Question

The enthalpy of combustion values of $C_{2}H_{4(g)}$, $C_{(graphite,s)}$ and $H_{2(g)}$ are respectively $-1411\ \text{kJ mol}^{-1}$, $-394\ \text{kJ mol}^{-1}$ and $-286\ \text{kJ mol}^{-1}$. What is the enthalpy of formation of $C_{2}H_{4(g)}$ in $\text{kJ mol}^{-1}$?

• A. -102
• B. -51
• C. +102
• D. +153
• E. +51

Hint:

Remember to multiply enthalpy values by their stoichiometric coefficients.

Answer 1

Answer: (E)

Step 1: Concept
$\Delta H_{f} = \sum \Delta H_{c}(\text{reactants}) - \sum \Delta H_{c}(\text{products})$.

Step 2: Meaning
Formation of $C_{2}H_{4}$ is: $2C + 2H_{2} \rightarrow C_{2}H_{4}$.

Step 3: Analysis
$\Delta H_{f} = [2(-394) + 2(-286)] - (-1411) = [-788 - 572] + 1411 = -1360 + 1411 = +51$.

Step 4: Conclusion
The enthalpy of formation is +51 kJ $mol^{-1}$. Final Answer: (E)