Question

For the reaction \( \text{H}_2 (g) + \frac{1}{2} \, \text{O}_2 (g) \longrightarrow \, \text{H}_2\text{O}(l) \), the following information is given: \[ \Delta H^\circ = -285 \, \text{kJ/mol}, \, S^\circ_{\text{H}_2\text{O}(l)} = 70 \, \text{J K}^{-1} \, \text{mol}^{-1}, \, S^\circ_{\text{O}_2(g)} = 204 \, \text{J K}^{-1} \, \text{mol}^{-1}, \, S^\circ_{\text{H}_2(g)} = 130 \, \text{J K}^{-1} \, \text{mol}^{-1} \] The value of \( \Delta S^\circ_{\text{universe}} \) for the reaction is ........... J K\(^{-1}\) mol\(^{-1}\). 
 

Hint:

The change in entropy of the universe for a reaction can be calculated using the equation: \[ \Delta S^\circ_{\text{universe}} = -\frac{\Delta H^\circ}{T} \]

Answer 1

Correct Answer: 786 - 790

Step 1: Formula for \( \Delta S^\circ_{\text{universe}} \). 
The change in entropy of the universe is given by: \[ \Delta S^\circ_{\text{universe}} = - \frac{\Delta H^\circ}{T} \] where \( \Delta H^\circ = -285 \, \text{kJ/mol} \) and \( T = 300 \, \text{K} \). 
 

Step 2: Calculating \( \Delta S^\circ_{\text{reaction}} \). 
The entropy change for the reaction, \( \Delta S^\circ_{\text{reaction}} \), is calculated as: \[ \Delta S^\circ_{\text{reaction}} = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \] For the products: \[ S^\circ_{\text{products}} = S^\circ_{\text{H}_2\text{O}(l)} = 70 \, \text{J K}^{-1} \, \text{mol}^{-1} \] For the reactants: \[ S^\circ_{\text{reactants}} = S^\circ_{\text{H}_2(g)} + \frac{1}{2} \, S^\circ_{\text{O}_2(g)} = 130 + \frac{1}{2} \times 204 = 232 \, \text{J K}^{-1} \, \text{mol}^{-1} \] Thus, \[ \Delta S^\circ_{\text{reaction}} = 70 - 232 = -162 \, \text{J K}^{-1} \, \text{mol}^{-1} \]

Step 3: Conclusion. 
The change in entropy of the universe is: \[ \Delta S^\circ_{\text{universe}} = \Delta S^\circ_{\text{reaction}} - \frac{\Delta H^\circ}{T} \] Substituting the values: \[ \Delta S^\circ_{\text{universe}} = -162 - \frac{-285000}{300} = -162 + 950 = 788 \, \text{J K}^{-1} \, \text{mol}^{-1} \]