If `DeltaG^(@)[HI(g)=-1.7kJ]`, the equilibrium constant for the reaction `2HI(g)hArr H_(2)(g)+I_(2)(g)` at `25^(@)C` is

Options

A. 24
B. 2
C. 3.6
D. 0.5

Answer

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 \( \Delta G^\circ = -1.7 \, \text{kJ} \), we can follow these steps:

### Step 1: Convert \( \Delta G^\circ \) to Joules
Since \( \Delta G^\circ \) is given in kilojoules, we need to convert it to joules for consistency with the gas constant \( R \).

\[ \Delta G^\circ = -1.7 \, \text{kJ} = -1.7 \times 10^3 \, \text{J} = -1700 \, \text{J} \]

### 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 (for 25°C, \( T = 25 + 273 = 298 \, \text{K} \))

### 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:

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

### Step 4: Calculate \( \ln K_p \)
Calculating the right side:

\[ \ln K_p = \frac{1700}{8.314 \times 298} \]

Calculating the denominator:

\[ 8.314 \times 298 \approx 2477.572 \]

Now substituting back:

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

### Step 5: Convert \( \ln K_p \) to \( K_p \)
To find \( K_p \), we take the exponential:

\[ K_p = e^{0.685} \]

Calculating \( K_p \):

\[ K_p \approx 1.985 \]

### Step 6: Round to appropriate significant figures
Rounding \( K_p \) gives us approximately:

\[ K_p \approx 2 \]

### Final Answer
The equilibrium constant \( K_p \) for the reaction at \( 25^\circ C \) is approximately \( 2 \).

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