If `DeltaG^(@)(HI, g)=+"1.7kJ/mole"`. What is the equilibrium constnat at `25^(@)C` for `2HI(g)hArrH_(2)(g)+I_(2)(g)`?

Options

A. `24.0`
B. 3.9
C. `2.0`
D. `0.5`

Answer

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

given that the standard Gibbs free energy change \( \Delta G^\circ \) for HI is \( +1.7 \, \text{kJ/mol} \).

### Step 1: Convert \( \Delta G^\circ \) to Joules
Since the universal gas constant \( R \) is typically expressed in J/(mol·K), we need to convert \( \Delta G^\circ \) from kJ to J:

\[ \Delta G^\circ = 1.7 \, \text{kJ/mol} = 1700 \, \text{J/mol} \]

### Step 2: Use the relationship between \( \Delta G^\circ \) and \( K_p \)
The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation:

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

Where:
- \( R \) is the universal gas constant \( = 8.314 \, \text{J/(mol·K)} \)
- \( T \) is the temperature in Kelvin. At \( 25^\circ C \), \( T = 298 \, \text{K} \)

### Step 3: Rearrange the equation to solve for \( K_p \)
Rearranging the equation gives:

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

Substituting the values we have:

\[ \ln K_p = -\frac{1700 \, \text{J/mol}}{(8.314 \, \text{J/(mol·K)})(298 \, \text{K})} \]

### Step 4: Calculate the right-hand side
Calculating the denominator:

\[ RT = 8.314 \times 298 = 2477.572 \, \text{J/mol} \]

Now substituting this back into the equation:

\[ \ln K_p = -\frac{1700}{2477.572} \approx -0.685 \]

### Step 5: Solve for \( K_p \)
To find \( K_p \), we take the exponential of both sides:

\[ K_p = e^{-0.685} \]

Calculating this gives:

\[ K_p \approx 0.503 \]

### Final Answer
The equilibrium constant \( K_p \) at \( 25^\circ C \) for the reaction is approximately:

\[ K_p \approx 0.503 \]