Using the data provided, calculate the multiple bond energy (`kJ "mol"^(-1)`) of a `C equiv C` bond in `C_(2)H_(2)`.That energy is (take the bond enrgy of a C-H bond as 350 kJ ` "mol"^(-1)`):
`2C(s) + H_(2)(g) to C_(2)H_(2)(g), DeltaH = 225 kJ "mol"^(-1)`
`2C(s) to 2C(g), DeltaH = 1410 kJ "mol"^(-1)`
`H_(2)(g) to 2H(g), DeltaH = 330 kJ "mol"^(-1)`

Options

A. 1165
B. 837
C. 865
D. 815

Explanation

To calculate the multiple bond energy of the `C≡C` bond in `C2H2`, we will follow these steps:

### Step 1: Write the given reactions and their enthalpy changes
1. **Formation of Ethyne**:
\[ 2C(s) + H_2(g) \rightarrow C_2H_2(g), \quad \Delta H_1 = 225 \, \text{kJ/mol} \]
2. **Sublimation of Carbon**:
\[ 2C(s) \rightarrow 2C(g), \quad \Delta H_2 = 1410 \, \text{kJ/mol} \]
3. **Dissociation of Hydrogen**:
\[ H_2(g) \rightarrow 2H(g), \quad \Delta H_3 = 330 \, \text{kJ/mol} \]

### Step 2: Write the target equation for bond energy calculation
To find the bond energy of the `C≡C` bond, we need to break down the `C2H2` molecule into its constituent atoms:
\[ C_2H_2(g) \rightarrow 2C(g) + 2H(g) \]

### Step 3: Combine the given reactions to derive the target equation
To derive the target equation, we can manipulate the given reactions:
- From the second reaction, we have `2C(s) → 2C(g)` which we will use as is.
- From the third reaction, we have `H2(g) → 2H(g)` which we will reverse to get `2H(g) → H2(g)` and change the sign of its enthalpy.
- The first reaction will be subtracted because we need to reverse it to break down `C2H2`.

Thus, we can combine the reactions as follows:
\[ \Delta H = \Delta H_2 + \Delta H_3 - \Delta H_1 \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ \Delta H = 1410 \, \text{kJ/mol} + 330 \, \text{kJ/mol} - 225 \, \text{kJ/mol} \]
Calculating this gives:
\[ \Delta H = 1410 + 330 - 225 = 1515 \, \text{kJ/mol} \]

### Step 5: Relate the enthalpy change to bond energies
The enthalpy change calculated represents the total energy required to break the bonds in `C2H2`. This includes:
- The bond energy of two `C-H` bonds (which is `2 * 350 \, \text{kJ/mol}`).
- The bond energy of one `C≡C` bond.

Let \( E_{C≡C} \) be the bond energy of the `C≡C` bond. Thus, we can write:
\[ 1515 = 2(350) + E_{C≡C} \]

### Step 6: Solve for the bond energy of `C≡C`
Now, substituting the values:
\[ 1515 = 700 + E_{C≡C} \]
\[ E_{C≡C} = 1515 - 700 = 815 \, \text{kJ/mol} \]

### Conclusion
The bond energy of the `C≡C` bond in `C2H2` is **815 kJ/mol**.

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