Calculate `Delta U` reaction for the hydrogenation of acetalence at constant volume and at `77^(@)C`.
Given that `-DeltaH_(f)(H_(2)O)= -678` kcal mole , `Delta H_("comb")(C_(2)H_(2))= -310.1 kcal//ms^(2)`

Explanation

To calculate the change in internal energy (ΔU) for the hydrogenation of acetylene (C2H2) at constant volume and at 77°C, we will follow these steps:

### Step 1: Write the Reaction
The hydrogenation of acetylene can be represented as:
\[ \text{C}_2\text{H}_2(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_4(g) \]

### Step 2: Identify Given Data
We have the following data:
- ΔH_f (H2O) = -678 kcal/mol (enthalpy of formation of water)
- ΔH_comb (C2H2) = -310.1 kcal/mol (enthalpy of combustion of acetylene)
- ΔH_comb (C2H4) = -337.2 kcal/mol (enthalpy of combustion of ethylene)

### Step 3: Write the Combustion Reactions
1. Combustion of C2H2:
\[ \text{C}_2\text{H}_2(g) + \frac{5}{2} \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + H_2O(l) \]
ΔH = -310.1 kcal

2. Combustion of C2H4:
\[ \text{C}_2\text{H}_4(g) + 3 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 2 H_2O(l) \]
ΔH = -337.2 kcal

### Step 4: Apply Hess's Law
To find the ΔH for the hydrogenation reaction, we can manipulate the combustion reactions:
- We will reverse the combustion of C2H2 and C2H4 and add them together.

The reaction we want is:
\[ \text{C}_2\text{H}_2(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_4(g) \]

Using Hess's Law:
\[ \Delta H_{\text{reaction}} = \Delta H_{\text{combustion of C2H2}} + \Delta H_{\text{formation of water}} - \Delta H_{\text{combustion of C2H4}} \]

### Step 5: Calculate ΔH for the Reaction
Substituting the values:
\[ \Delta H = (-310.1) + (-678) - (-337.2) \]
\[ \Delta H = -310.1 - 678 + 337.2 \]
\[ \Delta H = -651.9 \text{ kcal} \]

### Step 6: Relate ΔH to ΔU
Using the relation:
\[ \Delta H = \Delta U + \Delta n_g RT \]
Where:
- Δn_g = change in moles of gas = moles of products - moles of reactants = 1 (C2H4) - 2 (C2H2 + H2) = -1
- R = 1.987 cal/(mol·K)
- T = 77°C = 350 K

Calculating Δn_g RT:
\[ \Delta n_g RT = (-1)(1.987 \text{ cal/(mol·K)})(350 \text{ K}) \]
\[ \Delta n_g RT = -695.45 \text{ cal} \approx -0.695 \text{ kcal} \]

### Step 7: Solve for ΔU
Now substituting back into the equation:
\[ -651.9 = \Delta U - 0.695 \]
\[ \Delta U = -651.9 + 0.695 \]
\[ \Delta U = -651.205 \text{ kcal} \]

### Final Answer
\[ \Delta U \approx -651.2 \text{ kcal/mol} \]