Benzene burns according to the following equation at `300K (R=8.314J mol e^(-1)K^(-1))`
`2C_(6)H_(6)(l)+15O_(2)(g)rarr12CO_(2)(g)+6H_(2)O(l) " "DeltaH^(@)=-6542kJ //mol`
What is the `DeltaE^(@)` for the combustion of `1.5 mol` of benzene

Options

A. `-3271 kJ`
B. `-9813kJ`
C. `-4906.5kJ`
D. None of these

Explanation

To find the change in internal energy (ΔE) for the combustion of 1.5 moles of benzene, we can follow these steps:

### Step 1: Understand the Reaction
The combustion of benzene (C₆H₆) is given by the equation:
\[ 2C_{6}H_{6}(l) + 15O_{2}(g) \rightarrow 12CO_{2}(g) + 6H_{2}O(l) \]
The enthalpy change (ΔH) for this reaction is given as -6542 kJ per 2 moles of benzene.

### Step 2: Calculate Δn_g
Δn_g is the change in the number of moles of gas during the reaction:
- Moles of gaseous products = 12 (from CO₂)
- Moles of gaseous reactants = 15 (from O₂)

Thus,
\[ \Delta n_g = \text{Moles of products} - \text{Moles of reactants} = 12 - 15 = -3 \]

### Step 3: Use the Relationship Between ΔH and ΔE
The relationship between ΔH and ΔE is given by:
\[ \Delta E = \Delta H - \Delta n_g RT \]
Where:
- R = 8.314 J/mol·K
- T = 300 K

### Step 4: Substitute Values into the Equation
We need to convert ΔH from kJ to J for consistency:
\[ \Delta H = -6542 \text{ kJ} = -6542000 \text{ J} \]

Now, substituting the values into the equation:
\[ \Delta E = -6542000 \text{ J} - (-3)(8.314 \text{ J/mol·K})(300 \text{ K}) \]

### Step 5: Calculate the Second Term
Calculate the second term:
\[ -3 \times 8.314 \times 300 = -3 \times 2494.2 = -7482.6 \text{ J} \]

### Step 6: Combine the Terms
Now, substituting this back into the equation:
\[ \Delta E = -6542000 \text{ J} + 7482.6 \text{ J} \]
\[ \Delta E = -6542000 + 7482.6 = -6542000 + 7482.6 \approx -6542000 + 7482.6 = -6541517.4 \text{ J} \]

### Step 7: Convert ΔE back to kJ
Convert ΔE back to kJ:
\[ \Delta E \approx -6541.5174 \text{ kJ} \]

### Step 8: Adjust for 1.5 moles
Since ΔH is for 2 moles, we need to adjust for 1.5 moles:
\[ \Delta E \text{ for 1.5 moles} = \frac{1.5}{2} \times -6542 \text{ kJ} \]
\[ \Delta E \text{ for 1.5 moles} = -4915.5 \text{ kJ} \]

### Final Answer
Thus, the change in internal energy (ΔE) for the combustion of 1.5 moles of benzene is approximately:
\[ \Delta E \approx -4915.5 \text{ kJ} \]

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