The standard enthalpy of formation of gaseous `H_(2)O` at 298 K is `-241.82` kJ/mol. Calculate `DeltaH^(@)` at 373 K given the following values of the molar heat capacities at constant pressure :
`H_(2)O(g)=33.58 " JK"^(-1)" mol"^(-1), " "H_(2)(g)=29.84 " JK"^(-1)" mol"^(-1), " "O_(2)(g)=29.37 " JK"^(-1)" mol"^(-1)`
Assume that the heat capacities are independent of temperature :

Options

A. `-242.6 kJ//mol`
B. `-485.2 kJ//mol`
C. `-121.3 kJ//mol`
D. `-286.4 kJ//mol`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the change in heat capacity (ΔCp)

We need to calculate the change in heat capacity for the reaction:

\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(g) \]

The change in heat capacity (ΔCp) is given by:

\[ \Delta Cp = Cp_{\text{products}} - Cp_{\text{reactants}} \]

Where:
- \( Cp_{\text{products}} = Cp_{\text{H}_2O(g)} = 33.58 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( Cp_{\text{reactants}} = Cp_{\text{H}_2(g)} + \frac{1}{2} Cp_{\text{O}_2(g)} \)

Calculating \( Cp_{\text{reactants}} \):

\[ Cp_{\text{reactants}} = 29.84 \, \text{J K}^{-1} \text{mol}^{-1} + \frac{1}{2} \times 29.37 \, \text{J K}^{-1} \text{mol}^{-1} \]

Calculating \( \frac{1}{2} \times 29.37 \):

\[ \frac{1}{2} \times 29.37 = 14.685 \, \text{J K}^{-1} \text{mol}^{-1} \]

Now substituting back:

\[ Cp_{\text{reactants}} = 29.84 + 14.685 = 44.525 \, \text{J K}^{-1} \text{mol}^{-1} \]

Now we can find ΔCp:

\[ \Delta Cp = 33.58 - 44.525 = -10.945 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 2: Use Kirchhoff's equation to find ΔH at 373 K

Kirchhoff's equation is:

\[ \Delta H(T_2) - \Delta H(T_1) = \Delta Cp \cdot (T_2 - T_1) \]

Where:
- \( T_1 = 298 \, \text{K} \)
- \( T_2 = 373 \, \text{K} \)
- \( \Delta H(T_1) = -241.82 \, \text{kJ/mol} \)

Substituting the values into Kirchhoff's equation:

\[ \Delta H(373) - (-241.82) = -10.945 \cdot (373 - 298) \]

Calculating \( (373 - 298) \):

\[ 373 - 298 = 75 \, \text{K} \]

Now substituting this back into the equation:

\[ \Delta H(373) + 241.82 = -10.945 \cdot 75 \]

Calculating \( -10.945 \cdot 75 \):

\[ -10.945 \cdot 75 = -821.625 \, \text{J/mol} = -0.821625 \, \text{kJ/mol} \]

Now substituting this back into the equation:

\[ \Delta H(373) + 241.82 = -0.821625 \]

Solving for \( \Delta H(373) \):

\[ \Delta H(373) = -0.821625 - 241.82 \]

Calculating this:

\[ \Delta H(373) = -242.641625 \, \text{kJ/mol} \]

### Step 3: Round to appropriate significant figures

Rounding this to three significant figures gives:

\[ \Delta H(373) \approx -242.64 \, \text{kJ/mol} \]

### Final Answer

\[ \Delta H(373) \approx -242.64 \, \text{kJ/mol} \]

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