Calculate work done in chemical reaction (in kcal)
`A(s)+3B(g)rarrC(l)` at `227^(@)C` at 1 atm in closed vessel.

Answer

To calculate the work done in the chemical reaction \( A(s) + 3B(g) \rightarrow C(l) \) at \( 227^\circ C \) and 1 atm in a closed vessel, we can follow these steps:

### Step 1: Identify the Reaction and Conditions
The reaction involves solid \( A \), gaseous \( B \), and liquid \( C \). We need to calculate the work done at a temperature of \( 227^\circ C \) and a pressure of 1 atm.

### Step 2: Convert Temperature to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T = 227 + 273 = 500 \, K \]

### Step 3: Determine \(\Delta N_g\)
\(\Delta N_g\) is the change in the number of moles of gas during the reaction. It is calculated as:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
In this reaction:
- Moles of gaseous products = 0 (since \( C \) is a liquid)
- Moles of gaseous reactants = 3 (from \( 3B(g) \))

Thus,
\[ \Delta N_g = 0 - 3 = -3 \]

### Step 4: Use the Work Done Formula
The work done in a closed vessel is given by the formula:
\[ W = -\Delta N_g \cdot R \cdot T \]
Where:
- \( R \) is the gas constant. In kcal, \( R = 1.98 \times 10^{-3} \, \text{kcal} \, K^{-1} \, mol^{-1} \)
- \( T = 500 \, K \)

Substituting the values:
\[ W = -(-3) \cdot (1.98 \times 10^{-3}) \cdot (500) \]

### Step 5: Calculate the Work Done
Now, we can calculate:
\[ W = 3 \cdot (1.98 \times 10^{-3}) \cdot (500) \]
Calculating this step by step:
1. \( 3 \cdot 1.98 = 5.94 \)
2. \( 5.94 \cdot 500 = 2970 \)

Thus,
\[ W = 2970 \times 10^{-3} \, \text{kcal} = 2.97 \, \text{kcal} \]

### Final Answer
The work done in the reaction is approximately:
\[ \boxed{3 \, \text{kcal}} \]