Question

At 25°C and 1 atm pressure, the enthalpy of combustion of benzene (l) and acetylene (g) are \(3268 \, \text{kJ mol}^{-1}\) and \(1300 \, \text{kJ mol}^{-1}\), respectively. The change in enthalpy for the reaction 
is

• A. +324 kJ mol−1
• B. +632 kJ mol−1
• C. −632 kJ mol−1
• D. −732 kJ mol−1

Hint:

Hess’s Law states that the total enthalpy change for a reaction is independent of the pathway taken. The enthalpy change of a reaction can be calculated by summing the enthalpy changes of intermediate reactions.

Answer 1

The Correct Option is C

Step 1: Record Given Reactions and Enthalpies 1. Benzene Combustion:\[\text{C}_6\text{H}_6(\text{l}) + \frac{15}{2} \text{O}_2(\text{g}) \rightarrow 6 \, \text{CO}_2(\text{g}) + 3 \, \text{H}_2\text{O}(\text{l}), \quad \Delta H_1 = -3268 \, \text{kJ mol}^{-1}\]2. Acetylene Combustion:\[\text{C}_2\text{H}_2(\text{g}) + \frac{5}{2} \text{O}_2(\text{g}) \rightarrow 2 \, \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}), \quad \Delta H_2 = -1300 \, \text{kJ mol}^{-1}\]
Step 2: Apply Hess’s Law for Desired Reaction Enthalpy
The target reaction is:\[3 \, \text{C}_2\text{H}_2(\text{g}) \rightarrow \text{C}_6\text{H}_6(\text{l})\]
Hess’s Law yields the enthalpy change for the target reaction (\(\Delta H\)) as:\[\Delta H = \sum \Delta H (\text{Reactants}) - \sum \Delta H (\text{Products})\]\[\Delta H = 3 \times (-1300) - (-3268)\]\[\Delta H = -3900 + 3268 = -632 \, \text{kJ mol}^{-1}\]
Step 3: Identify Correct Option
The calculated enthalpy change for the reaction is \( -632 \, \text{kJ mol}^{-1} \), corresponding to option (C).Final Answer: The reaction's enthalpy change is \( -632 \, \text{kJ mol}^{-1} \).