Question

Calculate the standard enthalpy of formation of CH\(_{3}\)OH(l) from the following data: \[ \text{CH}_{3}\text{OH(l)} + \frac{3}{2} \text{O}_{2} (\text{g}) \rightarrow \text{CO}_{2} (\text{g}) + 2\text{H}_{2}\text{O(l)}, \quad \Delta H = -726 \, \text{kJ mol}^{-1}, \] \[ \text{C(graphite)} + \text{O}_{2} (\text{g}) \rightarrow \text{CO}_{2} (\text{g}), \quad \Delta H = -393 \, \text{kJ mol}^{-1}, \] \[ \text{H}_{2} (\text{g}) + \frac{1}{2} \text{O}_{2} \rightarrow \text{H}_{2}\text{O(l)}, \quad \Delta H = -286 \, \text{kJ mol}^{-1}. \]

Hint:

Understanding CPR is crucial for managing shared natural resources.

Answer 1

Step 1: The standard enthalpy of formation (\( \Delta H_f^\circ \)) of CH\(_{3}\)OH(l) is the enthalpy change for: \[ \text{C(graphite)} + 2\text{H}_{2} (\text{g}) + \frac{1}{2} \text{O}_{2} (\text{g}) \rightarrow \text{CH}_{3}\text{OH(l)}. \] Given: 1. \( \text{CH}_{3}\text{OH(l)} + \frac{3}{2} \text{O}_{2} \rightarrow \text{CO}_{2} + 2\text{H}_{2}\text{O(l)}, \quad \Delta H_1 = -726 \, \text{kJ mol}^{-1} \), 2. \( \text{C(graphite)} + \text{O}_{2} \rightarrow \text{CO}_{2}, \quad \Delta H_2 = -393 \, \text{kJ mol}^{-1} \), 3. \( \text{H}_{2} + \frac{1}{2} \text{O}_{2} \rightarrow \text{H}_{2}\text{O(l)}, \quad \Delta H_3 = -286 \, \text{kJ mol}^{-1} \).
Step 2: Use Hess’s law. Reverse equation 1: \[ \text{CO}_{2} + 2\text{H}_{2}\text{O(l)} \rightarrow \text{CH}_{3}\text{OH(l)} + \frac{3}{2} \text{O}_{2}, \quad \Delta H = +726 \, \text{kJ mol}^{-1}. \] Add equation 2: \[ \text{C(graphite)} + \text{O}_{2} \rightarrow \text{CO}_{2}, \quad \Delta H = -393 \, \text{kJ mol}^{-1}. \] Add equation 3 (multiplied by 2 for 2H\(_{2}\)): \[ 2\text{H}_{2} + \text{O}_{2} \rightarrow 2\text{H}_{2}\text{O(l)}, \quad \Delta H = 2 \times (-286) = -572 \, \text{kJ mol}^{-1}. \]
Step 3: Combine: \[ \text{C(graphite)} + \text{O}_{2} + 2\text{H}_{2} + \text{O}_{2} + \text{CO}_{2} + 2\text{H}_{2}\text{O(l)} \rightarrow \text{CO}_{2} + 2\text{H}_{2}\text{O(l)} + \text{CH}_{3}\text{OH(l)} + \frac{3}{2} \text{O}_{2}. \] Cancel common terms (CO\(_{2}\), 2H\(_{2}\)O, 1O\(_{2}\)) to get: \[ \text{C(graphite)} + 2\text{H}_{2} + \frac{1}{2} \text{O}_{2} \rightarrow \text{CH}_{3}\text{OH(l)}. \] Total \( \Delta H = 726 + (-393) + (-572) = 726 - 393 - 572 = -239 \, \text{kJ mol}^{-1} \). Thus, \( \Delta H_f^\circ (\text{CH}_{3}\text{OH(l)}) = -239 \, \text{kJ mol}^{-1} \).