Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

Answer:- **Given:** Given equations are: (i) \( \text{CH}_3\text{OH}_{(l)} + \frac{3}{2}\text{O}_2\text{(_g)} \rightarrow \text{CO}_2\text{(_g)} + 2\text{H}_2\text{O}_{(l)}; \Delta_r H^\circ = -726 \text{ kJ mol}^{-1}$ (ii) \( \text{C}_{(\text{graphite})} + \text{O}_2\text{(_g)} \rightarrow \text{CO}_2\text{(_g)}; \Delta_r H^\circ = -393 \text{ kJ mol}^{-1}$ (iii) \( \text{H}_2\text{(_g)} + \frac{1}{2}\text{O}_2\text{(_g)} \rightarrow \text{H}_2\text{O}_{(l)}; \Delta_r H^\circ = -286 \text{ kJ mol}^{-1}$ **To find:** The standard enthalpy of formation (\( \Delta_f H^\circ \)) of CH₃OH_(l) **Calculation:** The required equation is \( \text{C}_{(\text{graphite})} + 2\text{H}_{2(\text{g})} + \frac{1}{2}\text{O}_{2(\text{g})} \rightarrow \text{CH}_3\text{OH}_{(l)} \) 1. Multiply equation (iii) by 2 and add to equation (ii): \[ 2\text{H}_{2(\text{g})} + \text{O}_{2(\text{g})} \rightarrow 2\text{H}_2\text{O}_{(l)}, \Delta_r H^\circ = -575 \text{ kJ mol}^{-1} \] \[ \text{C}_{(\text{graphite})} + \text{O}_{2(\text{g})} \rightarrow \text{CO}_2\text{(_g)}, \Delta_r H^\circ = -393 \text{ kJ mol}^{-1} \] \[ \begin{array}{l} \text{C}_{(\text{graphite})} + 2\text{H}_{2(\text{g})} + 2\text{O}_{2(\text{g})} \rightarrow \text{CO}_2\text{(_g)} + 2\text{H}_2\text{O}_{(l)} \\ \Delta_r H^\circ = -572 - 393 = -965 \text{ kJ mol}^{-1} \end{array} \] 2. Reverse equation (i) and add to equation (iv): \[ \text{CO}_2\text{(_g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{CH}_3\text{OH}_{(l)} + \frac{3}{2}\text{O}_{2(\text{g})}, \Delta_r H^\circ = 726 \text{ kJ mol}^{-1} \] \[ \text{C}_{(\text{graphite})} + 2\text{H}_{2(\text{g})} + \frac{1}{2}2\text{O}_{2(\text{g})} \rightarrow \text{CH}_3\text{OH}_{(l)} \] \[ \Delta_f H^\circ = \Delta_r H^\circ = 726 - 965 = -239 \text{ kJ mol}^{-1} \] The standard enthalpy of formation (\( \Delta_f H^\circ \)) of CH₃OH_(l) from the given data is -239 kJ mol^-1.