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 use the following relationship:

\[ \Delta E^\circ = \Delta H^\circ - \Delta N_G RT \]

Where:
- \(\Delta H^\circ\) is the enthalpy change for the reaction.
- \(\Delta N_G\) is the change in the number of moles of gas.
- \(R\) is the universal gas constant (8.314 J/mol·K).
- \(T\) is the temperature in Kelvin (300 K).

### Step 1: Calculate ΔNG
From the balanced combustion reaction:
\[ 2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(l) \]
We can determine ΔNG:
- Moles of gaseous products = 12 (from CO2)
- Moles of gaseous reactants = 15 (from O2)

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

### Step 2: Calculate ΔH for 1.5 moles of benzene
Given that ΔH° for the combustion of 2 moles of benzene is -6542 kJ, we need to find ΔH° for 1.5 moles:

\[ \Delta H^\circ \text{ for 1.5 moles} = \frac{1.5}{2} \times (-6542 \text{ kJ}) = -4915.25 \text{ kJ} \]

### Step 3: Convert ΔH to Joules
Since ΔH is in kJ, we convert it to Joules:

\[ \Delta H^\circ = -4915.25 \text{ kJ} \times 1000 \text{ J/kJ} = -4915250 \text{ J} \]

### Step 4: Calculate ΔE°
Now we can substitute the values into the ΔE° equation:

\[ \Delta E^\circ = \Delta H^\circ - \Delta N_G RT \]

Substituting the values we have:

\[ \Delta E^\circ = -4915250 \text{ J} - (-3) \times (8.314 \text{ J/mol·K}) \times (300 \text{ K}) \]

Calculating the second term:

\[ \Delta N_G RT = 3 \times 8.314 \times 300 = 7482.6 \text{ J} \]

Now substituting this back into the equation:

\[ \Delta E^\circ = -4915250 \text{ J} + 7482.6 \text{ J} = -4907767.4 \text{ J} \]

### Final Result
Thus, the change in internal energy for the combustion of 1.5 moles of benzene is:

\[ \Delta E^\circ \approx -4907.77 \text{ kJ} \text{ (after converting back to kJ)} \]