Question

For the reaction given below at 25°C: \[ \text{A}_2 \rightleftharpoons 2\text{A} \] Find \( K_p \). Given \( \Delta G^\circ_A = -50.384 \, \text{kJ/mol} \) and \( \Delta G^\circ_{A_2} = -100 \, \text{kJ/mol} \).

• A. 0.43
• B. 0.23
• C. 0.31
• D. 0.53

Hint:

The equilibrium constant can be calculated from the standard Gibbs free energy change using the relation \( \Delta G^\circ = -RT \ln K \).

Answer 1

The Correct Option is C

Step 1: Relationship between \( K_p \) and \( \Delta G^\circ \).
The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation: \[ \Delta G^\circ = -RT \ln K \] where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. Step 2: Calculate \( \Delta G^\circ \) for the reaction.
We can calculate \( \Delta G^\circ_{\text{reaction}} \) by subtracting the \( \Delta G^\circ \) values of products and reactants: \[ \Delta G^\circ_{\text{reaction}} = \Delta G^\circ_{2A} - \Delta G^\circ_{A_2} = 2(-50.384) - (-100) = -100.768 + 100 = -0.768 \, \text{kJ/mol} \] Step 3: Calculate \( K_p \).
Now we can find \( K_p \) using the relation: \[ \Delta G^\circ = -RT \ln K_p \] Substituting the values, we get \( K_p = 0.31 \). Final Answer: \[ \boxed{0.31} \]