5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures.2NH₃(g) ⇌ 3H₂(g) + N₂(g) At equilibrium and a particular temperature, 1.00 mole of ammonia remains. Calculate K_c for the reaction.

Question

Quizplus · Accepted Answer

Kc = ([HS2][N2])/[NH3]^2

At equilibrium, there is 1 mole of NH3 remaining, therefore:

[NH3] = 1/5 = 0.2

[HS2] = [NS2] = 3x

Therefore, [HS2] + [NS2] = 6x

Substituting these values into the Kc expression:

Kc = (3x * 6x)/(0.2)^2 = 540x^2

Since the reaction is at equilibrium, the initial moles of NH3 will equal the sum of the moles of NS2 and HS2:

5.00 mol = 3x + 3x + 1.00 mol

Solving for x:

x = 0.834 mol

Substituting this value into the Kc expression:

Kc = 540(0.834)^2 = 303.5

Rounding to one decimal place:

Kc = 17.3

Therefore, the correct answer is B.

500 Mol of Ammonia Are Introduced into a 5