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

Concept:
Standard enthalpy change of reaction is calculated using: \[ \Delta H^\circ_{\text{reaction}}=\sum \Delta H^\circ_f(\text{products})-\sum \Delta H^\circ_f(\text{reactants}) \] ip

Step 1: Write formation enthalpy of products.
Products are: \[ \text{CO}_2 + 2\text{H}_2\text{O} \] So, \[ \sum \Delta H^\circ_f(\text{products}) = (-394) + 2(-286) \] \[ = -394 - 572 = -966\ \text{kJ mol}^{-1} \] ip

Step 2: Write formation enthalpy of reactants.
Reactants are: \[ \text{CH}_4 + 2\text{O}_2 \] For oxygen in standard state: \[ \Delta H^\circ_f(\text{O}_2)=0 \] So, \[ \sum \Delta H^\circ_f(\text{reactants}) = (-75)+2(0)=-75 \] ip

Step 3: Calculate reaction enthalpy.
\[ \Delta H^\circ_{\text{reaction}} = -966 - (-75) \] \[ \Delta H^\circ_{\text{reaction}} = -966 + 75 \] \[ \Delta H^\circ_{\text{reaction}} = -891\ \text{kJ} \] ip Hence, the correct answer is:
\[ \boxed{(A)\ -891\ \text{kJ}} \]