The standard enthalpy change for the following reaction is 436.4 kJ :
`H_(2)(g)rarrH(g)+H(g)`
What is the `Delta_(f)H^(@)` of atomic hydrogen (H)?

Options

A. 872.8 kJ/mol
B. 218.2 kJ/mol
C. `- 218.2` kJ/mol
D. `- 436.9` kJ/mol

Explanation

To find the standard enthalpy of formation (\( \Delta_f H^\circ \)) of atomic hydrogen (H), we start with the given reaction:

\[ H_2(g) \rightarrow H(g) + H(g) \]

The standard enthalpy change for this reaction is given as \( 436.4 \, \text{kJ} \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction shows that one mole of diatomic hydrogen gas (\( H_2 \)) dissociates into two moles of atomic hydrogen (H).

2. **Relating Enthalpy Change to Formation**:
The enthalpy change of \( 436.4 \, \text{kJ} \) corresponds to the dissociation of one mole of \( H_2 \) into two moles of H. To find the enthalpy of formation of one mole of atomic hydrogen, we need to consider that the formation of one mole of atomic hydrogen involves half of the dissociation of one mole of \( H_2 \).

3. **Calculating the Enthalpy of Formation**:
Since the enthalpy change for the dissociation of one mole of \( H_2 \) is \( 436.4 \, \text{kJ} \), the enthalpy change for the formation of one mole of atomic hydrogen (H) will be half of this value:
\[ \Delta_f H^\circ = \frac{436.4 \, \text{kJ}}{2} \]

4. **Performing the Calculation**:
\[ \Delta_f H^\circ = 218.2 \, \text{kJ/mol} \]

Thus, the standard enthalpy of formation of atomic hydrogen (H) is \( 218.2 \, \text{kJ/mol} \).

### Final Answer:
The \( \Delta_f H^\circ \) of atomic hydrogen (H) is \( 218.2 \, \text{kJ/mol} \).

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