If `CH_(3)COOH+OH^(-)toCH_(3)COO^(-)+H_(2)O+q_(1)and H^(+)+OH^(-)to H_(2)O+q_(2)`, then the enthalpy change for the reaction `CH_(3)COOHtoCH_(3)COO^(-)+H^(+)` is "equal to :

Options

A. `q_(1)+q_(2)`
B. `q_(1)-q_(2)`
C. `q_(2)-q_(1)`
D. `-q_(1)-q_(2)`

Answer

To find the enthalpy change for the reaction \( \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \), we can use the two given reactions and their associated enthalpy changes.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- The first reaction is:
\[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} + q_1 \]
The enthalpy change for this reaction is:
\[ \Delta H_1 = -q_1 \]
- The second reaction is:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} + q_2 \]
The enthalpy change for this reaction is:
\[ \Delta H_2 = -q_2 \]

2. **Write the Target Reaction:**
- The target reaction we want to analyze is:
\[ \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \]

3. **Manipulate the Given Reactions:**
- Keep the first reaction as it is:
\[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]
- Reverse the second reaction:
\[ \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \]
The enthalpy change for the reversed reaction is:
\[ \Delta H_2 = +q_2 \]

4. **Combine the Reactions:**
- Adding the first reaction and the reversed second reaction:
\[ \text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \]
\[ \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \]
- This gives:
\[ \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \]

5. **Calculate the Enthalpy Change:**
- The total enthalpy change for the combined reaction is:
\[ \Delta H = \Delta H_1 + \Delta H_2 \]
- Substituting the values:
\[ \Delta H = -q_1 + q_2 \]

6. **Final Expression:**
- Therefore, the enthalpy change for the reaction \( \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \) is:
\[ \Delta H = q_2 - q_1 \]

### Final Answer:
The enthalpy change for the reaction \( \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \) is equal to \( q_2 - q_1 \).