From the following thermochemical equations, calculate the standard enthalpy of formation of HCl(g) .
(A) `H_(2)(g) rarr 2H(g) , DeltaH^(0)= + 436.0 kJ mol^(-1) ` `(B) Cl_(2)(g) rarr 2Cl(g), DeltaH^(0)= + 242.7 kJ mol^(-1)`
(C ) `HCl(g) rarr H(g) + Cl(g) , DeltaH^(0) = + 431.8 kJ mol^(-1)`

Explanation

To calculate the standard enthalpy of formation of HCl(g) from the given thermochemical equations, we can follow these steps:

### Step 1: Write down the given equations and their enthalpy changes.
1. \( \text{(A)} \quad H_2(g) \rightarrow 2H(g), \quad \Delta H^{\circ} = +436.0 \, \text{kJ mol}^{-1} \)
2. \( \text{(B)} \quad Cl_2(g) \rightarrow 2Cl(g), \quad \Delta H^{\circ} = +242.7 \, \text{kJ mol}^{-1} \)
3. \( \text{(C)} \quad HCl(g) \rightarrow H(g) + Cl(g), \quad \Delta H^{\circ} = +431.8 \, \text{kJ mol}^{-1} \)

### Step 2: Modify the equations to derive the formation reaction of HCl.
The standard enthalpy of formation of HCl(g) can be represented by the reaction:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g) \]

### Step 3: Adjust the equations (A) and (B) to match the formation reaction.
- For equation (A), we need half of the reaction:
\[ \frac{1}{2} H_2(g) \rightarrow H(g) \]
The enthalpy change will be:
\[ \Delta H^{\circ} = \frac{436.0}{2} = +218.0 \, \text{kJ mol}^{-1} \]

- For equation (B), we also need half of the reaction:
\[ \frac{1}{2} Cl_2(g) \rightarrow Cl(g) \]
The enthalpy change will be:
\[ \Delta H^{\circ} = \frac{242.7}{2} = +121.35 \, \text{kJ mol}^{-1} \]

### Step 4: Combine the modified equations.
Now we can combine the modified equations (A) and (B) to get:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow H(g) + Cl(g) \]
The total enthalpy change for this step is:
\[ \Delta H = 218.0 + 121.35 = +339.35 \, \text{kJ mol}^{-1} \]

### Step 5: Use equation (C) to find the enthalpy of formation of HCl.
Now we will use equation (C) to find the enthalpy of formation of HCl:
\[ H(g) + Cl(g) \rightarrow HCl(g) \]
The enthalpy change for this reaction is:
\[ \Delta H^{\circ} = +431.8 \, \text{kJ mol}^{-1} \]

### Step 6: Apply Hess's law.
According to Hess's law, we can write:
\[ \Delta H_{\text{formation}} = \Delta H_{\text{from A and B}} - \Delta H_{\text{from C}} \]
Substituting the values:
\[ \Delta H_{\text{formation}} = 339.35 - 431.8 \]
\[ \Delta H_{\text{formation}} = -92.45 \, \text{kJ mol}^{-1} \]

### Final Answer:
The standard enthalpy of formation of HCl(g) is:
\[ \Delta H_f^{\circ} = -92.45 \, \text{kJ mol}^{-1} \]

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