The heat of combustion of naphthalene `(C_(10)H_(8)(s))` at constant volume was found to be - 5133 k J `mol^(-1)` . Calculate the value of enthalpy change.

Explanation

To calculate the enthalpy change (ΔH) from the heat of combustion at constant volume (ΔU), we can use the relationship between enthalpy and internal energy, which is given by the equation:

\[ \Delta H = \Delta U + \Delta nRT \]

Where:
- ΔH = Change in enthalpy
- ΔU = Change in internal energy (heat of combustion at constant volume)
- Δn = Change in the number of moles of gas during the reaction
- R = Universal gas constant (approximately 8.314 J/mol·K)
- T = Temperature in Kelvin

### Step 1: Identify the values
From the problem, we know:
- ΔU (heat of combustion) = -5133 kJ/mol = -5133000 J/mol (since 1 kJ = 1000 J)

### Step 2: Determine Δn
For the combustion of naphthalene (C10H8), the reaction can be represented as:

\[ C_{10}H_8(s) + O_2(g) \rightarrow CO_2(g) + H_2O(g) \]

To find Δn, we need to balance the combustion reaction. The balanced equation is:

\[ C_{10}H_8 + 12 O_2 \rightarrow 10 CO_2 + 4 H_2O \]

In this reaction:
- Moles of gaseous products = 10 (from CO2) + 4 (from H2O) = 14
- Moles of gaseous reactants = 12 (from O2)

Thus,

\[ \Delta n = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} = 14 - 12 = 2 \]

### Step 3: Calculate ΔH
Assuming the reaction occurs at standard temperature (298 K), we can substitute the values into the equation:

\[ \Delta H = \Delta U + \Delta nRT \]

Substituting the known values:

\[ \Delta H = -5133000 J/mol + (2) \times (8.314 J/mol·K) \times (298 K) \]

Calculating the second term:

\[ \Delta nRT = 2 \times 8.314 \times 298 = 4965.68 J/mol \]

Now substituting back into the ΔH equation:

\[ \Delta H = -5133000 J/mol + 4965.68 J/mol \]

\[ \Delta H = -5133000 J/mol + 4965.68 J/mol = -5133000 + 4.96568 \approx -5132995.03 J/mol \]

Converting back to kJ:

\[ \Delta H \approx -5132.995 kJ/mol \approx -5133 kJ/mol \]

### Final Answer:
The value of enthalpy change (ΔH) for the combustion of naphthalene is approximately -5133 kJ/mol.

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