37.8 g N₂O₅ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K
2 N₂O_(5( g))⇌ 2 N₂O_(4( g))+O_(2( g))
The total pressure at equilibrium was found to be 18.65 bar.
Then, Kp= _________ × 10⁻² [nearest integer]
Assume N₂O₅ to behave ideally under these conditions.
Given: R=0.082 bar Lmol⁻¹ K⁻¹

Solution

Chemistry ma’am · 4 steps

Turn grams into pressure first, then let the rise in total pressure tell you how far the reaction went.

1. Starting pressure. Molar mass of N₂O₅ is 108 g mol⁻¹, so n = 37.8/108 = 0.35 mol. In 1 L at 500 K, P₀ = nRT/V = 0.35 × 0.082 × 500 = 14.35 bar.
2. Pressure balance. Let the O₂ formed be x bar. Then N₂O₄ is 2x and N₂O₅ left is 14.35 - 2x. The total is 14.35 + x = 18.65, so x = 4.3 bar.
3. Partial pressures. p_(N₂O₅) = 14.35 - 8.6 = 5.75 bar, p_(N₂O₄) = 8.6 bar and p_(O₂) = 4.3 bar.

   Kₚ = (p_(N₂O₄)² p_(O₂))/p_(N₂O₅)² = ((8.6)² × 4.3)/(5.75)² = 318.03/33.06 = 9.62

4. So. Kₚ = 9.62 = 961.9 × 10⁻², which rounds to 962.

The common slip: 643. Leaving out the squares from the stoichiometric coefficients gives Kₚ = 8.6 × 4.3 / 5.75 = 6.43, that is 643, not 962.

The idea to keep. At fixed volume and temperature, pressure is proportional to moles. Do the whole equilibrium table in bar, and use the change in total pressure to find x.

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