Consider the reaction
N₂O₄(g)⇌ 2NO₂(g)
The temperature at which K C = 20.4 and K P = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K⁻¹ mol⁻¹]

Solution

Chemistry ma’am · 3 steps

Link K_P and K_C through (RT)^(Δ n). Here Δ n = 1, so T comes out in one division.

1. Find Δn. Count gaseous moles: 2 of NO₂ on the product side and 1 of N₂O₄ on the reactant side. So Δ n = 2 - 1 = 1.

   K_P = K_C (RT)^(Δ n) ⇒ T = K_P/(K_C R)

2. Put in numbers. The denominator is K_C R = 20.4 × 0.0831 = 1.69524. Then T = 600.1/1.69524 = 353.99 K.
3. So. To the nearest integer, the temperature is 354 K.

The idea to keep. Always get Δ n from gaseous moles only, products minus reactants. Then K_P = K_C(RT)^(Δ n). Use the R whose units match K_P, here L bar.

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