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NCERT Exemplar · Class 11 · Chemistry

Chapter 6: Thermodynamics

Practice each question, then open the answer or worked solution.

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Multiple Choice Questions (Type-I)

14 questions
Q1Written answer

Thermodynamics is not concerned about ________. 

(i) energy changes involved in a chemical reaction. 

(ii) the extent to which a chemical reaction proceeds.

(iii) the rate at which a reaction proceeds. 

(iv) the feasibility of a chemical reaction. 

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Worked solution

Ans: Thermodynamics is not concerned about how and at what rate these energy transformations are carried out but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Hence, option(iii) is the correct answer.


Q2Written answer

Which of the following statements is correct?

(i) The presence of reacting species in a covered beaker is an example of open system. 

(ii) There is an exchange of energy as well as matter between the system and the surroundings in a closed system. 

(iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system. 

(iv) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system. 

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Worked solution

Ans: The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system. Hence, option(iii) is the correct answer.


Q3Written answer

The state of a gas can be described by quoting the relationship between _____.

(i) pressure, volume, temperature

(ii) temperature, amount, pressure

(iii) amount, volume, temperature 

(iv) pressure, volume, temperature, amount 

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Worked solution

Ans: Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. Hence, option(i) is the correct answer.


Q4Written answer

The volume of gas is reduced to half from its original volume. The specific heat will _____. 

(i) reduce to half 

(ii) be doubled

(iii) remain constant 

(iv) increase four times 

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Worked solution

Ans: Specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one kelvin). That is why it is an intensive property which does not depend on mass. Hence, option(iii) is the correct answer.


Q5Written answer

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is 

\[\left( \mathbf{i} \right)\mathbf{ 2}{{\mathbf{C}}_{\mathbf{4}}}{{\mathbf{H}}_{\mathbf{10}}}\left( \mathbf{g} \right)\mathbf{ + 13}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\to \mathbf{8C}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\mathbf{ + 10}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\left( \mathbf{l} \right)\mathbf{ }{{\mathbf{\Delta }}_{\mathbf{c}}}\mathbf{H = 2658}\mathbf{.0 kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\mathbf{ }\]

\[\left( \mathbf{ii} \right)\mathbf{ }{{\mathbf{C}}_{\mathbf{4}}}{{\mathbf{H}}_{\mathbf{10}}}\left( \mathbf{g} \right)\mathbf{ + }{}^{\mathbf{13}}/{}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\mathbf{ }\to \mathbf{4C}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\mathbf{ + 5}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O }\left( \mathbf{g} \right)\mathbf{ }{{\mathbf{\Delta }}_{\mathbf{c}}}\mathbf{H= 1329}\mathbf{.0 kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\]

\[\left( \mathbf{iii} \right)\mathbf{ }{{\mathbf{C}}_{\mathbf{4}}}{{\mathbf{H}}_{\mathbf{10}}}\left( \mathbf{g} \right)\mathbf{ + }{}^{\mathbf{13}}/{}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\to \mathbf{4C}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\mathbf{ + 5}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O }\left( \mathbf{l} \right)\mathbf{ }{{\mathbf{\Delta }}_{\mathbf{c}}}\mathbf{H = 2658}\mathbf{.0 kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\]

\[\left( \mathbf{iv} \right)\mathbf{ }{{\mathbf{C}}_{\mathbf{4}}}{{\mathbf{H}}_{\mathbf{10}}}\left( \mathbf{g} \right)\mathbf{ + }{}^{\mathbf{13}}/{}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\to \mathbf{ 4C}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\mathbf{ + 5}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O }\left( \mathbf{l} \right)\mathbf{ }{{\mathbf{\Delta }}_{\mathbf{c}}}\mathbf{H= +2658}\mathbf{.0 kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\]

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Worked solution

Ans: Standard enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) of a substance when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature. Hence, option(iii) is the correct answer.


Q6Written answer

\[{{\mathbf{\Delta }}_{\mathbf{f}}}{{\mathbf{U}}^{\mathbf{e}}}\] of formation of \[\mathbf{C}{{\mathbf{H}}_{\mathbf{4}}}\] (g) at certain temperature is –393 kJ mol–1. The value of \[{{\mathbf{\Delta }}_{\mathbf{f}}}{{\mathbf{H}}^{\mathbf{e}}}\] is 

(i) zero

(ii) < \[{{\mathbf{\Delta }}_{\mathbf{f}}}{{\mathbf{U}}^{\mathbf{e}}}\] 

(iii) > \[{{\mathbf{\Delta }}_{\mathbf{f}}}{{\mathbf{U}}^{\mathbf{e}}}\] 

(iv) equal to \[{{\mathbf{\Delta }}_{\mathbf{f}}}{{\mathbf{U}}^{\mathbf{e}}}\]

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Worked solution

Ans: The balanced equation for the combustion of methane is: 

\[\text{C}{{\text{H}}_{\text{4}}}\left( \text{g} \right)\text{+2}{{\text{O}}_{\text{2}}}\left( \text{g} \right)\to \text{C}{{\text{O}}_{\text{2}}}\left( \text{g} \right)\text{+2}{{\text{H}}_{\text{2}}}\text{O}\left( \text{l} \right)\]

Therefore, \[\text{ }\!\!\Delta\!\!\text{ }{{\text{n}}_{\text{g}}}\] =1−3=−2

\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{o}}}\] = \[\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}_{\text{o}}}\] + \[\text{ }\!\!\Delta\!\!\text{ }{{\text{n}}_{\text{g}}}\]​RT

\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{o}}}\]=−393−2RT

∴ \[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{o}}}\]< \[\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}_{\text{o}}}\]

Hence, option (ii) is the correct answer.


Q7Written answer

In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following. 

(i) q = 0, ∆T ≠ 0, w = 0 

(ii) q ≠ 0, ∆T = 0, w = 0 

(iii) q = 0, ∆T = 0, w = 0 

(iv) q = 0, ∆T < 0, w ≠ 0 

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Worked solution

Ans: In free expansion, w=0 because volume is constant, as the process is adiabatic q = 0 and from first law of thermodynamics.

\[\text{ }\!\!\Delta\!\!\text{ U=q+w}\]

This means that internal energy remains constant.

Hence, option(iii) is the correct answer.


Q8Written answer

The pressure-volume work for an ideal gas can be calculated by using the expression \[\mathbf{w=-}\int\limits_{{{\mathbf{V}}_{\mathbf{i}}}}^{{{\mathbf{V}}_{\mathbf{f}}}}{{{\mathbf{P}}_{\mathbf{ex}}}}\mathbf{dV}\]

The work can also be calculated from the pV plot by using the area under the curve within the specified limits. When an ideal gas is compressed 

(a) reversibly or (b) irreversibly from volume \[{{\mathbf{V}}_{\mathbf{i}}}\] to \[{{\mathbf{V}}_{\mathbf{f}}}\]. 

choose the correct option. 

(i) w (reversible) = w (irreversible) 

(ii) w (reversible) < w (irreversible) 

(iii) w (reversible) > w (irreversible) 

(iv) w (reversible) = w (irreversible) + \[{{\mathbf{P}}_{\mathbf{ex}}}\] ∆V

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Worked solution

Ans: As the area under the curve is always more in irreversible compression and this can be explained by the below curves- 


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Hence, option(ii) is the correct answer.


Q9Written answer

The entropy change can be calculated by using the expression \[\mathbf{\Delta S =}{{\mathbf{q}}_{\mathbf{rev}}}\mathbf{T}\] . When water freezes in a glass beaker, choose the correct statement amongst the following : 

(i) ∆S (system) decreases but ∆S (surroundings) remains the same. 

(ii) ∆S (system) increases but ∆S (surroundings) decreases.

(iii) ∆S (system) decreases but ∆S (surroundings) increases.

(iv) ∆S (system) decreases and ∆S (surroundings) also decreases.

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Worked solution

Ans: (iii) ∆S (system) decreases but ∆S (surroundings) increases.


Q10Written answer

On the basis of thermochemical equations (a), (b) and (c), find out which of the algebraic relationships given in options (i) to (iv) is correct. 

\[\left( \mathbf{a} \right)\mathbf{ C}{{\mathbf{ }}_{\left( \mathbf{graphite} \right)\mathbf{ }}}\mathbf{+ }{{\mathbf{O}}_{\mathbf{2 }}}\left( \mathbf{g} \right)\mathbf{ }\to \mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\mathbf{ ; }{{\mathbf{\Delta }}_{\mathbf{r}}}\mathbf{H = x kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\]

\[\left( \mathbf{b} \right)\mathbf{ C}{{\mathbf{ }}_{\left( \mathbf{graphite} \right)}}\mathbf{ + }\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\to \mathbf{CO }\left( \mathbf{g} \right)\mathbf{ ; }{{\mathbf{\Delta }}_{\mathbf{r}}}\mathbf{H  = y kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\]

\[\left( \mathbf{c} \right)\mathbf{ CO }\left( \mathbf{g} \right)\mathbf{ + }\frac{\mathbf{1}}{\mathbf{2}}\mathbf{ }{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\to \mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\left( \mathbf{g} \right)\mathbf{ ; }{{\mathbf{\Delta }}_{\mathbf{r}}}\mathbf{H = z kJ mo}{{\mathbf{l}}^{\mathbf{1}}}\]

(i) z = x + y

(ii) x = y – z

(iii) x = y + z 

(iv) y = 2z – x

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Worked solution

Ans:  When we add equations (b) and (c) we will get equation(a) hence, Algebraic sum of y and z will give x, x =y + z. Hence, option(iii) is the correct answer.