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NCERT Solutions Class 11 Chemistry Chapter 6 Thermodynamics– Here are all the NCERT solutions for Class 11 Chemistry Chapter 6. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Thermodynamics taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Chemistry, then you must come across chapter 6 Thermodynamics After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics in one place.

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NCERT Solutions Class 11 Chemistry Chapter 6 Thermodynamics

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Chemistry for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 6 Thermodynamics , Chemistry, Class 11.

Class 11
Subject Chemistry
Book Chemistry Part I
Chapter Number 6
Chapter Name

Thermodynamics

NCERT Solutions Class 11 Chemistry chapter 6 Thermodynamics

Class 11, Chemistry chapter 6, Thermodynamics solutions are given below in PDF format. You can view them online or download PDF file for future use.

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics Download

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Question & Answer

Q.1: Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes 
(ii) whose value is independent of path 
(iii) used to determine pressure volume work 
(iv) whose value depends on temperature only

Ans : A thermodynamic state function is a quantity Whose value is independent of a path. Functions like p, V, T etc. depend only on the state of a system and not on the path. 
Hence, alternative (ii) is correct. 
Q.2: For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0 
(ii) ∆p = 0 
(iii) q = 0 
(iv) w = 0

Ans : A system is said to be under adiabatic conditions if there is no exchange Of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0. 
Therefore, alternative (iii) is correct, 
Q.3: The enthalpies of all elements in their standard states are: 
(i) unity 
(ii) zero 
(iii) < 0 
(iv) different for each element

Ans : The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct 
Q.4: \(\Delta U^{\ominus}{\text { of combustion of methane is }-\mathrm{X} \text { kJ } \mathrm{mol}^{-1} . \text { The value of } \Delta H^{\ominus} \text { is }}\) 
(i) \(=\Delta U^{\ominus}\)
(ii) > \(\Delta U^{\ominus}\) 
(iii) < \(=\Delta U^{\ominus}\) 
(iv) = 0

Ans : Since \(\Delta H^{\theta}=\Delta U^{\theta}+\Delta n_{g} R T \text { and } \Delta U^{\theta}=-X \mathrm{k} 3 \mathrm{mol}^{-1}\)
\(\Delta H^{\theta}=(-X)+\Delta n_{g} R T\)
\(\Rightarrow \triangle H^{\theta}<\Delta U^{\theta}\)
Therefore, alternative (iii) is correct. 
Q.5: The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 \(\mathrm{k} J \mathrm{mol}^{-1}-393.5 \mathrm{kJ} \mathrm{mol}^{-1}, \text { and }-285.8 \mathrm{kJ} \mathrm{mol}^{-1}\) respectively. Enthalpy of formation of CH4  will be 
(i) \(-74.8 \mathrm{kJ} \mathrm{mol}^{-1}\) 
(ii) \(-52.27 \mathrm{kJ} \mathrm{mol}^{-1}\) 
(iii) \(+74.8 \mathrm{kJ} \mathrm{mol}^{-1}\) 
(iv) \(+52.26 \mathrm{kJ} \mathrm{mol}^{-1}\)

Ans : According to the question, 
(i) \(\begin{aligned} \mathrm{CH}_{4(\mathrm{g})}+2 \mathrm{O}_{2(\mathrm{g})} & \longrightarrow \mathrm{CO}_{2(\mathrm{z})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \\ \Delta H &=-890.3 \mathrm{kJ} \mathrm{mol}^{-1} \end{aligned}\)
(ii) \(\begin{aligned} \mathrm{C}_{(x)}+\mathrm{O}_{2(y)} \longrightarrow & \mathrm{CO}_{2(g)} \\ \Delta H &=-393.5 \mathrm{kJ} \mathrm{mol}^{-1} \end{aligned}\)
(iii) \(2 \mathrm{H}_{2(g)}+\mathrm{O}_{2(z)} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(g)}\)
\(\Delta H=-285.8 \mathrm{kJ} \mathrm{mol}^{-1}\)
Thus, the desired equation is the one that represents the formation of CH4(g) i.e..,

\(\begin{array}{l}{\therefore \text { Enthalpy of formation of } \mathrm{CH}_{4(\mathrm{g})}=-74.8 \mathrm{kJ} \mathrm{mol}^{-1}} \\ {\text { Hence, alternative (i) is correct. }}\end{array}\)

NCERT / CBSE Book for Class 11 Chemistry

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