The difference between `Delta H and Delta E ` for the reaction `2C_(6)H_(6)(l) + 15O_(2)(g) rightarrow 12 CO_(2)(g) + 6H_(2)O(l) at 25^(@) C` in kJ is

Options

A. `-7.43 kJ`
B. `+ 3.72 kJ`
C. `-3.72 kJ`
D. `+ 7.43 kJ`

Answer

To find the difference between ΔH and ΔE for the reaction

\[ 2C_{6}H_{6}(l) + 15O_{2}(g) \rightarrow 12 CO_{2}(g) + 6H_{2}O(l) \]

at 25°C, we can use the relationship:

\[ \Delta H = \Delta E + \Delta N_{g}RT \]

where:
- \( \Delta H \) is the change in enthalpy,
- \( \Delta E \) is the change in internal energy,
- \( \Delta N_{g} \) is the change in the number of moles of gas (products - reactants),
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin.

### Step 1: Calculate \( \Delta N_{g} \)

1. **Identify the number of gaseous moles on the product side:**
- From the products: \( 12 \, CO_{2}(g) \) contributes 12 moles of gas.
- \( 6 \, H_{2}O(l) \) does not contribute to gaseous moles.

Total gaseous moles in products = 12.

2. **Identify the number of gaseous moles on the reactant side:**
- From the reactants: \( 15 \, O_{2}(g) \) contributes 15 moles of gas.
- \( 2 \, C_{6}H_{6}(l) \) does not contribute to gaseous moles.

Total gaseous moles in reactants = 15.

3. **Calculate \( \Delta N_{g} \):**
\[ \Delta N_{g} = \text{(moles of gas in products)} - \text{(moles of gas in reactants)} = 12 - 15 = -3 \]

### Step 2: Substitute values into the equation

Now, we can substitute the values into the equation \( \Delta H = \Delta E + \Delta N_{g}RT \).

1. **Convert temperature to Kelvin:**
\[ T = 25°C + 273.15 = 298.15 \, K \]

2. **Use the universal gas constant:**
\[ R = 8.314 \, J/(mol \cdot K) = 0.008314 \, kJ/(mol \cdot K) \]

3. **Calculate \( \Delta N_{g}RT \):**
\[ \Delta N_{g}RT = (-3) \times (0.008314 \, kJ/(mol \cdot K)) \times (298.15 \, K) \]
\[ = -3 \times 0.008314 \times 298.15 \approx -7.43 \, kJ \]

### Step 3: Find the difference between \( \Delta H \) and \( \Delta E \)

From the equation:

\[ \Delta H - \Delta E = \Delta N_{g}RT \]

Thus,

\[ \Delta H - \Delta E = -7.43 \, kJ \]

### Final Answer

The difference between \( \Delta H \) and \( \Delta E \) for the reaction is:

\[ \Delta H - \Delta E = -7.43 \, kJ \]