For the reaction taking place at certain temperature `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)` if equilibrium pressure is 3 X bar then `Delta_(r)G^(@)` would be

Options

A. `-"RT ln 9"-"3RT ln X"`
B. RT ln 4 `-3RT` ln X
C. `-3RT` ln X
D. None of these

Explanation

To find the standard Gibbs free energy change (Δ_rG°) for the reaction:

\[ \text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g) \]

given that the equilibrium pressure is \(3X\) bar, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (K)
For the reaction, the equilibrium constant \(K\) can be expressed in terms of the partial pressures of the gaseous products:

\[ K = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} \]

Since the solid does not contribute to the equilibrium expression, we only consider the gases.

### Step 2: Determine the partial pressures at equilibrium
Let the partial pressure of \(\text{NH}_3\) be \(P\) and the partial pressure of \(\text{CO}_2\) be \(P\). Since there are 2 moles of \(\text{NH}_3\) produced, the total pressure at equilibrium is:

\[ P_{\text{total}} = 2P + P = 3P \]

Given that \(P_{\text{total}} = 3X\) bar, we can equate:

\[ 3P = 3X \implies P = X \]

### Step 3: Substitute the partial pressures into the equilibrium constant expression
Now substituting \(P\) into the expression for \(K\):

\[ K = \frac{(X)^2 \cdot (X)}{1} = X^3 \]

### Step 4: Use the Gibbs free energy equation
The standard Gibbs free energy change can be calculated using the equation:

\[ \Delta_rG° = -RT \ln K \]

Substituting \(K\):

\[ \Delta_rG° = -RT \ln(X^3) \]

### Step 5: Simplify the expression
Using the logarithmic property \(\ln(a^b) = b \ln(a)\):

\[ \Delta_rG° = -RT \cdot 3 \ln(X) = -3RT \ln(X) \]

### Final Result
Thus, the expression for the standard Gibbs free energy change is:

\[ \Delta_rG° = -3RT \ln(X) \]

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