Question

Thank you to anyone who answers correctly ;w; (and if you tried)

What is the Kp for the decomposition of ammonia (if the reaction is at 500°C)?

2 NH3(g) N2(g) + 3 H2(g)​

Answer 1

Kp = 4.5 x 10⁻⁹

Step-by-step explanation:

from equilibrium of gas phase rxns, Kp = Kc(R·T)^-Δn

2NH₃(g) ⇄ 3H₂(g) + N₂(g)

2Vm 3Vm 1Vm

Δn = ∑Vm(products) - ∑Vm(reactants) = (3Vm + 1Vm) - 2Vm = 2Vm = +2

Kc = Kb = [H₂]³[N₂]/[NH₃]² = 1.8 x 10⁻⁵M²

R = 0.08206 L·atm/mol·K

T = 500°C = (500 + 273)K = 773K

Kp = Kb(R·T)^-Δn = 1.8 x 10⁻⁵M²[(0.08206L·atm/mol·K)(773K)]⁻⁽⁺²⁾

= (1.8 x 10⁻⁵mol²/L²)(63.43atm/mol)⁻²

= 4.5 x 10⁻⁹atm

Hope this helps. :-)