Question

The combustion of benzene (L) gives CO\(_2\) (g) and H\(_2\)O (L). Given that heat of combustion of benzene at constant volume is -3263.9 kJ/mol at 25°C, heat of combustion (in kJ/mol) of benzene at constant pressure will be: (R = 8.314 J\(K^{-1}\) \(mol^{-1}\))

• A. 4152.6
• B. 452.46
• C. 3260
• D. -3267.6

Hint:

For reactions involving gases, the heat of combustion at constant pressure is adjusted by the work done due to changes in gas volume, which is linked to the change in the number of moles of gas.

Answer 1

The Correct Option is D

Step 1: {Understanding the Problem}
The given reaction is the combustion of benzene. We are given the heat of combustion at constant volume, and we need to find the heat of combustion at constant pressure. For the combustion of benzene: \[ {C}_6{H}_6(l) + \frac{15}{2} {O}_2(g) \rightarrow 6{CO}_2(g) + 3{H}_2{O}(l) \] Step 2: {Using the Relation between Heat at Constant Volume and Pressure}
The change in enthalpy is related to the change in internal energy by the equation: \[ \Delta H = \Delta U + \Delta n_{{g}} RT \] where \( \Delta n_{{g}} \) is the change in the number of moles of gas between products and reactants. In the given reaction, we calculate the change in the number of moles of gas: \[ \Delta n_{{g}} = (6 { mol CO}_2) - \left( \frac{15}{2} { mol O}_2 \right) = 6 - 7.5 = -1.5 \] Thus, we have: \[ \Delta H = \Delta U + (-1.5) \times (8.314 \times 10^{-3} \times 298) \] \[ \Delta H = -3263.9 + (-1.5) \times 2.478 \approx -3267.6 \, {kJ/mol} \] Thus, the correct answer is (D).

Answer 2

Step 1: Understand the difference between constant volume and constant pressure combustion
Heat at constant volume is \( q_v = \Delta U \) (change in internal energy)
Heat at constant pressure is \( q_p = \Delta H \) (change in enthalpy)
The relation between them is:
ΔH = ΔU + Δn_gas × R × T

Step 2: Write the balanced chemical equation for benzene combustion
C₆H₆(l) + (15/2) O₂(g) → 6CO₂(g) + 3H₂O(l)
Only gaseous species are considered for Δn_gas
Reactant gases = 7.5 mol (O₂)
Product gases = 6 mol (CO₂)
Δn_gas = 6 − 7.5 = −1.5

Step 3: Apply the formula
Given:
ΔU = −3263.9 kJ/mol
R = 8.314 J·mol⁻¹·K⁻¹ = 0.008314 kJ·mol⁻¹·K⁻¹
T = 25°C = 298 K

Now calculate:
ΔH = ΔU + Δn_gas × R × T
ΔH = −3263.9 + (−1.5 × 0.008314 × 298)
ΔH = −3263.9 − 3.7 ≈ −3267.6 kJ/mol

Final Answer: −3267.6 kJ/mol