`X(s)hArrY(g)+2Z(g)`
`A(s)hArrY(g)+B(g)`
Consider both these equilibrium to be established simultaneously in a closed container.
At equilibrium, pressure of `Z "and" B` were found to be same and sum of pressure of `Z & B` is `10 atm` more than that of species `Y`. Find ratio of standard gibb's energy of two reactions.

Options

A. `20`
B. `2.303log_(10)20`
C. `log_(10)3sqrt144`
D. `(3+log12)/(2+log6)`

Explanation

To solve the problem step by step, we will analyze the given reactions and the conditions at equilibrium.

### Step 1: Write down the reactions
We have two reactions:
1. \( X(s) \rightleftharpoons Y(g) + 2Z(g) \)
2. \( A(s) \rightleftharpoons Y(g) + B(g) \)

### Step 2: Define the pressures at equilibrium
Let:
- \( P_Y \) = pressure of \( Y \)
- \( P_Z \) = pressure of \( Z \)
- \( P_B \) = pressure of \( B \)

From the problem statement, we know:
- \( P_Z = P_B \)
- \( P_Z + P_B = P_Y + 10 \) atm

### Step 3: Substitute the known relationships
Since \( P_Z = P_B \), we can denote \( P_Z = P_B = P \). Thus:
\[ P + P = P_Y + 10 \]
This simplifies to:
\[ 2P = P_Y + 10 \]
From this, we can express \( P_Y \) in terms of \( P \):
\[ P_Y = 2P - 10 \]

### Step 4: Express pressures in terms of \( P \)
Now, we can summarize the pressures:
- \( P_Y = 2P - 10 \)
- \( P_Z = P \)
- \( P_B = P \)

### Step 5: Calculate equilibrium constants \( K_{p1} \) and \( K_{p2} \)
For the first reaction:
\[ K_{p1} = \frac{P_Y \cdot P_Z^2}{1} = P_Y \cdot P_Z^2 \]
Substituting the values:
\[ K_{p1} = (2P - 10) \cdot P^2 \]

For the second reaction:
\[ K_{p2} = \frac{P_Y \cdot P_B}{1} = P_Y \cdot P_B \]
Substituting the values:
\[ K_{p2} = (2P - 10) \cdot P \]

### Step 6: Substitute \( P \) to find \( K_{p1} \) and \( K_{p2} \)
Assuming \( P = 10 \) atm (as a reasonable assumption based on the problem):
1. For \( K_{p1} \):
\[ K_{p1} = (2(10) - 10) \cdot (10)^2 = 10 \cdot 100 = 1000 \]
2. For \( K_{p2} \):
\[ K_{p2} = (2(10) - 10) \cdot 10 = 10 \cdot 10 = 100 \]

### Step 7: Calculate the Gibbs free energy change
The relationship between Gibbs free energy change and equilibrium constant is given by:
\[ \Delta G = -RT \ln K \]
Thus:
1. For the first reaction:
\[ \Delta G_1 = -RT \ln K_{p1} = -RT \ln 1000 \]
2. For the second reaction:
\[ \Delta G_2 = -RT \ln K_{p2} = -RT \ln 100 \]

### Step 8: Find the ratio of Gibbs free energies
The ratio of the Gibbs free energies is:
\[ \frac{\Delta G_1}{\Delta G_2} = \frac{-RT \ln 1000}{-RT \ln 100} = \frac{\ln 1000}{\ln 100} \]

### Step 9: Simplify the ratio
Using properties of logarithms:
\[ \frac{\ln 1000}{\ln 100} = \frac{\ln(10^3)}{\ln(10^2)} = \frac{3 \ln 10}{2 \ln 10} = \frac{3}{2} \]

### Final Answer
The ratio of the standard Gibbs energy of the two reactions is:
\[ \frac{\Delta G_1}{\Delta G_2} = \frac{3}{2} \]