Question

Calculate the standard enthalpy change of the following reaction: 
\[ \text{CH}_4{}_{\text{(g)}} + 2\text{O}_2{}_{\text{(g)}} \rightarrow \text{CO}_2{}_{\text{(g)}} + 2\text{H}_2\text{O}_{\text{(l)}} \] 
If: 
\[ \Delta_f H^\circ (\text{CH}_4) = -75 \, \text{kJ mol}^{-1} \] \[ \Delta_f H^\circ (\text{CO}_2) = -394 \, \text{kJ mol}^{-1} \] \[ \Delta_f H^\circ (\text{H}_2\text{O}) = -286 \, \text{kJ mol}^{-1} \]

• A. -891 kJ
• B. -1041 kJ
• C. -966 kJ
• D. -1782 kJ

Hint:

Use: \[ \Delta H^\circ_{\text{reaction}}= \sum \Delta H^\circ_f(\text{products})- \sum \Delta H^\circ_f(\text{reactants}) \] and remember that elemental standard states have \(\Delta H^\circ_f=0\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The standard enthalpy of a reaction is the sum of standard enthalpies of formation of products minus the sum of standard enthalpies of formation of reactants.
Step 2: Key Formula or Approach:
\[ \Delta_{\text{r}}\text{H}^\circ = \sum \Delta_{\text{f}}\text{H}^\circ (\text{products}) - \sum \Delta_{\text{f}}\text{H}^\circ (\text{reactants}) \] Step 3: Detailed Explanation:
Reaction: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)
Note: \(\Delta_{\text{f}}\text{H}^\circ (\text{O}_2) = 0\) (element in standard state).
\[ \Delta_{\text{r}}\text{H}^\circ = [ \Delta_{\text{f}}\text{H}^\circ (\text{CO}_2) + 2 \times \Delta_{\text{f}}\text{H}^\circ (\text{H}_2\text{O}) ] - [ \Delta_{\text{f}}\text{H}^\circ (\text{CH}_4) + 2 \times \Delta_{\text{f}}\text{H}^\circ (\text{O}_2) ] \] \[ \Delta_{\text{r}}\text{H}^\circ = [ (-394) + 2 \times (-286) ] - [ (-75) + 0 ] \] \[ \Delta_{\text{r}}\text{H}^\circ = [ -394 - 572 ] - [ -75 ] \] \[ \Delta_{\text{r}}\text{H}^\circ = -966 + 75 = -891 \text{ kJ} \] Step 4: Final Answer:
The enthalpy change is -891 kJ.