Important Questions for CBSE Class 11 Chemistry Chapter 5 - Thermodynamics

1 Marks Question

1. Define a system

Ans: A system in thermodynamics is a part of the universe within a specified boundary.

2. Define surroundings

Ans: The remaining of the universe which might be in a position to exchange energy and matter with the system is termed its surroundings.

3. State the first law of thermodynamics

Ans: The first law of thermodynamics states that energy can neither be created nor be destroyed but can be transformed from one form to another.

4. What kind of system is the coffee held in a cup?

Ans: Coffee held in a cup is an open system because it can exchange matter (water vapour and energy (heat) with the surroundings.

5. Give an example of an isolated system

Ans: Coffee held in a thermos flask is an example of an isolated system because it can neither exchange energy nor matter with the surroundings.

6. Name the different types of the system.

Ans: There are three types of system 

1. Open system

2. Closed system

3. Isolated system

7. What will happen to internal energy if work is done by the system?

Ans: There is a decrease in internal energy of the system if work is done by the system.

8. From a thermodynamic point of view, to which system the animals and plants belong?

Ans: Open system

9. How may the state of the thermodynamic system be defined?

Ans: The state of the thermodynamic system can be defined by specifying values of state variables like temperature, pressure and volume.

10. Define enthalpy

Ans:  It is defined as the total heat content of the system.

11. Give the mathematical expression of enthalpy.

Ans: Mathematically $H = U + PV$ where $U$ is internal energy

12. When is enthalpy change $\Delta H$- 

(i) positive (ii) negative.

Ans:

1. In an endothermic reaction it absorbs heat from the surroundings so enthalpy change in positive 

2. In an exothermic reaction heat is evolved so enthalpy change is negative.

13. Give the expression for

 (i) isothermal irreversible change, 

and isothermal reversible change

Ans: For isothermal reversible change 

$Q=\text{ }\!\!~\!\!\text{ }-w$

$Q={{p}_{ext}}({{V}_{f}}-{{V}_{i}})\text{ }\!\!~\!\!\text{ }$

For isothermal reversible change 

$Q=\text{ }\!\!~\!\!\text{ }-w$

$Q=2.303nRT\log \left( \frac{{{V}_{f}}}{{{V}_{i}}} \right)\text{ }\!\!~\!\!\text{ }$

14. Define Heat capacity

Ans: Specific heat /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).

15. Define specific heat.

Ans: Specific heat /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).

16. Give the mathematical expression of heat capacity.

Ans: The mathematical expression of heat capacity is as follows

$q = mc\Delta T$

If,

 $m = 1$

$q = C\Delta T$

17. Define reaction enthalpy.

Ans: The enthalpy change concerning a reaction is termed as reaction enthalpy.

18. Define standard enthalpy.

Ans: The standard enthalpy of reaction is defined as the enthalpy change for a reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.

19. The standard heat of formation of ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is ${\text{824}}{\text{.2}}\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ . Calculate heat change for the reaction. $4{\text{Fe}}(s) + 3{{\text{O}}_{\text{2}}}(g) \to 2{\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s)$ 

Ans: 

$\Rightarrow \Delta {{H}_{R}}=[2\times \Delta {{H}_{f}}^{o}(\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s))-[4\Delta {{H}_{f}}^{o}(\text{Fe}(s))+3\times \Delta {{H}_{f}}^{o}({{\text{O}}_{\text{2}}})]$ $\Rightarrow \Delta {{H}_{R}}=[2\times \Delta {{H}_{f}}^{o}(\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s))-[4\Delta {{H}_{f}}^{o}(\text{Fe(}s))+3\times \Delta {{H}_{f}}^{o}({{\text{O}}_{\text{2}}})]$$\Rightarrow \Delta {{H}_{R}}=[2\times \Delta {{H}_{f}}^{o}(\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s))-[4\Delta {{H}_{f}}^{o}(\text{Fe}(s))+3\times \Delta {{H}_{f}}^{o}({{\text{O}}_{\text{2}}})]$

20. Define spontaneous process.

Ans: A process which occurs without use of an external agent is termed as spontaneous process.

21. Define non-spontaneous processes.

Ans:  A process which occurs with the use of an external agent is termed as spontaneous process.

22. What is the sign of enthalpy of formation of a highly stable compound?

Ans: Negative

23. Predict the sign of ${{\Delta S}}$for the following reaction

${\text{CaC}}{{\text{O}}_{\text{3}}}(s) \to {\text{CaO}}(s) + {\text{C}}{{\text{O}}_{\text{2}}}(g)$

Ans: $\Delta S$ is positive

24. Two ideal gases under the same pressure and temperature are allowed to mix in an isolated system – what will be a sign of entropy change?

Ans: Entropy change is positive since degree of disorderness decreases on mixing.

2 Marks Question

1. Change in internal energy is a state function while work is not, why?

Ans: In a process, the change in internal energy depends upon the initial and final state of the system. Therefore, it is a state function. But work is dependent on the path followed. Therefore, it is not a state function.

2. With the help of first law of thermodynamics and ${\text{H = U + pV}}$ prove ${{\Delta H = }}{{\text{q}}_{\text{p}}}$

Ans: Enthalpy is defined as follows

$H=U+pV$

$\Delta H=\Delta (U+pV)$

$\Delta H=\Delta U+\Delta (pV)$

$\Delta H=\Delta U+p\Delta V+V\Delta p$……(i)

From first law of thermodynamics,

$ \Delta U=q+w$ 

 $ =q-p\Delta V$

From equation (i) and (ii), we get

$\Delta H=q-p\Delta V+p\Delta V+V\Delta p$

$\Delta H=q+V\Delta p$

At constant pressure,

$\Delta p=0$

$V\Delta p=0$

$\Delta H={{q}_{p}}$ at constant pressure

Therefore, $\Delta H={{q}_{p}}$

3. Why is the difference between $\Delta H$ and $\Delta U$ not significant for solids or liquids?

Ans: The difference between $\Delta H$ and $\Delta U$ is not significant for solids or liquids because systems made up entirely of solids and/or liquids do not experience significant volume changes when heated, the difference between and is usually insignificant.

4. What is an extensive and intensive property?

Ans: Extensive property is defined as the property which depends on the quantity or size of the matter present in the system.

Intensive property is defined as the property which depends on the quantity or size of matter present in the system.

5. Show that for an ideal gas ${C_P} - {C_V} = R$

Ans: At constant pressure, when gas is heated, heat is required for increase in the temperature of gas and for doing mechanical work against expansion

At constant volume, heat capacity is written as ${C_V}$ and at constant pressure it is written as ${C_P}$

At constant volume ${q_v} = {C_V}\Delta T$ which is equal to $\Delta U$

At constant pressure ${q_P} = {C_P}\Delta T$ which is equal to $\Delta H$

For one mole ideal gas $\Delta H = \Delta U + \Delta (PV)$

$\Delta H = \Delta U + \Delta (RT)$

$\Delta H = \Delta U + R\Delta T$

$\Delta H = \Delta U + R\Delta T$

On substituting values of ${{\Delta H,\Delta U}}$ , we have

${C_P}\Delta T = {C_V}\Delta T + R\Delta T$

${{C}_{P}}={{C}_{V}}+R$ 

 ${{C}_{P}}-{{C}_{V}}=R $

6. Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to $\dfrac{3}{2}R$.

Ans: For an ideal gas, the average kinetic energy per mole of the gas at any temperature is given by ${E_k} = \dfrac{3}{2}RT$

Hence increase in average kinetic energy of gas for ${{\text{1}}^{\text{o}}}{\text{C}}$rise in temperature is

$\Delta \overrightarrow {{E_k}}  = \dfrac{3}{2}R(T + 1) - \dfrac{3}{2}RT$

By definition,$\overrightarrow {{E_K}} $is the molar heat capacity of gas at constant volume ${C_V}$

$\therefore {C_V} = \dfrac{3}{2}R$

7. A $1.25$ g sample of octane (${C_{18}}{H_{18}}$) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from $294.05$ to $300.78$${\text{K}}$. If heat capacity of the calorimeter is $8.93\;{\text{KJ}}\;{{\text{K}}^{{\text{ - 1}}}}$. find the heat transferred to calorimeter.

Ans: Mass of octane,

$M = 1.250\;{\text{g}}$

$M = 0.00125$

Heat capacity, $c = 8.93\;{\text{KJ}}\;{{\text{K}}^{{\text{ - 1}}}}$

Increase in temperature,$\Delta T = 300.78 - 294.05$

$\Delta T = 6.73\;K$

So, the heat transferred to calorimeter is

$mc\Delta T = 0.00125 \times 8.93 \times 6.73$

$ = 0.075\;{\text{KJ}}$

8. Calculate the heat of combustion of ethylene (gas) to from ${\text{C}}{{\text{O}}_{\text{2}}}$ (gas) and ${{\text{H}}_{\text{2}}}{\text{O}}$ (gas) at ${\text{298}}\;{\text{K}}$ and 1 atmospheric pressure. The heats of formation of ${\text{C}}{{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{\text{O}}$ and${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$  are $ - 393.7$,$ - 241.8$ , $ + 52.3$${\text{kJ}}$ per mole respectively.

Ans: The combustion reaction of ethylene gas is as follows

${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\, + \,\,3{{\text{O}}_{\text{2}}} \to 2{\text{C}}{{\text{O}}_{\text{2}}}(g)\,\, + \,\,2{{\text{H}}_{\text{2}}}{\text{O}}(g)$

$\Delta {H_f}({\text{C}}{{\text{O}}_{\text{2}}}) =  - 393.7\;{\text{KJ}}$

$\Delta {H_f}({{\text{H}}_{\text{2}}}{\text{O}}) =  - 241.8\;{\text{KJ}}$

$\Delta {H_f}({{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}) =  + 52.7\;{\text{KJ}}$

$\Delta {H_R} = \sum {\Delta {H_f}^o(p) - \sum {\Delta {H_f}^o(r)} } $

$\Delta {H_R} = [2 \times \Delta {H_f}^o({\text{C}}{{\text{O}}_{\text{2}}}) + 2 \times \Delta {H_f}^o({{\text{H}}_{\text{2}}}{\text{O}})] - [\Delta {H_f}^o({{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}) + 3 \times \Delta {H_f}^o({{\text{O}}_{\text{2}}})]$

$\Delta {H_R} = [2 \times  - 393.7 + 2 \times  - 241.8] - [523 + 0]$

Enthalpy of formation is zero for elementary change

$\Delta {H_R} =  - 1323.3\;{\text{KJ}}$

9. Give two examples of reactions which are driven by enthalpy change.

Ans: The examples of reactions which are driven by enthalpy change is as follows

For a highly exothermic process its enthalpy change is negative and large value but entropy change is negative is driven by enthalpy change

1. ${{\text{H}}_{\text{2}}}(g) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}(g) \to {{\text{H}}_{\text{2}}}{\text{O}}(l)$

$\Delta {H_f}^o =  - 285.8\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

2. ${{\text{N}}_{\text{2}}}(g) + 3{{\text{H}}_{\text{2}}}(g) \to 2{\text{N}}{{\text{H}}_{\text{3}}}(g)$

$\Delta H_f^o =  - 92\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

10. Will the heat released in the following two reactions be equal? Give reasons in support of your answer

1. ${{\text{H}}_{\text{2}}}(g) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}(g) \to {{\text{H}}_{\text{2}}}{\text{O}}(g)$

2. ${{\text{H}}_{\text{2}}}(g) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}(g) \to {{\text{H}}_{\text{2}}}{\text{O}}(l)$

Ans: The heats released in the two reactions are not equal. The heat released in a reaction depends on the reactants, products and the physical states.

In reaction:

1. Water is produced in the gaseous state whereas in 

2. It is in liquid state.

Also, when water vapors condensed to form water latent heat of vaporization is released. Therefore, more heat is released in reaction (ii).

11. What is the relation between the enthalpy of reaction and bond enthalpy?

Ans: In a chemical reaction the breaking of bonds and formation of new bonds in products takes place. The heat of reaction is dependent on the values needed to break the bond formation. Thus

Heat of reaction = Heat required for breaking of bonds in reactants$ - $Heat required for breaking of bonds in products 

1. $\Delta {H^o} = $ Bond energy is required to break the bonds - Bond energy required to form the bonds = Bond energy of reactants – Bond energy of products.

12. The reaction ${\text{C}}(graphite) + {{\text{O}}_{\text{2}}}(g) \to {\text{C}}{{\text{O}}_{\text{2}}}(g) + 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ represents the formation of ${\text{C}}{{\text{O}}_{\text{2}}}$ and also combustion of carbon. Write the $\Delta {H^o}$ values of the two processes.

Ans: The standard enthalpy of formation of ${\text{C}}{{\text{O}}_{\text{2}}}$ is $ - 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$

$\Delta H_f^o({\text{C}}{{\text{O}}_{\text{2}}}(g)) =  - 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

The standard enthalpy of formation of combustion of carbon is ${\text{ - 393}}{\text{.5}}\;{\text{kJ}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

$\Delta {H_{comb}}({\text{C}}{{\text{O}}_{\text{2}}}(g)) = \; - 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

13. Explain how is enthalpy related to spontaneity of a reaction?

Ans: Most of the exothermic reactions are spontaneous due to an increase in energy.

Burning a substance is a spontaneous process.

${\text{C}}(s) + {{\text{O}}_{\text{2}}}(g) \to {\text{C}}{{\text{O}}_{\text{2}}}(g)$, $\Delta H = \; - 394\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

Many spontaneous reactions start with the absorption of heat. Conversion of water into water vapour is an endothermic spontaneous reaction. Hence the change in enthalpy is not the only criteria for deciding the spontaneity of the reaction.

14. The $\Delta H$ and $\Delta S$ for $2{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}(s) \to 4{\text{Ag}}(s) + {{\text{O}}_{\text{2}}}(g)$ are given $ + \,61.17\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$and $ + \,132\;{\text{J}}{{\text{k}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively. Above what temperature will the reaction be spontaneous?

Ans: From the above question we have $\Delta H$and $\Delta S$ for $2{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}(s) \to 4{\text{Ag}}(s) + {{\text{O}}_{\text{2}}}(g)$ are given as $ + \,61.17\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ and $ + \,132\;{\text{J}}{{\text{k}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively.

As per the Gibbs Helmholtz equation, $\Delta G = \Delta H - T\Delta S$

Here, $\Delta G$ is the change in Gibbs free energy, $\Delta H$is the change in the enthalpy, $T$ is the absolute temperature.

For the reaction $2{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}(s) \to 4{\text{Ag}}(s) + {{\text{O}}_{\text{2}}}(g)$

It is spontaneous when $\Delta G$is negative.

Here $\Delta H,\Delta S$is positive. So $\Delta G$is negative when $\Delta H < T\Delta S$

$T > \dfrac{{\Delta H}}{{\Delta S}}$

$T = \dfrac{{61170\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}}{{132\;{\text{J}}{{\text{k}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}}$

$T = 463.4K$

Therefore, the process is spontaneous above a temperature of ${\text{T}} = {\text{463}}{\text{.4K}}$.

3 Marks Question

1. Give the relationship between ${{\Delta H}}$ and ${{\Delta U}}$ for gases.

Ans: For an appreciable volume change,

Let ${V_A}$ represent the volume of gaseous reactants

Let ${V_B}$ represent the volume of gaseous products

Let ${n_A}$ be the moles of reactant

Let ${n_B}$ be the moles of product

At constant pressure and temperature

$p{V_A} = {n_A}RT$

$p{V_B} = {n_B}RT$

$p{V_B} - p{V_A} = ({n_B} - {n_A})RT$

$p\Delta V = \Delta {n_g}RT$

$\Delta {n_g} = {n_B} - {n_A}$

Now substituting the value of $p\Delta V$ , we get

$\Delta H = \Delta U + \Delta {n_g}RT$

Heat change at constant pressure,$\Delta H = {q_p}$

Heat change at constant temperature, $\Delta H = {q_V}$

For gaseous system

${q_P} = {q_V} + \Delta {n_g}RT$

2. It has been found that $221.4\;{\text{J}}$ is needed to heat $30\;{\text{g}}$ of ethanol from${\text{1}}{{\text{5}}^{\text{o}}}{\text{C}}$  to ${\text{1}}{{\text{8}}^{\text{o}}}{\text{C}}$. calculate (a) specific heat capacity, and (b) molar heat capacity of ethanol.

Ans: 

1. From the question we have $221.4\;{\text{J}}$of energy is needed to heat $30\;{\text{g}}$ of ethanol from${\text{1}}{{\text{5}}^{\text{o}}}{\text{C}}$  to ${\text{1}}{{\text{8}}^{\text{o}}}{\text{C}}$.

Since, specific heat capacity is given by the formula,

$c = \dfrac{q}{{m\Delta T}}$

$q = 221.4\;{\text{J}}$

$m = 30\,{\text{g}}$

$C = \dfrac{{221.1{\text{J}}}}{{30{\text{g}}({{18}^ \circ }{\text{C}} - {{15}^ \circ }{\text{C}})}}$

$C = \dfrac{{221.4}}{{30 \times 3}}$

$C = 2.46\;{\text{J}}{{\text{g}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{c}}^{{\text{ - 1}}}}$

As ${{\text{1}}^{\text{o}}}{\text{C}}$is equal to ${\text{1K}}$ ,therefore specific heat capacity of ethanol is $C = 2.46\;{\text{J}}{{\text{g}}^{ - 1}}{\,^ \circ }{{\text{c}}^{ - 1}}$

2. Molar heat capacity is product of specific heat and molar mass

${C_m} = 2.46 \times 46$

${C_m} = 113.2{\text{J}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{C}}^{{\text{ - 1}}}}$

Therefore, molar heat capacity of ethanol is ${C_m} = 113.2\,{\text{J}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{C}}^{{\text{ - 1}}}}$

The Following Topics Are Covered in Class 11 Chemistry Chapter 5 Important Questions

• Thermodynamics state

• Applications 

• Measurement of ΔU & ΔH: Calorimetry

• Enthalpy change, ΔH of a reaction

• Enthalpies for different types of reactions

• Spontaneity 

• Gibbs energy change and Equilibrium

After going through all these topics, student will be able to understand the system (close, open & isolated system) and surroundings, internal energy, work, heat, the first law of thermodynamics, state function, enthalpy change, Hess’s law, the difference between extensive and intensive properties, spontaneous and nonspontaneous processes and Gibbs free energy. On Vedantu, there are some essential Class 11 Chemistry Chapter 5 Extra Questions that clarify the concepts of this chapter in an applicational approach.

Conclusion

In exams like IIT or NEET, thermodynamics is one of the most important topics because it has a weightage of around 3%. Most students find it difficult to understand the concepts of thermodynamics. Students can learn Class 11 Chemistry Chapter 5 Important Questions on Vedantu online as well as offline. Vedantu is one of the leading online learning websites, that enables students to learn the concepts with a better understanding with the help of free PDFs. Our team of experts has developed all of the study materials in accordance with the existing CBSE board syllabus. Students are advised to practice more and more questions and answers to excel in their examinations.

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