To find the standard enthalpy of formation of acetylene (C2H2) at 25°C, we can use the combustion enthalpies of carbon, hydrogen, and acetylene provided in the question. The combustion reactions and their respective enthalpy changes are as follows:
1. **Combustion of Carbon:**
\[
C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_c = -393.5 \, \text{kJ/mol}
\]
2. **Combustion of Hydrogen:**
\[
H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \quad \Delta H_c = -285.8 \, \text{kJ/mol}
\]
3. **Combustion of Acetylene:**
\[
2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g) \quad \Delta H_c = -1309.5 \, \text{kJ/mol}
\]
Next, we need to write the formation reaction for acetylene:
\[
2C(s) + H_2(g) \rightarrow C_2H_2(g)
\]
To find the standard enthalpy of formation (\(\Delta H_f\)) of acetylene, we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We will manipulate the combustion reactions to arrive at the formation reaction.
### Step 1: Reverse the combustion of acetylene
When we reverse the combustion of acetylene, we change the sign of the enthalpy:
\[
4CO_2(g) + 2H_2O(g) \rightarrow 2C_2H_2(g) + 5O_2(g) \quad \Delta H = +1309.5 \, \text{kJ/mol}
\]
### Step 2: Multiply the combustion of carbon
Since we need 2 moles of carbon in the formation reaction, we multiply the combustion of carbon by 2:
\[
2C(s) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = 2 \times (-393.5) = -787.0 \, \text{kJ/mol}
\]
### Step 3: Use the combustion of hydrogen
We will use the combustion of hydrogen as it is:
\[
H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \quad \Delta H = -285.8 \, \text{kJ/mol}
\]
### Step 4: Combine the equations
Now, we can add these equations together:
1. From step 1:
\[
4CO_2(g) + 2H_2O(g) \rightarrow 2C_2H_2(g) + 5O_2(g) \quad (+1309.5 \, \text{kJ/mol})
\]
2. From step 2:
\[
2C(s) + O_2(g) \rightarrow 2CO_2(g) \quad (-787.0 \, \text{kJ/mol})
\]
3. From step 3:
\[
H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \quad (-285.8 \, \text{kJ/mol})
\]
Adding these together:
\[
\Delta H_f = 1309.5 - 787.0 - 285.8
\]
\[
\Delta H_f = 236.7 \, \text{kJ/mol}
\]
Thus, the standard enthalpy of formation of acetylene (C2H2) is:
\[
\Delta H_f = +236.7 \, \text{kJ/mol}
\]
