Given that -
`2C(s)+2O_(2)(g)rarr 2CO_(2)(g) Delta H = -787 KJ`
`H_(2)(g)+ 1//2O_(2)(g)rarr H_(2)O(l) Delta =-286 KJ`
`C_(2)H_(2)(g)+(5)/(2)O_(2)(g)rarr H_(2)O(l) + 2CO_(2)Delta H=-1310 KJ`
Heat of formation of acetylene is :-

Options

A. `+1802` KJ
B. `-1802` KJ
C. `-800` KJ
D. `+237` KJ

Answer

To determine the heat of formation of acetylene (C₂H₂), we will manipulate the given reactions to derive the desired reaction. The heat of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.

### Given Reactions:
1. \( 2C(s) + 2O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -787 \, \text{kJ} \)
2. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H = -286 \, \text{kJ} \)
3. \( C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow H_2O(l) + 2CO_2(g) \quad \Delta H = -1310 \, \text{kJ} \)

### Step-by-Step Solution:

1. **Reverse the Third Reaction**:
We need to express the formation of acetylene from its elements. Thus, we reverse the third reaction:
\[ H_2O(l) + 2CO_2(g) \rightarrow C_2H_2(g) + \frac{5}{2}O_2(g) \]
When we reverse the reaction, the sign of \(\Delta H\) also changes:
\[ \Delta H = +1310 \, \text{kJ} \]

2. **Use the First Reaction as it is**:
The first reaction is already in the desired form for our calculations:
\[ 2C(s) + 2O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -787 \, \text{kJ} \]

3. **Use the Second Reaction as it is**:
The second reaction is also in the desired form:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H = -286 \, \text{kJ} \]

4. **Combine the Reactions**:
Now we will combine the three reactions. The water produced in the second reaction will cancel with the water in the reversed third reaction. The carbon dioxide produced in the first reaction will also cancel with the carbon dioxide in the reversed third reaction.

The overall reaction will be:
\[ 2C(s) + H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_2(g) \]

5. **Calculate the Total Enthalpy Change**:
Now we sum the enthalpy changes:
\[ \Delta H_{formation} = \Delta H_{reversed \, 3rd} + \Delta H_{1st} + \Delta H_{2nd} \]
\[ \Delta H_{formation} = (+1310 \, \text{kJ}) + (-787 \, \text{kJ}) + (-286 \, \text{kJ}) \]
\[ \Delta H_{formation} = 1310 - 787 - 286 = 237 \, \text{kJ} \]

### Final Answer:
The heat of formation of acetylene (C₂H₂) is \( \Delta H = 237 \, \text{kJ} \).

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