Question

A student ran the following reaction in the laboratory at 691 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 3.69x10-2 moles of N2(g) and 6.11x10-2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of H2(8) to be 5.87*10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Ko

Answer 1

The balanced equilibrium reaction is as follows:

N₂(g) + 3H₂(g) $\rightleftharpoons$ 2NH₃(g)

The initial concentrations of N₂and H₂ are as follows:

[N₂] = 0.0369 mol / 1 L = 0.0369 M

[H₂] = 0.0611 mol / 1 L = 0.0611 M

The equilibrium concentration of H₂ is as follows:

[H₂]_eq = 0.0587 M

Draw an ICE chart as follows:

	N2(g)	3H2(g)	2NH3(g)
I(M)	0.0369	0.0611	0
C(M)	-x	-3x	+2x
E(M)	0.0369-x	0.0611-3x	2x

As per the ICE chart,

The equilibrium concentration of H₂ gas is as follows:

0.0611-3x = 0.0587

Solve the equation for "x" as follows:

x = 0.0008

Determine the equilibrium concentrations of N₂, and NH₃ as follows:

[N₂] = 0.0369-x = 0.0361 M

[NH₃] = 2x = 0.0016 M

Also given,

[H₂] = 0.0587 M

The K_c expression for the equilibrium is as follows:

K_c = [NH₃]² / [N₂] [H₂]³

Now,

K_c = [0.0016]² / [0.0361] [0.0587]³

K_c = 0.3506

Hence, K_c for the equilibrium reaction is 0.351. [3 S.F]