The enthalpy change for first proton neutralisation of `H_(2)S` is `-37.1kJ mol^(-1)`. What is the enthalpy change for first ionisation of `H_(2)S.' in KJ/mol.

Options

A. `94.2`
B. `-20.0`
C. `20.0`
D. `-94.2`

Explanation

To find the enthalpy change for the first ionization of \( H_2S \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- The first reaction is the ionization of \( H_2S \):
\[ H_2S \rightarrow H^+ + HS^- \]
Let the enthalpy change for this reaction be \( \Delta H_1 \).

- The second reaction is the neutralization of \( H^+ \) by a strong base (e.g., \( OH^- \)):
\[ H^+ + OH^- \rightarrow H_2O \]
The enthalpy change for this reaction (enthalpy of neutralization) is known to be \( \Delta H_2 = -57.1 \, \text{kJ/mol} \).

2. **Write the Overall Reaction:**
- The overall reaction combining both steps is:
\[ H_2S + OH^- \rightarrow H_2O + HS^- \]
- The enthalpy change for this overall reaction is given as \( \Delta H = -37.1 \, \text{kJ/mol} \).

3. **Apply Hess's Law:**
- According to Hess's law:
\[ \Delta H = \Delta H_1 + \Delta H_2 \]
- Plugging in the known values:
\[ -37.1 = \Delta H_1 + (-57.1) \]

4. **Solve for \( \Delta H_1 \):**
- Rearranging the equation gives:
\[ \Delta H_1 = -37.1 + 57.1 \]
- Calculating this:
\[ \Delta H_1 = 20 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy change for the first ionization of \( H_2S \) is \( 20 \, \text{kJ/mol} \).

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