The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25°C; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : [R = 8.314 JK–1 mol

Engineering

Chemistry

Enthalpy and Internal Energy Relation

Hess Law

Calorimetry

Question

The combustion of benzene (l) gives CO₂(g) and H₂O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol^–1 at 25°C; heat of combustion (in kJ mol^–1) of benzene at constant pressure will be : [R = 8.314 JK^–1 mol^–1]

3260

4152.6

– 3267.6

– 452.46

$∵$ Heat at constant volume ⇒ $∆$ U (Given)

and heat at constant pressure ⇒ $∆$ H (Asked)

For a reaction,

$∆$ H = $∆$ U + $∆$ n_g.RT

Combustion reaction of C₆H₆ (ℓ)

C₆H₆ (ℓ) + $\frac{15}{2} O_{2} (s)$ → 6 CO₂ (s) + 3H₂O (ℓ)

$∆$ H = (–3263.9 kJ) + $\frac{(- 1 .5) \times 8 .314 \times 298}{1000} kJ$

= – 3267.6 kJ

The question involves finding the heat of combustion of benzene at constant pressure given the heat at constant volume. This relates to the difference between enthalpy change (ΔH) and internal energy change (ΔU) for a reaction.

Key Concept: For any reaction, the relationship between ΔH (enthalpy change at constant pressure) and ΔU (internal energy change at constant volume) is given by:

Δ H = Δ U + Δ n_{g} R T

where:

Δn_g = change in number of moles of gaseous products minus gaseous reactants
R = universal gas constant
T = temperature in Kelvin

Step 1: Write the balanced combustion reaction for benzene.

Benzene (C₆H₆) combusts with O₂ to give CO₂(g) and H₂O(l). The balanced equation is:

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)

Step 2: Calculate Δn_g, the change in moles of gas.

Only gaseous species contribute to Δn_g.

Gaseous reactants: O₂ = 15/2 = 7.5 moles
Gaseous products: CO₂ = 6 moles
H₂O is liquid, so not counted.

Therefore,

Δ n_{g} = n_{g, p r o d u c t s} - n_{g, r e a c t a n t s} = 6 - \frac{15}{2} = \frac{12}{2} - \frac{15}{2} = - \frac{3}{2}

Step 3: Convert temperature to Kelvin and ensure unit consistency.

T = 25°C = 25 + 273 = 298 K

R = 8.314 J K^-1 mol^-1 = 0.008314 kJ K^-1 mol^-1 (since the heat is given in kJ)

Step 4: Apply the formula ΔH = ΔU + Δn_gRT

Given: ΔU = -3263.9 kJ mol^-1

Δ H = (- 3263.9) + (- \frac{3}{2}) \times (0.008314) \times (298)

First, calculate Δn_gRT:

Δ n_{g} R T = (- 1.5) \times (0.008314) \times (298) \approx (- 1.5) \times (2.477572) \approx - 3.716358

Now, add this to ΔU:

Δ H = - 3263.9 + (- 3.716358) \approx - 3267.616358

Rounding to one decimal place (as per the given ΔU), the heat of combustion at constant pressure is approximately -3267.6 kJ mol^-1.

Final Answer: The correct option is –3267.6 kJ mol^-1.

Detailed Solution

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Detailed Solution

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