Question

The answers are below, I just want to know how they are found.

1.	Reactant A decomposes into product B according to the equation A(g) ® B(g), for which Kc = 0.156 at 400ºC. 0.550 mol of A was injected into a 2.00 L reaction vessel at 400ºC. Calculate the concentration of A at equilibrium.
	D)	0.238 M
2.	For the nitrogen fixation reaction 3H2(g) + N2(g) ® 2NH3(g), Kc = 6.0 ´ 10–2 at 500°C. If 0.250 M H2 and 0.050 M NH3 are present at equilibrium, what is the equilibrium concentration of N2?
	C)	2.7 M

Answer 1

Kc = Concentration of products at equilibrium /  Concentration of reactants at equilibrium

1) A ------------> B Kc = 0.156

Initial [A]= moles / volume of vessel = 0.55 mol / 2 L = 0.275 M

A ------------> B

Initial 0.275 M 0

at equilibrium 0.275 - X X

Kc = [B]/[A]

0.156 = X / 0.275 -X

0.0429 -0.156 X = X

X = 0.0429 / 1.156 = 0.037 M

Hence, Concentration of A at equilibrium = 0.275 -X = 0.275 - 0.037 = 0.238 M

[A]_eq = 0.238 M

2) 3H2(g) + N2(g) -------------> 2NH3(g) Kc= 6.0 x 10^–2 =0.06

Given that 0.250 M H2 and 0.050 M NH3 are present at equilibrium.

Kc = [NH₃]² / [H₂]³ [N2]

0.06 = [0.05]² / [0.25]³ [N2]

[N2] = 2.7 M

Therefore, equilibrium concentration of N2 = 2.7 M