The standard enthalpy of formation of `H_(2)(g)` and `Cl_(2)(g)` and `HCl(g)` are `218 kJ//mol`, `121.88 kJ//mol` and `-93.31 kJ//mol` respectively. Calculate standard enthalpy change in `kJ` for
`(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g)`

Options

A. `+431.99
B. `-262.14`
C. `-431.99`
D. `+247.37`

Explanation

To calculate the standard enthalpy change for the reaction

\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g) \]

we will use the standard enthalpy of formation values provided for \(H_2(g)\), \(Cl_2(g)\), and \(HCl(g)\).

### Step 1: Write down the standard enthalpy of formation values
- \( \Delta H_f^\circ (H_2(g)) = 218 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (Cl_2(g)) = 121.88 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (HCl(g)) = -93.31 \, \text{kJ/mol} \)

### Step 2: Apply the formula for standard enthalpy change
The standard enthalpy change for the reaction can be calculated using the formula:

\[ \Delta H_{reaction} = \sum \Delta H_f^\circ \, \text{(products)} - \sum \Delta H_f^\circ \, \text{(reactants)} \]

### Step 3: Calculate the enthalpy of the products
For the products, we have:
- 1 mole of \(HCl(g)\):
\[ \Delta H_f^\circ (HCl) = -93.31 \, \text{kJ/mol} \]

Thus, the total for products is:
\[ \Delta H_{products} = -93.31 \, \text{kJ/mol} \]

### Step 4: Calculate the enthalpy of the reactants
For the reactants, we have:
- \(\frac{1}{2}\) mole of \(H_2(g)\):
\[ \Delta H_f^\circ (H_2) = 218 \, \text{kJ/mol} \Rightarrow \frac{1}{2} \times 218 = 109 \, \text{kJ} \]
- \(\frac{1}{2}\) mole of \(Cl_2(g)\):
\[ \Delta H_f^\circ (Cl_2) = 121.88 \, \text{kJ/mol} \Rightarrow \frac{1}{2} \times 121.88 = 60.94 \, \text{kJ} \]

Thus, the total for reactants is:
\[ \Delta H_{reactants} = 109 \, \text{kJ} + 60.94 \, \text{kJ} = 169.94 \, \text{kJ} \]

### Step 5: Calculate the standard enthalpy change for the reaction
Now substituting the values into the enthalpy change formula:

\[ \Delta H_{reaction} = -93.31 \, \text{kJ} - 169.94 \, \text{kJ} \]

\[ \Delta H_{reaction} = -93.31 - 169.94 = -263.25 \, \text{kJ} \]

### Final Answer
The standard enthalpy change for the reaction is:

\[ \Delta H_{reaction} = -263.25 \, \text{kJ} \]

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