For the reaction N₂O_(4( g))⇄ 2NO_(2( g)),Kₚ=0.492 atm at 300 K.K_c for the reaction at same temperature is _________ × 10⁻².
(Given : R=0.082 L atm mol⁻¹ K⁻¹)

Solution

Chemistry ma’am · 4 steps

Gas moles go from 1 to 2, so Δ n = 1: divide Kₚ by RT once and you have K_c.

1. Link Kp and Kc. For a gas-phase equilibrium, Kₚ and K_c differ by a factor of RT raised to Δ n. Here Δ n is moles of gaseous products minus moles of gaseous reactants.

   Kₚ = K_c (RT)^(Δ n) ⇒ K_c = Kₚ/(RT)^(Δ n)

2. Find Δn. One mole of N₂O₄ gives two moles of NO₂, so Δ n = 2 - 1 = 1.
3. Put in numbers. RT = 0.082 × 300 = 24.6 L atm mol⁻¹. So K_c = 0.492/24.6 = 0.02 mol L⁻¹.
4. So. K_c = 0.02 = 2 × 10⁻², so the blank is 2.

The common slip: 0.02. Entering the full value 0.02 ignores the × 10⁻² already printed in the blank; the number to fill in is 2.

The idea to keep. Always find Δ n from gaseous moles first. If products have more gas moles, Kₚ is bigger than K_c, so you divide Kₚ by (RT)^(Δ n).

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