Chapter 6: Thermodynamics

Q15Written answer

Thermodynamics mainly deals with

(i) interrelation of various forms of energy and their transformation from one form to another.

(ii) energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.

(iii) how and at what rate these energy transformations are carried out.

(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.

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Ans: The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Hence, option (i) an (iv) is the correct answer.

Q16Written answer

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. For such system

(i) \[{{\mathbf{q}}_{\mathbf{p}}}\] will be negative

(ii) \[{{\mathbf{\Delta }}_{\mathbf{r}}}\mathbf{H }\] will be negative

(iii) \[{{\mathbf{q}}_{\mathbf{p}}}\] will be positive

(iv) \[{{\mathbf{\Delta }}_{\mathbf{r}}}\mathbf{H }\] will be positive

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Ans: In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, \[{{\text{q}}_{\text{p}}}\] will be negative and \[{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}\text{H }\] will also be negative. Hence, option(i) and (ii) is the correct answer.

Q17Written answer

The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are

(i) flow of heat from colder to warmer body.

(ii) gas in a container contracting into one comer.

(iii) gas expanding to fill the available volume.

(iv) burning carbon in oxygen to give carbon dioxide.

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Ans: Gas expands to fill the available space spontaneously and burning of carbon-to-carbon dioxide is spontaneous. Hence, option(iii) and (iv) is the correct answer.

Q18Written answer

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression \[\mathbf{w = nRT ln }\frac{{{\mathbf{V}}_{\mathbf{f}}}}{{{\mathbf{V}}_{\mathbf{i}}}}\] .A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.

(i) Work done at 600 K is 20 times the work done at 300 K.

(ii) Work done at 300 K is twice the work done at 600 K.

(iii) Work done at 600 K is twice the work done at 300 K.

(iv) ∆U = 0 in both cases.

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Ans: For isothermal reversible change, we know

\[\text{q=-w=nRTIn}\frac{{{\text{V}}_{\text{f}}}}{{{\text{V}}_{\text{i}}}}\text{=2}\text{.303nRTlog}\frac{{{\text{V}}_{\text{f}}}}{{{\text{V}}_{\text{i}}}}\]

\[\frac{\begin{align} & \\ & {{\text{W}}_{\text{600K}}} \\ \end{align}}{{{\text{W}}_{\text{300K}}}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ 300KIn}\frac{\text{1}}{\text{10}}}{\text{1 }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ 600KIn}\frac{\text{1}}{\text{10}}}\text{=}\frac{\text{600}}{\text{300}}\text{=2}\]

For isothermal expansion of ideal gases, ΔU=0

Since temperature is constant this means there is no change in internal energy.
Therefore, ΔU=0. Hence, option(iii) and (iv) is the correct answer.

Q19Written answer

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:

\[\mathbf{2}\text{ }\mathbf{Zn}\text{ }\left( \mathbf{s} \right)+{{\mathbf{O}}_{\mathbf{2}}}\text{ }\left( \mathbf{g} \right)\text{ }\to \mathbf{2}\text{ }\mathbf{ZnO}\text{ }\left( \mathbf{s} \right)\text{ };\text{ }\Delta \mathbf{H}=\text{ }\mathbf{693}.\mathbf{8}\text{ }\mathbf{kJ}\text{ }\mathbf{mo}{{\mathbf{l}}^{\mathbf{1}}}\]

(i) The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.

(ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.

(iii) 693.8 kJ mol–1 energy is evolved in the reaction.

(iv) 693.8 kJ mol–1 energy is absorbed in the reaction.

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Ans: As we know from thermodynamics,

\[{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{o}}}{{\text{=}}_{\text{i}}}\sum{{{\text{a}}_{\text{i}}}}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\odot }}\left( \text{products} \right)\text{-}\sum{{{\text{a}}_{\text{i}}}}{{\text{b}}_{\text{i}}}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\odot }}\left( \text{reactants} \right)\]
Since, the above reaction is an exothermic reaction, the enthalpy of reactant is more than the enthalpy of product. Hence, option(i) and (iii) is the correct answer.

III. Short Answer Type

Q20Written answer

18.0 g of water completely vaporises at \[\mathbf{10}{{\mathbf{0}}^{\mathbf{\bullet }}}\mathbf{C}\] and 1 bar pressure and the enthalpy change in the process is 40.79 kJ \[\mathbf{Mo}{{\mathbf{l}}^{\mathbf{-1}}}\]. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

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Ans: Enthalpy of a reaction is defined as the energy change per mole for the process. Therefore,

18 g of \[{{\text{H}}_{\text{2}}}\text{O}\]= 1 mole (\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{Vap}}}\] = 40.79 kJ \[\text{Mo}{{\text{l}}^{\text{-1}}}\])

Enthalpy change for vaporising 2 moles of \[{{\text{H}}_{\text{2}}}\text{O}\] = 2 x 40.79 = 81.58 kJ

\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{Vap}}}\] = 40.79 kJ \[\text{Mo}{{\text{l}}^{\text{-1}}}\].

Q21Written answer

One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vaporization?

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Ans: Due to weak force of attraction between molecules, acetone requires less heat to vaporise. Hence, water has higher enthalpy of vaporization.

Q22Written answer

Standard molar enthalpy of formation, $\Delta $fH- is just a special case of enthalpy of reaction, $\Delta $rH- Is the $\Delta $rH- forhe following reaction same as $\Delta $fH-? Give reason for your answer.

CaO(s)+CO2(g)$\to $CaCO3(s) $\Delta $fH-= -178.3KJ/mol

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Ans: Standard molar enthalpy of formation,rH1- is just a special case of fH2-, where

one mole of a compound is formed from its constituent elements. In the above equation, enthalpy of formation and enthalpy of reaction is not the same.

Q23Written answer

The value of $\Delta {{H}^{\Theta }}$ for NH3 is – 91.8 kJ mol-1. Calculate the enthalpy change for the following reaction :

2NH3(g) $\to $N2(g) + 3H2(g)

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Ans: The enthalpy change for the given reaction can be calculated as: -

N2(g) + 3H2(g)$\to $2NH3 (g)

$\Delta $fH=-91.8kg/mol

The enthalpy of the reaction is +91.8 kg/mol. This happens as with the reversal of reaction, the value of $\Delta $fH also gets reversed.

Q24Written answer

Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is $\Delta $rH and $\Delta $rH1, $\Delta $rH2, $\Delta $rH3……represent enthalpies of intermediate reactions leading to product B. What will be the relation between $\Delta $rH for overall reaction and$\Delta $rH1, $\Delta $rH2, $\Delta $rH3 etc for intermediate reactions.

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Ans: According to Hess’s law, $\Delta $rH = $\Delta $rH1+ $\Delta $rH2+$\Delta $rH3

This is so because during the reaction A$\to $ B, B’s formation undergoes various intermediate reactions, with the overall value of the enthalpy being $\Delta $rH.

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Multiple Choice Questions (Type-II)