For the reaction
`NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)`
in a closed flask, the equilibrium pressure is `P` atm. The standard free energy of the reaction would be:

Options

A. `-RT ln p`
B. `-RT (ln p-ln 2)`
C. `-2 RT ln p`
D. `-2 RT (ln p-ln 2)`

Explanation

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

\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

in a closed flask where the equilibrium pressure is \( P \) atm, we can follow these steps:

### Step 1: Write the expression for ΔG
We know the relationship between the Gibbs free energy change (ΔG) and the standard Gibbs free energy change (ΔG°) is given by:

\[ \Delta G = \Delta G^\circ + RT \ln K \]

where:
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin,
- \( K \) is the equilibrium constant.

### Step 2: Determine the equilibrium constant (Kp)
For the reaction, the equilibrium constant \( K_p \) in terms of partial pressures is given by:

\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{P_{\text{NH}_4\text{HS}}} \]

Since NH4HS is a solid, its activity is 1, and we only consider the gaseous products:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]

### Step 3: Relate the partial pressures to the total pressure
At equilibrium, the total pressure \( P \) is the sum of the partial pressures of NH3 and H2S. Since both gases are produced in a 1:1 ratio, we can express their partial pressures as:

\[ P_{\text{NH}_3} = \frac{P}{2} \quad \text{and} \quad P_{\text{H}_2\text{S}} = \frac{P}{2} \]

Thus, we can substitute these into the expression for \( K_p \):

\[ K_p = \left(\frac{P}{2}\right) \cdot \left(\frac{P}{2}\right) = \frac{P^2}{4} \]

### Step 4: Substitute Kp into the ΔG equation
Now we can substitute \( K_p \) back into the ΔG equation. Since at equilibrium \( \Delta G = 0 \):

\[ 0 = \Delta G^\circ + RT \ln K_p \]

This leads to:

\[ \Delta G^\circ = -RT \ln K_p \]

Substituting \( K_p \):

\[ \Delta G^\circ = -RT \ln \left(\frac{P^2}{4}\right) \]

### Step 5: Simplify the expression
Using properties of logarithms:

\[ \Delta G^\circ = -RT \left(\ln P^2 - \ln 4\right) \]

This can be further simplified:

\[ \Delta G^\circ = -RT (2 \ln P - \ln 4) \]

\[ \Delta G^\circ = -2RT \ln P + RT \ln 4 \]

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

\[ \Delta G^\circ = -2RT \ln P + RT \ln 4 \]