed move randomly, i.e., the degree of randomness ofthe molecules increases.

Extraction (Crystallisation) of salt from seawater: In seawater, the attractive forces between Na+ and Cl- ions are weak as the distance between these ions is large. Hence, these ions are virtually free to move randomly. NaCl extracted from the seawater is in a solid state with a crystal structure in which Na+ and Cl- are arranged in a definite order. Hence, the extraction of salt from seawater is associated with a decrease in the degree of randomness.

Condensation of water vapor: In water vapor, the intermolecular forces of attraction between H2O molecules are weak, so the molecules remain in a state of randomness. However, water obtained by condensation of water vapor has less freedom of motion and hence less degree of randomness because of the stronger intermolecular forces of attraction compared to water vapour. Hence, in the case of condensation of water vapour, the degree of randomness decreases.

Crystallization of a solid from its Aqueous solutions:

In an aqueous solution, the solute molecules exist in a state of random motion. When the solute is crystallised from its solution, the solute molecules in the die crystal remain at fixed positions and become almost motionless. So, the crystallization of a solid from its solution causes a decrease in randomness.

⇒ \(\mathrm{Cr}^{3+}+6 \mathrm{H}_2 \mathrm{O}(a q) \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)\): In this process, one Cr3+ ion combines with six water molecules to form a single complex ion, [Cr(H20)6]3+. Consequently, the number of particles in the system reduces (from 7 to 1 for each combination), thereby decreasing the degree of randomness.

⇒ \(\mathrm{NH}_4 \mathrm{NO}_2(s) \xrightarrow{\Delta} \mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}):\) On decomposition, 1 formula unit of solid ammonium nitrite produces 1 molecule of N2(g) and 2 molecules of water vapour i.e., H20(g). In this process, a solid (in which the molecules are orderly arranged) converts into gaseous substances. Furthermore, the number of molecules also increases. Naturally, this process increases the randomness ofthe system.

Thermodynamics Important Questions Class 11

Question 9. Mention if the entropy of the system increases or decreases In each of the following cases: Bolling of water, Sublimation of solid iodine,

⇒ \(\begin{aligned}
& 2 \mathrm{O}_3(g) \rightarrow 3 \mathrm{O}_2(g) \\
& \mathrm{NH}_4 \mathrm{Cl}(s) \rightarrow \mathrm{NH}_3(g)+\mathrm{HCl}(g) \\
& \mathrm{Hg}(l) \rightarrow \mathrm{Hg}(g) \quad \text { (6) } \mathrm{I}_2(g) \rightarrow \mathrm{I}_2(s) \\
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \text { (8) } 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \\
& \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2(g) \\
& \mathrm{PCl}_5(g) \rightarrow \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \\
& \text { Haemoglobin }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \text { Oxyhaemoglobin } \\
& \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightarrow \mathrm{Ni}(\mathrm{CO})_4(g) \\
& 4 \mathrm{Fe}(s)+3 \mathrm{O}_2(g) \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(s) \\
& \mathrm{C}(\text { diamond }) \rightarrow \mathrm{C} \text { (graphite) } \\
&
\end{aligned}\)

Answer: The change in entropy in a phase transition or a reaction depends on the following factors.

For a given amount of substance, the entropy of the substance in different physical states varies in the order: of S_gas > S_liquid > S_soild where S is the molar entropy. Because of this, the entropy of the system increases in the phase changes such as liquid → vapour, solid → vapour, while it decreases in the phase changes such as liquid → solid, vapour → liquid and vapour → solid.

In a reaction, indie reactants are solids or liquids or in a dissolved state in solution and they convert into gaseous products, then the entropy of the die system increases, (b) If the number of gaseous particles (atoms or molecules) in a reaction increases or decrease, then the entropy of the system increaser or decreases.

Increases (liquid — vapour transition), Increases (solid-gas transition), Increases (the number of gaseous particles increases) Increases (gaseous substances are formed from a solid reactant) Increases (liquid — gas transition) Decreases (gas-solid transition) Increases (the number of gaseous particles increases).

Decreases (the number of gaseous molecules decreases) Increases (gaseous substance is formed through the reaction of the reactants, one of which is solid and the other is in a dissolved state in solution) Increases (the number of gaseous molecules increases) Decreases (the number of gaseous molecules decreases)

Thermodynamics Important Questions Class 11

Decreases (the number of gaseous molecules decreases)

Decreases (the number of gaseous molecules decreases) (g) Increases (the crystal structure of diamond is more compact than that of graphite. As a result, the molecular randomness in graphite is more than that in diamond.

Question 10. At 25°C and 1 atm pressure, for the reaction 3O₂(g) → 20₂(g); H = 286kJ and AS = -137.2 J.K-1. Is this reaction spontaneous? Does the spontaneity of this reaction depend on temperature? Is the reverse reaction spontaneous? If so, then why? Does the spontaneity of the reverse reaction depend on temperature?
Answer: No. (2) For the given reaction ΔH > 0 and ΔS < 0. So, according to the equation ΔG = ΔH-TΔS, AG will be positive at any temperature. Hence, the spontaneity of this reaction is independent of temperature.

The reverse reaction is spontaneous.

In the reverse reaction [20₃(g)→ 3O₂(g)] ; AH = —286 kj and AS = +137.2 J K- . So, according to the equation, AG = AH- TAS, AG is negative.

The spontaneity of the reverse reaction is also independent of temperature as AG is negative.

H<0 and S> 0 at any temperature.