NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

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. 

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, 

Ans : The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct 

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. 

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..,

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