© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Chemical Reaction
The reaction given is:
A2(g) ↔ 2 A(g)
This reaction shows the dissociation of A2 into two moles of A.
Step 2: Determine the Initial and Equilibrium Amounts
Let the initial concentration of A2 be 1 M (for simplicity). The problem states that A2 is 20% dissociated at equilibrium, meaning 20% of the initial A2 converts into A.
Initially, [A2] = 1 M and [A] = 0 M.
20% of A2 is dissociated. Therefore, the amount of A2 that dissociates is 0.2 M.
Hence, the amount of A2 left at equilibrium is 0.8 M.
Because 1 mole of A2 forms 2 moles of A, the concentration of A formed is 2 × 0.2 = 0.4 M.
Step 3: Calculate the Equilibrium Constant (K)
The equilibrium constant expression for the reaction A2(g) ↔ 2 A(g) is:
$K = \frac{[A]^2}{[A_2]}$
Substitute the equilibrium concentrations:
$[A_2] = 0.8\ \text{M}, \quad [A] = 0.4\ \text{M}$
$K = \frac{(0.4)^2}{0.8} = \frac{0.16}{0.8} = 0.2$
Step 4: Relate the Equilibrium Constant to Standard Free Energy Change
The relationship between the standard free energy change ($\Delta G^\circ$) and the equilibrium constant K is given by:
$\Delta G^\circ = - R T \ln K$
where R = 8.314 J K-1 mol-1 and T = 320 K.
Step 5: Compute $\Delta G^\circ$
Substitute the values into the formula:
$\Delta G^\circ = - (8.314\ \text{J K}^{-1}\text{mol}^{-1})(320\ \text{K})\ln(0.2)$
Now, ln(0.2) is the same as ln(2 × 10-1), or ln(2) + ln(10-1). Numerically, ln(0.2) is approximately -1.609. We can use the given logarithms, but the direct approximate value is often known to be about -1.609.
Carrying out the calculation gives:
$\Delta G^\circ \approx - (8.314 \times 320) \times (-1.609) \approx 4281\ \text{J mol}^{-1}$
Step 6: State the Final Answer
Thus, the standard free energy change at 320 K and 1 atm is approximately 4281 J mol-1. This matches the given correct answer.
