To calculate the multiple bond energy of a C≡C bond in C₂H₂ using the provided data, we will follow these steps:
### Step 1: Write down the given data
- The enthalpy change for the formation of C₂H₂ from its elements is given as:
\[
2C_{(s)} + H_{2(g)} \rightarrow C_{2}H_{2(g)}, \Delta H = 225 \, \text{kJ mol}^{-1}
\]
- The bond energy of a C-H bond is given as:
\[
\text{Bond energy of C-H} = 350 \, \text{kJ mol}^{-1}
\]
- The enthalpy change for the sublimation of carbon is:
\[
2C_{(s)} \rightarrow 2C_{(g)}, \Delta H = 1410 \, \text{kJ mol}^{-1}
\]
- The enthalpy change for the dissociation of hydrogen is:
\[
H_{2(g)} \rightarrow 2H_{(g)}, \Delta H = 330 \, \text{kJ mol}^{-1}
\]
### Step 2: Set up the equation for the reaction
Using Hess's law, we can express the enthalpy change of the reaction in terms of bond energies:
\[
\Delta H = \text{Bond energy of reactants} - \text{Bond energy of products}
\]
### Step 3: Calculate the bond energy of reactants
For the reactants:
- We have 2 moles of solid carbon turning into gaseous carbon, which contributes:
\[
\Delta H_{\text{reactants}} = 1410 \, \text{kJ} + 330 \, \text{kJ} = 1740 \, \text{kJ}
\]
### Step 4: Set up the bond energy of products
For the products:
- C₂H₂ has one C≡C bond and two C-H bonds. Therefore, the bond energy of products can be expressed as:
\[
\Delta H_{\text{products}} = \text{Bond energy of C≡C} + 2 \times 350 \, \text{kJ}
\]
Let \( x \) be the bond energy of the C≡C bond. Thus:
\[
\Delta H_{\text{products}} = x + 700 \, \text{kJ}
\]
### Step 5: Substitute into the equation
Now substituting into the enthalpy change equation:
\[
225 \, \text{kJ} = 1740 \, \text{kJ} - (x + 700 \, \text{kJ})
\]
### Step 6: Solve for x
Rearranging gives:
\[
225 = 1740 - x - 700
\]
\[
225 = 1040 - x
\]
\[
x = 1040 - 225
\]
\[
x = 815 \, \text{kJ mol}^{-1}
\]
### Final Answer
The multiple bond energy of the C≡C bond in C₂H₂ is:
\[
\boxed{815 \, \text{kJ mol}^{-1}}
\]
