The difference between hea of reaction at constant pressure and at constant volume for the reaction `C_6H_12(l) + 90_2(g) rightarrow 6CO_2(g)+6H_2O(l), at 27^oC` is

Options

A. `- 14.96 kJ`
B. `- 7.48 kJ`
C. `+ 7.43 kJ`
D. `- 10.23 kJ`

Answer

To find the difference between the heat of reaction at constant pressure (ΔH) and at constant volume (ΔU) for the reaction:

\[ C_6H_{12}(l) + 9O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \]

at 27°C, we can use the relationship:

\[ \Delta H - \Delta U = \Delta N_g RT \]

where:
- \( \Delta H \) is the heat of reaction at constant pressure,
- \( \Delta U \) is the heat of reaction at constant volume,
- \( \Delta N_g \) 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.

### Step 1: Calculate \( \Delta N_g \)

To find \( \Delta N_g \), we need to determine the number of moles of gaseous reactants and products:

- **Reactants**:
- \( C_6H_{12}(l) \) contributes 0 moles of gas.
- \( 9O_2(g) \) contributes 9 moles of gas.

Total moles of gas in reactants = 9.

- **Products**:
- \( 6CO_2(g) \) contributes 6 moles of gas.
- \( 6H_2O(l) \) contributes 0 moles of gas.

Total moles of gas in products = 6.

Now, we can calculate \( \Delta N_g \):

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 6 - 9 = -3 \]

### Step 2: Convert temperature to Kelvin

The temperature given is 27°C. To convert this to Kelvin:

\[ T(K) = 27 + 273.15 = 300.15 \approx 300 \text{ K} \]

### Step 3: Calculate \( \Delta H - \Delta U \)

Using the formula:

\[ \Delta H - \Delta U = \Delta N_g RT \]

Substituting the values we have:

\[ \Delta H - \Delta U = (-3)(8.314 \, \text{J/mol·K})(300 \, \text{K}) \]

Calculating this gives:

\[ \Delta H - \Delta U = -3 \times 8.314 \times 300 = -7494.6 \, \text{J} \approx -7.5 \, \text{kJ} \]

### Final Answer

The difference between the heat of reaction at constant pressure and at constant volume for the given reaction is approximately:

\[ \Delta H - \Delta U \approx -7.5 \, \text{kJ} \]

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