The combustion enthalpies of carbon, hydrogen and ethyne are `-393.5 kJ mol^(-1), - 285.8 kJ mol^(-1)` and `-1309.5 kJ mol^(-1)` respectively at `25^(@)C`. The value of standard enthalpy of formation of acetylene at this temperature is

Options

A. `236.2 kJ mol^(-1)`
B. `298.8 kJ mol^(-1)`
C. `74.7 kJ mol^(-1)`
D. `-107.7 kJ mol^(-1)`

Answer

To find the standard enthalpy of formation of acetylene (C₂H₂) at 25°C, we can use the combustion enthalpies of carbon, hydrogen, and ethyne. Here’s the step-by-step solution:

### Step 1: Write the combustion reactions
1. **Combustion of Carbon (C)**:
\[ C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -393.5 \, \text{kJ/mol} \]

2. **Combustion of Hydrogen (H₂)**:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H = -285.8 \, \text{kJ/mol} \]

3. **Combustion of Acetylene (C₂H₂)**:
\[ C_2H_2(g) + \frac{5}{2} O_2(g) \rightarrow 2 CO_2(g) + H_2O(l) \quad \Delta H = -1309.5 \, \text{kJ/mol} \]

### Step 2: Write the formation reaction for Acetylene
The formation reaction for acetylene is:
\[ 2C(s) + H_2(g) \rightarrow C_2H_2(g) \]

### Step 3: Use Hess's Law
To find the enthalpy of formation of acetylene, we can manipulate the combustion reactions using Hess's Law. We need to reverse the combustion reaction of acetylene and adjust the coefficients of the other reactions accordingly.

1. **Reverse the combustion of acetylene**:
\[ 2 CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2} O_2(g) \quad \Delta H = +1309.5 \, \text{kJ/mol} \]

2. **Multiply the combustion of carbon by 2**:
\[ 2C(s) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = 2 \times (-393.5) = -787.0 \, \text{kJ/mol} \]

3. **Use the combustion of hydrogen as is**:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H = -285.8 \, \text{kJ/mol} \]

### Step 4: Combine the equations
Now, we add these three equations together:
- From the reversed combustion of acetylene:
\[ 2 CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2} O_2(g) \quad (+1309.5 \, \text{kJ}) \]
- From the modified combustion of carbon:
\[ 2C(s) + O_2(g) \rightarrow 2CO_2(g) \quad (-787.0 \, \text{kJ}) \]
- From the combustion of hydrogen:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad (-285.8 \, \text{kJ}) \]

Combining these gives:
\[ 2C(s) + H_2(g) + \frac{5}{2} O_2(g) \rightarrow C_2H_2(g) \]

### Step 5: Calculate the total enthalpy change
Now sum the enthalpy changes:
\[ \Delta H_{formation} = 1309.5 - 787.0 - 285.8 \]
\[ \Delta H_{formation} = 1309.5 - 1072.8 = 236.7 \, \text{kJ/mol} \]

### Final Answer
The standard enthalpy of formation of acetylene (C₂H₂) at 25°C is approximately:
\[ \Delta H_f^\circ (C_2H_2) = 236.7 \, \text{kJ/mol} \]