Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

answer:- **Given:** Enthalpy change (\(\Delta H\)) = -620 J Volumes of reactants: \(C_2H_4\) = 100 mL, \(H_2\) = 100 mL Pressure (\(P_{\text{ext}}\)) = 1 bar **To find:** Pressure-volume work (\(W\)) and change in internal energy (\(\Delta U\)) for the given reaction **Formulas:** 1) \(W = -P_{\text{ext}} \Delta V\) 2) \(\Delta H = \Delta U + P_{\text{ext}} \Delta V\) **Calculation:** According to the equation of the reaction, 1 mole of \(C_2H_4\) reacts with 1 mole of \(H_2\) to produce 1 mole of \(C_2H_6\). Hence, 100 mL of \(C_2H_4\) would react with 100 mL of \(H_2\) to produce 100 mL of \(C_2H_6\). \(V_1\) = 100 mL + 100 mL = 200 mL = 0.2 dm\(^3\) \(V_2\) = 100 mL = 0.1 dm\(^3\) From formula (1), \(W = -P_{\text{ext}} \Delta V\) \(= -1 \text{ bar} \cdot (0.1 \text{ dm}^3 - 0.2 \text{ dm}^3)\) \(= 0.10 \text{ dm}^3\text{ bar}\) \(= 0.10 \text{ dm}^3\text{ bar} \cdot 100 \frac{\text{J}}{\text{dm}^3 \text{bar}} = +10.00 \text{ J}\) Therefore, \(-P_{\text{ext}} \Delta V = 10.00 \text{ J}\) Therefore, \(P_{\text{ext}} \Delta V = -10.00 \text{ J}\) From formula (2), \(\Delta H = \Delta U + P_{\text{ext}} \Delta V\) Therefore, \(\Delta U = \Delta H - P_{\text{ext}} \Delta V\) \(= -620 \text{ J} - (-10.00 \text{ J}) = -610 \text{ J}\) Pressure-volume work (\(W\) = +10.00 J) and \(\Delta U\) = -610 J.