Question

The standard enthalpies of formation of $C_6H_6(l)$, $CO_2(g)$ and $H_2O(l)$ are respectively $+49\text{kJ mol}^{-1}$, $-394\text{kJ mol}^{-1}$ and $-286\text{kJ mol}^{-1}$ respectively. What is the value of standard enthalpy of combustion of $C_6H_6(l)$?

• A. -3222 $\text{kJ mol}^{-1}$
• B. -3173 $\text{kJ mol}^{-1}$
• C. -3271 $\text{kJ mol}^{-1}$
• D. +3173 $\text{kJ mol}^{-1}$

Hint:

Remember: "Combustion is always exothermic", so the final value must be negative. Always pay close attention to the stoichiometry of the combustion reaction.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The enthalpy of combustion can be calculated using the enthalpies of formation of the products and the reactants.
Reaction: $C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$.
Key Formula or Approach:
$\Delta H^0_{comb} = [ \sum \Delta H^0_f (\text{products}) ] - [ \sum \Delta H^0_f (\text{reactants}) ]$.

Step 2: Detailed Explanation:
1. Sum of product enthalpies:
$= 6 \times \Delta H^0_f(CO_2) + 3 \times \Delta H^0_f(H_2O)$
$= 6(-394) + 3(-286) = -2364 - 858 = -3222 \text{ kJ}$.
2. Sum of reactant enthalpies:
$= \Delta H^0_f(C_6H_6) + \frac{15}{2} \Delta H^0_f(O_2)$
$= 49 + 0 = 49 \text{ kJ}$ (Enthalpy of formation of an element in standard state is 0).
3. Combustion enthalpy:
$\Delta H^0_{comb} = -3222 - 49 = -3271 \text{ kJ mol}^{-1}$.

Step 3: Final Answer:
The standard enthalpy of combustion is $-3271\text{ kJ mol}^{-1}$.