The following are the heats of reactions -
(i) `Delta H_(f)^(@)` of `H_(2)O_((l))=-68.3 "K cal mol"^(-1)`
(ii) `Delta H_("comb.")^(@)` of `C_(2)H_(2)=-337.2 "K cal mol"^(-1)`
(iii) `Delta H_("comb.")^(@)` of `C_(2)H_(4)=-363.7 "K cal mol"^(-1)`
Then heat change for the reaction `C_(2)H_(2)+H_(2)rarr C_(2)H_(4)` is -

Options

A. `-716.1` K cal
B. `+337.2` K cal
C. `-41.8` K cal
D. `-579.5` K cal

Explanation

To find the heat change for the reaction \( C_2H_2 + H_2 \rightarrow C_2H_4 \), we can use the heats of combustion provided in the question. The heat change for the reaction can be calculated using the following steps:

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction we are considering is:
\[ C_2H_2 + H_2 \rightarrow C_2H_4 \]

2. **Identify the Heats of Combustion**:
We have the following heats of combustion:
- \( \Delta H_{\text{comb}}(C_2H_2) = -337.2 \, \text{kcal/mol} \)
- \( \Delta H_{\text{comb}}(C_2H_4) = -363.7 \, \text{kcal/mol} \)

3. **Use the Heats of Combustion**:
The heat change for the reaction can be calculated using the formula:
\[ \Delta H = \left( \Delta H_{\text{comb}}(C_2H_2) + \Delta H_{\text{comb}}(H_2) \right) - \Delta H_{\text{comb}}(C_2H_4) \]
Since the heat of combustion for \( H_2 \) is zero (as it is in its elemental form), we can simplify the equation:
\[ \Delta H = \left( -337.2 \, \text{kcal/mol} + 0 \right) - (-363.7 \, \text{kcal/mol}) \]

4. **Calculate the Heat Change**:
Now substituting the values:
\[ \Delta H = -337.2 - (-363.7) = -337.2 + 363.7 \]
\[ \Delta H = 26.5 \, \text{kcal/mol} \]

5. **Final Result**:
The heat change for the reaction \( C_2H_2 + H_2 \rightarrow C_2H_4 \) is:
\[ \Delta H = 26.5 \, \text{kcal/mol} \]