`C_(2)H_(2) + 5/2 O_2(2) rarr 2CO_(2) + H_(2)O , Delta H = -310 kcal`
`C + O_(2) rarr CO_(2) , " "Delta H = -94 kcal`
`H_(2) + 1/2 O_(2) rarr H_(2)O, " " Delta H = -68 kcal`
On the basis of the above equations, `DeltaH_(f)` (enthalpy of formation) of `C_2H_2` will be :

Options

A. (a) `-148 kcal`
B. (b) `+54 kcal`
C. (c) `-54 kcal`
D. (d) `+80 kcal`

Explanation

To find the enthalpy of formation (ΔH_f) of C₂H₂ (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 of the reaction.

### Given Reactions:
1. \( C_2H_2 + \frac{5}{2} O_2 \rightarrow 2 CO_2 + H_2O \), \( \Delta H = -310 \) kcal
2. \( C + O_2 \rightarrow CO_2 \), \( \Delta H = -94 \) kcal
3. \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \), \( \Delta H = -68 \) kcal

### Step 1: Reverse the first reaction
To find the enthalpy of formation of C₂H₂, we need to reverse the first reaction because we want to express the formation of C₂H₂ from its elements:
\[ 2 CO_2 + H_2O \rightarrow C_2H_2 + \frac{5}{2} O_2 \]
The enthalpy change for this reversed reaction will be:
\[ \Delta H = +310 \text{ kcal} \]

### Step 2: Use the second and third reactions
We will use the second and third reactions as they are, to express the formation of the products \( CO_2 \) and \( H_2O \) from their elements:
- For \( 2 \) moles of \( CO_2 \):
\[ 2(C + O_2 \rightarrow CO_2) \Rightarrow 2C + O_2 \rightarrow 2CO_2 \]
The enthalpy change for this reaction is:
\[ \Delta H = 2 \times (-94) \text{ kcal} = -188 \text{ kcal} \]

- For \( H_2O \):
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \]
The enthalpy change for this reaction is:
\[ \Delta H = -68 \text{ kcal} \]

### Step 3: Combine the reactions
Now we can combine all the reactions to find the overall reaction:
1. \( 2 CO_2 + H_2O \rightarrow C_2H_2 + \frac{5}{2} O_2 \) (reversed first reaction, \( +310 \) kcal)
2. \( 2C + O_2 \rightarrow 2CO_2 \) (from second reaction, \( -188 \) kcal)
3. \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \) (from third reaction, \( -68 \) kcal)

### Step 4: Calculate the total ΔH
Now we can add the enthalpy changes:
\[ \Delta H_f (C_2H_2) = (+310) + (-188) + (-68) \]
\[ \Delta H_f (C_2H_2) = 310 - 188 - 68 \]
\[ \Delta H_f (C_2H_2) = 310 - 256 = 54 \text{ kcal} \]

### Final Answer:
The enthalpy of formation (ΔH_f) of C₂H₂ is **54 kcal**.

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