Calculate the heat of hydrogenation ,
`C_(2)H_(4)(g)+H_(2)(g)toC_(2)H_(6)(g)`
Given that, the heats of combustion of ethylene, hydrogen and ethane are -337.0, -68.4 and -373.0 kcal respectively.

Options

A. `-32.4` kcal
B. `-36.3` kcal
C. `-37.8` kcal
D. `39.2` kcal

Answer

To calculate the heat of hydrogenation for the reaction:

\[ C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) \]

we can use the heats of combustion provided:

- Heat of combustion of ethylene (\(C_2H_4\)): \(-337.0 \, \text{kcal}\)
- Heat of combustion of hydrogen (\(H_2\)): \(-68.4 \, \text{kcal}\)
- Heat of combustion of ethane (\(C_2H_6\)): \(-373.0 \, \text{kcal}\)

### Step-by-Step Solution:

1. **Write the formula for heat of hydrogenation**:
The heat of hydrogenation can be calculated using the following formula:
\[ \Delta H_{\text{hydrogenation}} = \Delta H_{\text{combustion of } C_2H_4} + \Delta H_{\text{combustion of } H_2} - \Delta H_{\text{combustion of } C_2H_6} \]

2. **Substitute the values into the formula**:
Substitute the given values into the formula:
\[ \Delta H_{\text{hydrogenation}} = (-337.0 \, \text{kcal}) + (-68.4 \, \text{kcal}) - (-373.0 \, \text{kcal}) \]

3. **Simplify the equation**:
Simplifying the equation gives:
\[ \Delta H_{\text{hydrogenation}} = -337.0 - 68.4 + 373.0 \]

4. **Calculate the result**:
Now, perform the arithmetic:
\[ \Delta H_{\text{hydrogenation}} = -337.0 - 68.4 + 373.0 = -32.4 \, \text{kcal} \]

5. **Final Answer**:
The heat of hydrogenation for the reaction is:
\[ \Delta H_{\text{hydrogenation}} = -32.4 \, \text{kcal} \]