Value of K P for the equilibrium reaction
N₂O 4(g) ⇌ 2NO 2(g) at 288 K is 47.9. The K C for this reaction at same temperature is ____________. (Nearest integer)
(R = 0.083 L bar K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

One mole of gas becomes two, so Δ n_g = 1. Divide K_P by RT just once.

1. Link Kp and Kc. For a gas-phase equilibrium, the two constants are related through Δ n_g, the moles of gaseous products minus the moles of gaseous reactants.

   K_P = K_C (RT)^(Δ n_g)

2. Find Δn. Here 1 mol of N₂O₄ gives 2 mol of NO₂, so Δ n_g = 2 - 1 = 1. That gives K_C = K_P/RT.
3. Put in numbers. RT = 0.083 × 288 = 23.904. Then K_C = 47.9/23.904 = 2.004.
4. So. K_C ≈ 2.004, which rounds to 2.

The common slip: 1145. Multiplying K_P by RT instead of dividing gives about 1145, not 2.

The idea to keep. Always find Δ n_g from the balanced equation first. If gas moles increase, K_P is bigger than K_C, so you divide by (RT)^(Δ n_g) to get K_C.

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