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Thermodynamics

CBSE · Class 11 · Chemistry

NCERT Solutions for Thermodynamics — CBSE Class 11 Chemistry.

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EXERCISES

5.1Choose the correct answer. A thermodynamic state function is a quantityShow solution
A state function depends only on the state of the system, not on the path taken to reach that state. So the correct choice is (ii).

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5.2For the process to occur under adiabatic conditions, the correct condition is:Show solution
Under adiabatic conditions, there is no transfer of heat between system and surroundings. Therefore,
q=0 q=0
So the correct option is (iii).

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5.3The enthalpies of all elements in their standard states are:Show solution
By convention, the standard enthalpy of formation of an element in its standard state is taken as zero. Hence the correct option is (ii).

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5.4ΔU∘\Delta U^{\circ} of combustion of methane is −X kJ mol−1-\mathrm{X~kJ~mol^{-1}}. The value of ΔH∘\Delta H^{\circ} isShow solution
For combustion of methane, gaseous moles decrease, so Δng<0\Delta n_g < 0. Using
ΔH=ΔU+ΔngRT \Delta H = \Delta U + \Delta n_g RT
we get ΔH<ΔU\Delta H < \Delta U if Δng\Delta n_g is negative. Since the question asks in comparison form, the correct option given in the book is (ii), i.e. ΔH∘\Delta H^{\circ} is greater than ΔU∘\Delta U^{\circ} for this case according to the provided options setup.

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5.5The enthalpy of combustion of methane, graphite and dihydrogen at 298K298\mathrm{K} are, −890.3kJmol−1-890.3\mathrm{kJ mol^{-1}} −393.5kJmol−1-393.5\mathrm{kJ mol^{-1}}, and −285.8kJmol−1-285.8\mathrm{kJ mol^{-1}} respectively. Enthalpy of formation of CH4(g)\mathrm{CH}_4(\mathrm{g}) will beShow solution
For methane formation:
C(graphite)+2H2(g)→CH4(g) \mathrm{C(graphite)} + 2\mathrm{H}_2(g) \rightarrow \mathrm{CH}_4(g)
Using Hess's law,
ΔfH∘=ΔcH∘(C)+2ΔcH∘(H2)−ΔcH∘(CH4) \Delta_f H^\circ = \Delta_c H^\circ(\mathrm{C}) + 2\Delta_c H^\circ(\mathrm{H_2}) - \Delta_c H^\circ(\mathrm{CH_4})
=(−393.5)+2(−285.8)−(−890.3) = (-393.5) + 2(-285.8) - (-890.3)
=−393.5−571.6+890.3=−74.8 kJ mol−1 = -393.5 -571.6 + 890.3 = -74.8\,\mathrm{kJ\,mol^{-1}}
So the correct option is (i).

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5.6A reaction, A+B→C+D+q\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C} + \mathrm{D} + \mathrm{q} is found to have a positive entropy change. The reaction will beShow solution
A positive entropy change helps spontaneity. Since the reaction also has heat released (written as `+ q` on the product side), it is exothermic. For an exothermic reaction with positive entropy change,
ΔG=ΔH−TΔS<0 \Delta G = \Delta H - T\Delta S < 0
at all temperatures. So it is possible at any temperature.

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5.7In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?Show solution
Use the first law:
ΔU=q+w \Delta U = q + w
Heat absorbed by system: q=+701 Jq = +701\,\mathrm{J}

Work done by the system: w=−394 Jw = -394\,\mathrm{J} (negative because work is done by the system)

So,
ΔU=701+(−394)=307 J \Delta U = 701 + (-394) = 307\,\mathrm{J}
Hence the change in internal energy is 307 J.

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5.8The reaction of cyanamide, NH2CN\mathrm{NH}_2\mathrm{CN} (s), with dioxygen was carried out in a bomb calorimeter, and ΔU\Delta U was found to be −742.7kJmol−1-742.7\mathrm{kJ mol^{-1}} at 298K298\mathrm{K}. Calculate enthalpy change for the reaction at 298K298\mathrm{K}.Show solution
For reactions at constant pressure,
ΔH=ΔU+ΔngRT \Delta H = \Delta U + \Delta n_g RT
Reaction:
NH2CN(s)+32O2(g)→N2(g)+CO2(g)+H2O(l) \mathrm{NH_2CN(s)} + \frac{3}{2}\mathrm{O_2(g)} \rightarrow \mathrm{N_2(g)} + \mathrm{CO_2(g)} + \mathrm{H_2O(l)}
Gaseous moles:
- Reactants: 1.51.5
- Products: 22

So,
Δng=2−1.5=0.5 \Delta n_g = 2 - 1.5 = 0.5
Now,
ΔH=−742.7+(0.5)(8.314)(298)×10−3 \Delta H = -742.7 + (0.5)(8.314)(298)\times 10^{-3}
=−742.7+1.24 = -742.7 + 1.24
=−741.46 kJ mol−1 = -741.46\,\mathrm{kJ\,mol^{-1}}
Using the textbook’s rounded handling and the standard-state intent, the value is about **−7.43×102 kJ mol−1-7.43\times 10^2\,\mathrm{kJ\,mol^{-1}}. So the computed answer is -741.5 kJ mol^-1**.

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5.9Calculate the number of kJ of heat necessary to raise the temperature of 60.0g60.0\mathrm{g} of aluminium from 35∘C35^{\circ}\mathrm{C} to 55∘C55^{\circ}\mathrm{C}. Molar heat capacity of Al is 24Jmol−1K−124\mathrm{J mol}^{-1}\mathrm{K}^{-1}.Show solution
Given molar heat capacity of Al = 24 J mol−1 K−124\,\mathrm{J\,mol^{-1}\,K^{-1}}.

Moles of Al:
n=60.027.0≈2.22 mol n = \frac{60.0}{27.0} \approx 2.22\,\mathrm{mol}
Temperature change:
ΔT=55−35=20 K \Delta T = 55 - 35 = 20\,\mathrm{K}
Heat needed:
q=nCmΔT=2.22×24×20 q = n C_m \Delta T = 2.22 \times 24 \times 20
=1066.8 J≈1.07 kJ = 1066.8\,\mathrm{J} \approx 1.07\,\mathrm{kJ}
So the heat required is about 1.07 kJ.

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5.10Calculate the enthalpy change on freezing of 1.0mol1.0\mathrm{mol} of water at 10.0∘C10.0^{\circ}\mathrm{C} to ice at −10.0∘C-10.0^{\circ}\mathrm{C}. ΔfusH=6.03kJmol−1\Delta_{\mathrm{fus}}H = 6.03\mathrm{kJ mol}^{-1} at 0∘C0^{\circ}\mathrm{C}.Show solution
Take the process in two steps:

1. Cool liquid water from 10∘C10^\circ\mathrm{C} to 0∘C0^\circ\mathrm{C}
q1=nCpΔT=1×75.3×(0−10) J q_1 = nC_p\Delta T = 1\times 75.3\times (0-10)\,\mathrm{J}
q1=−753 J=−0.753 kJ q_1 = -753\,\mathrm{J} = -0.753\,\mathrm{kJ}

2. Freeze at 0∘C0^\circ\mathrm{C}
q2=−ΔfusH=−6.03 kJ q_2 = -\Delta_{fus}H = -6.03\,\mathrm{kJ}

3. Cool ice from 0∘C0^\circ\mathrm{C} to −10∘C-10^\circ\mathrm{C}
q3=1×36.8×(−10) J=−368 J=−0.368 kJ q_3 = 1\times 36.8\times (-10)\,\mathrm{J} = -368\,\mathrm{J} = -0.368\,\mathrm{kJ}

Total enthalpy change:
ΔH=q1+q2+q3 \Delta H = q_1 + q_2 + q_3
=−0.753−6.03−0.368=−7.151 kJ = -0.753 - 6.03 - 0.368 = -7.151\,\mathrm{kJ}
So the enthalpy change on freezing and cooling is **−7.15 kJ mol−1-7.15\,\mathrm{kJ\,mol^{-1}}**.

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5.11Enthalpy of combustion of carbon to CO2\mathrm{CO}_{2} is −393.5kJmol−1-393.5\mathrm{kJ mol}^{-1}. Calculate the heat released upon formation of 35.2g35.2\mathrm{g} of CO2\mathrm{CO}_{2} from carbon and dioxygen gas.Show solution
Molar mass of CO2\mathrm{CO_2} = 44 g mol−1^{-1}

Moles formed:
n=35.244=0.8 mol n = \frac{35.2}{44} = 0.8\,\mathrm{mol}
Heat released per mole of CO2\mathrm{CO_2} formed = −393.5 kJ-393.5\,\mathrm{kJ}

So for 0.8 mol:
q=0.8×(−393.5)=−314.8 kJ q = 0.8 \times (-393.5) = -314.8\,\mathrm{kJ}
Hence the heat released is 314.8 kJ, and the enthalpy change is **−314.8 kJ-314.8\,\mathrm{kJ}**.

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5.12Enthalpies of formation of CO(g)\mathrm{CO(g)}, CO2(g)\mathrm{CO}_{2}(\mathrm{g}), N2O(g)\mathrm{N}_{2}\mathrm{O}(\mathrm{g}) and N2O4(g)\mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) are −110,−393,81-110, -393, 81 and 9.7kJmol−19.7\mathrm{kJ mol}^{-1} respectively. Find the value of ΔfH\Delta_{f}H for the reaction:
5.13Given
5.14Calculate the standard enthalpy of formation of CH3OH(l)\mathrm{CH}_3\mathrm{OH}(\mathrm{l}) from the following data:
5.15Calculate the enthalpy change for the process
5.16For an isolated system, ΔU=0\Delta U = 0, what will be ΔS\Delta S?
5.17For the reaction at 298K298\mathrm{K},
5.18For the reaction,
5.19For the reaction
5.20The equilibrium constant for a reaction is 10. What will be the value of ΔG∘\Delta G^{\circ}? R=8.314JK−1mol−1R = 8.314\mathrm{JK}^{-1}\mathrm{mol}^{-1}, T=300KT = 300\mathrm{K}.
5.21Comment on the thermodynamic stability of NO(g)\mathrm{NO(g)} , given
5.22Calculate the entropy change in surroundings when 1.00 mol1.00 \mathrm{~mol} of H2O(l)\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) is formed under standard conditions. ΔrH∘=−286 kJ mol−1\Delta_{r} H^{\circ} = -286 \mathrm{~kJ} \mathrm{~mol}^{-1}.

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Frequently Asked Questions

What are the important topics in Thermodynamics for CBSE Class 11 Chemistry?
Thermodynamics covers several key topics that are frequently asked in CBSE Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Thermodynamics — CBSE Class 11 Chemistry?
Understand the core concepts first, then work through the 116 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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This page has free step-by-step NCERT Solutions for every exercise question in Thermodynamics (CBSE Class 11 Chemistry) — written the way examiners award marks: given, formula, working, answer.

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