Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

**Given:** 1) Oxidation of 1 mole HCl_(g) Temperature = \(T = 200^\circ\)C = 473 K 2) Decomposition of one mole of NO Temperature = \(T = 300^\circ\)C = 573 K **To find:** Work done **Formula:** \[W = -\Delta n_g RT\] **Calculations:** 1) The given reaction is for 4 moles of HCl. For 1 mole of HCl, the reaction is given as follows: \[HCl_{(g)} + \frac{1}{4}O_{2(g)} \rightarrow \frac{1}{2}Cl_{2(g)} + \frac{1}{2}H_{2}O_{(g)}\] Now, \[\Delta n_g = (\text{moles of product gases}) - (\text{moles of reactant gases})\] \[\Delta n_g = \left(\frac{1}{2} + \frac{1}{2}\right) - \left(1 + \frac{1}{4}\right) = -0.25 \text{ mol}\] Hence, \[W = -\Delta n_g RT = -(-0.25 \text{ mol}) \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 473 \text{ K} = +983 \text{ J}\] 2) The given reaction is for 2 moles of NO. For 1 mole of NO, the reaction is given as follows: \[NO_{(g)} \rightarrow \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}\] Now, \[\Delta n_g = (\text{moles of product gases}) - (\text{moles of reactant gases})\] \[\Delta n_g = \left(\frac{1}{2} + \frac{1}{2}\right) - 1 = 0 \text{ mol}\] Hence, \[W = -\Delta n_g RT = -0 \text{ mol} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 573 \text{ K} = 0 \text{ kJ}\] No work is done (since \(W = 0\)). Therefore: - The work done is +983 J. The work is done on the system. - The work done is 0 kJ. There is no work done.