Change in enthalpy and change in internal energy are state functions. The value of `DeltaH, DeltaU` can be determined by using Kirchoff's equation.
I mole of naphthalene `(C_(10)H_(8))` was burnt is oxygen gas at `25^(@)C` at constant volume. The heat evolved was found to be 5138.8kJ. Calculate the heat of reaction at constant pressure

Options

A. 4770.9kJ
B. 5143.8kJ
C. 6796.6kJ
D. 5791.2kJ

Answer

To solve the problem of calculating the heat of reaction at constant pressure for the combustion of naphthalene (C₁₀H₈), we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of naphthalene.
The combustion of naphthalene can be represented as:
\[ C_{10}H_8 + 12 O_2 \rightarrow 10 CO_2 + 4 H_2O \]

### Step 2: Calculate the change in the number of moles of gas (Δn_g).
Δn_g is calculated as the number of moles of gaseous products minus the number of moles of gaseous reactants.
- Gaseous products: 10 moles of CO₂
- Gaseous reactants: 12 moles of O₂

Thus,
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 10 - 12 = -2 \]

### Step 3: Use the given value of ΔU.
The heat evolved at constant volume (ΔU) is given as:
\[ \Delta U = 5138.8 \, \text{kJ} \]

### Step 4: Use the formula to calculate ΔH.
The relationship between ΔH and ΔU is given by Kirchhoff's equation:
\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]

Where:
- R (universal gas constant) = 8.314 J/(mol·K) or 0.008314 kJ/(mol·K)
- T = temperature in Kelvin = 25°C + 273 = 298 K

### Step 5: Substitute the values into the equation.
Now substituting the values into the equation:
\[ \Delta H = 5138.8 \, \text{kJ} + (-2) \cdot (0.008314 \, \text{kJ/(mol·K)}) \cdot (298 \, \text{K}) \]

Calculating the second term:
\[ -2 \cdot 0.008314 \cdot 298 = -4.955 \, \text{kJ} \]

Now substituting this back into the equation for ΔH:
\[ \Delta H = 5138.8 \, \text{kJ} - 4.955 \, \text{kJ} \]
\[ \Delta H = 5133.845 \, \text{kJ} \]

### Step 6: Round off the final answer.
Thus, the heat of reaction at constant pressure is approximately:
\[ \Delta H \approx 5133.8 \, \text{kJ} \]

### Final Answer:
The heat of reaction at constant pressure for the combustion of 1 mole of naphthalene is:
\[ \Delta H = 5133.8 \, \text{kJ} \]

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