Consider the reaction at `300K`
`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr 6CO_(2)(g)+3H_(2)O(l),DeltaH=-3271`kJ
What is `DeltaU` for the combustion of `1.5` mole of benzene at `27^(@)C` ?

Options

A. `-3267.25kJ`
B. `-4900.88kJ`
C. `-4906.5kJ`
D. `-3274.75kJ`

Explanation

To solve for the change in internal energy (ΔU) for the combustion of 1.5 moles of benzene at 27°C, we can follow these steps:

### Step 1: Understand the given reaction and data
The combustion reaction of benzene is given as:
\[ C_{6}H_{6}(l) + \frac{15}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(l) \]
We are provided with the enthalpy change (ΔH) for the combustion of 1 mole of benzene:
\[ \Delta H = -3271 \text{ kJ} \]

### Step 2: Convert the temperature
The temperature is given as 27°C, which needs to be converted to Kelvin:
\[ T = 27 + 273.15 = 300.15 \text{ K} \]

### Step 3: Use the relation between ΔH and ΔU
The relationship between change in enthalpy (ΔH) and change in internal energy (ΔU) is given by:
\[ \Delta H = \Delta U + \Delta n_{gas} \cdot R \cdot T \]
Where:
- Δn_gas = change in the number of moles of gaseous species
- R = universal gas constant = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)

### Step 4: Calculate Δn_gas
To find Δn_gas, we need to calculate the difference between the moles of gaseous products and reactants:
- Moles of gaseous products (CO₂) = 6
- Moles of gaseous reactants (O₂) = 7.5 (from \(\frac{15}{2}\))

Thus,
\[ \Delta n_{gas} = 6 - 7.5 = -1.5 \]

### Step 5: Substitute values into the equation
Now, substituting the values into the equation:
\[ -3271 \text{ kJ} = \Delta U + (-1.5) \cdot (0.008314 \text{ kJ/(mol·K)}) \cdot (300.15 \text{ K}) \]

### Step 6: Calculate the term involving Δn_gas
Calculating the term:
\[ -1.5 \cdot 0.008314 \cdot 300.15 \approx -3.747 \text{ kJ} \]

### Step 7: Solve for ΔU
Now, substituting this back into the equation:
\[ -3271 \text{ kJ} = \Delta U - 3.747 \text{ kJ} \]
\[ \Delta U = -3271 + 3.747 \]
\[ \Delta U \approx -3267.253 \text{ kJ} \]

### Step 8: Calculate ΔU for 1.5 moles of benzene
Since ΔU is for 1 mole of benzene, for 1.5 moles:
\[ \Delta U_{1.5} = -3267.253 \text{ kJ} \times 1.5 \]
\[ \Delta U_{1.5} \approx -4900.8795 \text{ kJ} \]

### Conclusion
Thus, the change in internal energy (ΔU) for the combustion of 1.5 moles of benzene at 27°C is approximately:
\[ \Delta U \approx -4900.88 \text{ kJ} \]

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