The following are heats of reactions `:`
`(A)` `DeltaH_(f)^(@)` of `H_(2)O(l)=-68.3kcal` `mol^(-1)`
`(B)` `DeltaH_(comb)^(@)` of `C_(2)H_(2)=-337.2 kcal` `mol^(-1)`
`(C)` `DeltaH_(comb)^(@)` of `C_(2)H_(4)=-363.7Kal` `mol_(-1)`
Then the heat charge for the reaction `:-`
`C_(2)H_(2)+H_(2)rarrC_(2)H_(4)`

Options

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

Answer

To find the heat change for the reaction \( C_2H_2 + H_2 \rightarrow C_2H_4 \), we will use the heats of combustion provided for the reactants and products. The heat change (\( \Delta H \)) for the reaction can be calculated using the following formula:

\[ \Delta H = \sum \Delta H_{\text{combustion of reactants}} - \sum \Delta H_{\text{combustion of products}} \]

### Step-by-Step Solution:

1. **Identify the given heats of combustion**:
- For \( C_2H_2 \): \( \Delta H_{\text{comb}}(C_2H_2) = -337.2 \, \text{kcal/mol} \)
- For \( C_2H_4 \): \( \Delta H_{\text{comb}}(C_2H_4) = -363.7 \, \text{kcal/mol} \)
- The heat of formation for \( H_2 \) is not needed since it is in its elemental form and its heat of formation is zero.

2. **Write the equation for heat change**:
\[ \Delta H = \Delta H_{\text{comb}}(C_2H_2) + \Delta H_{\text{comb}}(H_2) - \Delta H_{\text{comb}}(C_2H_4) \]
Since \( \Delta H_{\text{comb}}(H_2) = 0 \):
\[ \Delta H = \Delta H_{\text{comb}}(C_2H_2) - \Delta H_{\text{comb}}(C_2H_4) \]

3. **Substitute the values**:
\[ \Delta H = (-337.2 \, \text{kcal/mol}) - (-363.7 \, \text{kcal/mol}) \]

4. **Calculate the heat change**:
\[ \Delta H = -337.2 + 363.7 = 26.5 \, \text{kcal/mol} \]

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