Question

consider the following reaction:

2NH₃ (g) <=> N₂ (g) + 3H₂ (g)

if 7.92 x 10^-4 moles of NH_3, 0.336 moles of N₂ and 0.287 moles of H₂ are at equilibrium in a 10.2 L container at 884 K, the value of the equilibrium, K_p, is _

Answer 1

At first, we need to calculate molar concentrations of gases in a vessel of known volume(10.2 L) at equilibrium

[N₂] = 0.336 mol/10.2L = 0.0328 M

[H₂] = 0.287 mol/10.2L = 0.0281 M

[NH₃] = 7.92 x 10^-4 mol/10.2 L = 0.00007 M

Equilibrium constant (K_c) = [product] / [reactant]

= [N₂][H₂]³ / [NH₃]²

= 0.0328 x ( 0.0281)² / (0.00007)² = 5285.55

Big number,it is impossible, so consider NH₃ as product and calculate K_c.

K_c = (0.00007)²/0.0328 x (0.0281)² = 1.17 x 10^-10

Now, we can calculate K_P from the equation as

Kp = K_c(0.0821T)ⁿ where t is absolute temperature and n = no.of moles of gaseous products - moles of gaseous reactants.

n = 2-4 = -2

K_p = 1.17 x 10^-10(0.0821L.atm.K x 881k)^-2

= 2.236 x 10^-14