The heat of combustion of acetylene is 312 Kcal . If heat of formation of `CO_(2) & H_(2)O` are -94 & -68 Kcal respectively . Given that heat of atomisation of C& H are 150 & 50 Kcal respectively and C-H bond energy is 93 Kcal . Calculate `(DeltaH_(C=C))/(6)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the combustion reaction of acetylene
The combustion of acetylene (C2H2) can be represented as:
\[ \text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \]
The heat of combustion (ΔH_comb) is given as -312 kcal.

### Step 2: Use the heat of formation to find ΔH_f for C2H2
The heat of combustion can be expressed in terms of the heats of formation:
\[ \Delta H_{\text{comb}} = \sum \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \]
Substituting the known values:
\[ -312 = [2(-94) + (-68)] - \Delta H_f (\text{C}_2\text{H}_2) \]
Calculating the right side:
\[ -312 = [-188 - 68] - \Delta H_f (\text{C}_2\text{H}_2) \]
\[ -312 = -256 - \Delta H_f (\text{C}_2\text{H}_2) \]
Rearranging gives:
\[ \Delta H_f (\text{C}_2\text{H}_2) = -256 + 312 = 56 \text{ kcal} \]

### Step 3: Write the formation reaction of C2H2 from its elements
The formation of acetylene from its elements can be represented as:
\[ 2\text{C (s)} + \text{H}_2 (g) \rightarrow \text{C}_2\text{H}_2 (g) \]
The heat of formation (ΔH_f) for this reaction is:
\[ \Delta H_f = 0 - [2 \times \Delta H_{\text{atomization of C}} + \Delta H_{\text{atomization of H}} - 2 \times \text{C-H bond energy}] \]
Substituting the values:
\[ 56 = 0 - [2 \times 150 + 50 - 2 \times 93] \]
Calculating the right side:
\[ 56 = -[300 + 50 - 186] \]
\[ 56 = -[164] \]
Thus:
\[ 56 = 164 - 2x \]
Where \( x \) is the ΔH for the C=C bond formation. Rearranging gives:
\[ 2x = 164 - 56 \]
\[ 2x = 108 \]
\[ x = 54 \text{ kcal} \]

### Step 4: Calculate ΔH_C=C / 6
Now we need to find:
\[ \frac{\Delta H_{C=C}}{6} = \frac{54}{6} = 9 \text{ kcal} \]

### Final Answer
\[ \frac{\Delta H_{C=C}}{6} = 9 \text{ kcal} \]

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