`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. `-148 kcal`
B. `+54 kcal`
C. `-54 kcal`
D. `+80 kcal`

Explanation

To find the enthalpy of formation (ΔHf) of C₂H₂ (acetylene), we will 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. We will rearrange the given reactions to derive the formation reaction for C₂H₂.

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

### Step 1: Write the formation reaction for C₂H₂
The formation reaction for C₂H₂ from its elements in their standard states is:
\[ 2C (s) + H_2 (g) \rightarrow C_2H_2 (g) \]

### Step 2: Reverse the first reaction
We need to reverse the first reaction to have C₂H₂ on the reactant side:
\[ 2 CO_2 + H_2O \rightarrow C_2H_2 + \frac{5}{2} O_2 \]
When we reverse the reaction, the sign of ΔH changes:
ΔH = +310 kcal

### Step 3: Adjust the second reaction
We need 2 moles of carbon (C) in our formation reaction. Therefore, we multiply the second reaction by 2:
\[ 2C + O_2 \rightarrow 2CO_2 \]
The ΔH for this reaction becomes:
ΔH = 2 × (-94 kcal) = -188 kcal

### Step 4: Use the third reaction as is
The third reaction can be used as is:
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \]
The ΔH for this reaction is:
ΔH = -68 kcal

### Step 5: Combine the reactions
Now we will add all the reactions together:
1. \( 2 CO_2 + H_2O \rightarrow C_2H_2 + \frac{5}{2} O_2 \) (ΔH = +310 kcal)
2. \( 2C + O_2 \rightarrow 2CO_2 \) (ΔH = -188 kcal)
3. \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \) (ΔH = -68 kcal)

When we add these reactions, we can cancel out the common species:
- The \( 2CO_2 \) from the second reaction cancels with \( 2CO_2 \) from the first reaction.
- The \( H_2O \) from the first reaction cancels with \( H_2O \) from the third reaction.

The resulting reaction is:
\[ 2C + H_2 + \frac{5}{2} O_2 \rightarrow C_2H_2 \]

### Step 6: Calculate the total ΔH
Now we can sum the ΔH values:
\[ ΔH = +310 kcal - 188 kcal - 68 kcal \]
\[ ΔH = +310 - 188 - 68 = +54 kcal \]

### Conclusion
The enthalpy of formation (ΔHf) of C₂H₂ is:
\[ ΔHf = +54 \text{ kcal} \]