`A(s)hArrB(g)+C(g) K_(P)=40atm^(2)`
`X(s)hArrB(g)+E(g)`
Above equilibrium is allowed to attain in a closed container and pressure of `B` was found to be `10` atm. Calculate standard Gibb's free energy change for `X(s)hArrB(g)+E(g) "at" 300K` (take`R=2 cal//k//mol`)

Options

A. `3.5 Kcal//mol`
B. `3 Kcal//mol`
C. `2.5 Kcal//mol`
D. `2 Kcal//mol`

Answer

To solve the problem step by step, we will go through the reactions and calculations systematically.

### Step 1: Write down the reactions and given information
We have two reactions:
1. \( A(s) \rightleftharpoons B(g) + C(g) \) with \( K_P = 40 \, \text{atm}^2 \)
2. \( X(s) \rightleftharpoons B(g) + E(g) \)

Given that the pressure of \( B \) at equilibrium is \( 10 \, \text{atm} \).

### Step 2: Analyze the first reaction
For the first reaction, the equilibrium constant expression is:
\[ K_P = P_B \cdot P_C \]
Given \( K_P = 40 \, \text{atm}^2 \), we can express this as:
\[ P_B \cdot P_C = 40 \]

Let \( P_B = X + Y \) (the total pressure of \( B \)) and \( P_C = X \) (the pressure of \( C \)). From the problem, we know:
\[ X + Y = 10 \, \text{atm} \]

### Step 3: Substitute into the equilibrium expression
Substituting \( P_B \) and \( P_C \) into the equilibrium expression:
\[ (10) \cdot X = 40 \]
Solving for \( X \):
\[ X = \frac{40}{10} = 4 \, \text{atm} \]

### Step 4: Calculate \( Y \)
Using the equation \( X + Y = 10 \):
\[ 4 + Y = 10 \implies Y = 10 - 4 = 6 \, \text{atm} \]

### Step 5: Analyze the second reaction
For the second reaction, the equilibrium constant expression is:
\[ K_{P2} = P_B \cdot P_E \]
Where \( P_B = 10 \, \text{atm} \) and \( P_E = Y = 6 \, \text{atm} \):
\[ K_{P2} = 10 \cdot 6 = 60 \, \text{atm}^2 \]

### Step 6: Calculate the standard Gibbs free energy change
The standard Gibbs free energy change (\( \Delta G \)) can be calculated using the formula:
\[ \Delta G = -RT \ln K_{P2} \]
Where:
- \( R = 2 \, \text{cal/K/mol} \)
- \( T = 300 \, \text{K} \)
- \( K_{P2} = 60 \)

Substituting the values:
\[ \Delta G = -2 \cdot 300 \cdot \ln(60) \]

### Step 7: Calculate \( \ln(60) \)
Using a calculator, we find:
\[ \ln(60) \approx 4.094 \]

### Step 8: Substitute \( \ln(60) \) back into the equation
\[ \Delta G = -2 \cdot 300 \cdot 4.094 \]
\[ \Delta G = -2456.8 \, \text{cal} \]

### Step 9: Convert to kilocalories
\[ \Delta G \approx -2.4568 \, \text{kcal} \approx -2.5 \, \text{kcal} \]

### Final Answer
The standard Gibbs free energy change for the reaction \( X(s) \rightleftharpoons B(g) + E(g) \) at 300 K is approximately:
\[ \Delta G \approx -2.5 \, \text{kcal/mol} \]