Question

Calculate the standard enthalpy change of following reaction:
$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)}$
if $\Delta_{f}H^{\circ}(CH_{4}) = -75~kJ~mol^{-1}$, $\Delta_{f}H^{\circ}(CO_{2}) = -390~kJ~mol^{-1}$, $\Delta_{f}H^{\circ}(H_{2}O) = -286~kJ~mol^{-1}$

• A. -887.00 kJ
• B. -1325.00 kJ
• C. -1035.00 kJ
• D. -1770.00 kJ

Hint:

Always remember that standard enthalpy of formation for elements in their natural state is zero.

Answer 1

The Correct Option is A

Step 1: Formula
$\Delta H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{Products}) - \sum \Delta_{f}H^{\circ}(\text{Reactants})$

Step 2: Analysis
- $\text{Products} = [(-390) + 2(-286)] = -390 - 572 = -962~kJ$ - $\text{Reactants} = [(-75) + 2(0)] = -75~kJ$ (Note: $\Delta_{f}H^{\circ}$ for $O_{2}$ is zero).

Step 3: Calculation
- $\Delta H^{\circ} = -962 - (-75) = -962 + 75 = -887~kJ$

Step 4: Conclusion
Hence, the enthalpy change is -887.00 kJ. Final Answer: (A)