The gas phase reaction 2A(g)⇌ A₂(g) at 400 K has ΔGᵒ = + 25.2 kJ mol⁻¹.
The equilibrium constant K C for this reaction is ________ × 10⁻². (Round off to the Nearest Integer).
[Use : R = 8.3 J mol⁻¹ K⁻¹, ln 10 = 2.3 log₁₀ 2 = 0.30, 1 atm = 1 bar]
[antilog (-0.3) = 0.501]

Solution

Chemistry ma’am · 4 steps

The K you get from ΔG° for a gas reaction is Kp, so convert it to Kc before you round.

1. K from ΔG°. For a gas reaction the standard state is 1 bar, so Δ G° = -RT ln K gives Kₚ. Write the log in base 10 using ln 10 = 2.3.

   log Kₚ = -(Δ G°)/(2.3 RT) = -25200/(2.3 × 8.3 × 400) = -25200/7636 = -3.3

2. Take the antilog. Kₚ = 10^(-3.3) = 10⁻³ × 10^(-0.3) = 0.501 × 10⁻³ = 5.01 × 10⁻⁴.
3. Convert to Kc. Kₚ = K_c(RT)^(Δ n) with Δ n = 1 - 2 = -1, so K_c = Kₚ × RT. Use R = 0.083 L bar K⁻¹ mol⁻¹ here, which gives RT = 0.083 × 400 = 33.2.
4. So. K_c = 5.01 × 10⁻⁴ × 33.2 = 1.66 × 10⁻², which rounds to 2.

The common slip: 0. Treating the K from Δ G° as K_c gives 5.01 × 10⁻⁴ = 0.05 × 10⁻², which rounds to 0. The (RT)^(Δ n) conversion is still needed.

The idea to keep. For gases, Δ G° = -RT ln K gives Kₚ. To get K_c, use Kₚ = K_c(RT)^(Δ n), with R in L bar units in that step.

Answer from the NTA final answer key. Solution by PrepWiser.