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 \(3 \, \text{C}_2\text{H}_2(\text{g}) \rightarrow \text{C_6\text{H}_6(\text{l})\), is} (A) \( +324 \, \text{kJ mol}^{-1} \)
(B) \( +632 \, \text{kJ mol}^{-1} \)
(C) \( -632 \, \text{kJ mol}^{-1} \)
(D) \( -732 \, \text{kJ mol}^{-1} \)

• 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: Write the Given Reactions and Their Enthalpies 1. Combustion of Benzene: \[ \text{C}_6\text{H}_6(\text{l}) + \frac{15}{2} \text{O}_2(\text{g}) \rightarrow 6 \, \text{CO}_2(\text{g}) + 3 \, \text{H}_2\text{O}(\text{l}), \quad \Delta H_1 = -3268 \, \text{kJ mol}^{-1} \] 2. Combustion of Acetylene: \[ \text{C}_2\text{H}_2(\text{g}) + \frac{5}{2} \text{O}_2(\text{g}) \rightarrow 2 \, \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}), \quad \Delta H_2 = -1300 \, \text{kJ mol}^{-1} \]
Step 2: Use Hess’s Law to Find the Enthalpy Change for the Desired Reaction
The desired reaction is: \[ 3 \, \text{C}_2\text{H}_2(\text{g}) \rightarrow \text{C}_6\text{H}_6(\text{l}) \]
Using Hess’s Law, the enthalpy change for the desired reaction (\(\Delta H\)) is: \[ \Delta H = \sum \Delta H (\text{Reactants}) - \sum \Delta H (\text{Products}) \] \[ \Delta H = 3 \times (-1300) - (-3268) \] \[ \Delta H = -3900 + 3268 = -632 \, \text{kJ mol}^{-1} \]
Step 3: Determine the Correct Option
The change in enthalpy for the reaction is \( -632 \, \text{kJ mol}^{-1} \), which matches option (C). Final Answer: The enthalpy change for the reaction is \( -632 \, \text{kJ mol}^{-1} \).