The molar enthalpies of combustion of `C_(2)H_(2)(g), C("graphite")` and `H_(2)(g)`are `-1300,-394`, and `-286 kJ mol^(-1)`, respectively. The standard enthalpy of formation of `C_(2)H_(2)(g)` is

Options

A. `-226 kJ mol^(-1)`
B. `-626 kJ mol^(-1)`
C. `226 kJ mol^(-1)`
D. `626 kJ mol^(-1)`

Explanation

To find the standard enthalpy of formation of \( C_2H_2(g) \), we will use the given molar enthalpies of combustion of \( C_2H_2(g) \), graphite, and hydrogen. The enthalpy of formation can be determined using 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. **Write the combustion reactions**:
- For \( C \) (graphite):
\[
C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -394 \, \text{kJ/mol}
...