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 \( K \) for the reaction:

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

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

### Step 1: Convert Gibbs Free Energy Change to Joules
Since the Gibbs free energy is given in kilojoules, we need to convert it to joules for consistency with the gas constant \( R \).

\[ \Delta G^\circ = 1.7 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 1700 \, \text{J/mol} \]

### Step 2: Use the Relationship Between Gibbs Free Energy and Equilibrium Constant
The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation:

\[ \Delta G^\circ = -R T \ln K \]

Where:
- \( R = 8.314 \, \text{J/(mol K)} \) (the gas constant)
- \( T = 25^\circ C = 298 \, \text{K} \)

### Step 3: Rearrange the Equation to Solve for \( K \)
Rearranging the equation to solve for \( K \):

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

### Step 4: Substitute the Values
Now we can substitute the values into the equation:

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

Calculating the denominator:

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

Now substituting back into the equation:

\[ \ln K = -\frac{1700}{2477.572} \approx -0.686 \]

### Step 5: Calculate \( K \) from \( \ln K \)
To find \( K \), we take the exponential of both sides:

\[ K = e^{-0.686} \approx 0.503 \]

### Conclusion
Thus, the equilibrium constant \( K \) at \( 25^\circ C \) for the reaction is approximately:

\[ K \approx 0.5 \]