Question

In a given experiment 4.0 moles of pure NOCl was placed in an otherwise empty 2.0 L container. Equilibrium was established given the following reaction: 2NOCl(g) 2NO(g) + Cl2(g) K = 1.6 × 10-5 Complete the following table. Use numerical values in the Initial row and values containing the variable "x" in both the Change and Equilibrium rows. Let x = the amount of Cl2 needed to reach equilibrium. Units are understood to be M and do not need to be included in your answers. NOCl NO Cl2 Initial Change Equilibrium 2.0 - 2x Determine the equilibrium concentrations of each substance in the system. [NOCl] = M [NO] = M [Cl2] = M

Answer 1

we know that

concentration = moles / volume

so

intially

[NOCL] = 4 /2 = 2

now

consider the given reaction

2NOCl ---> 2N0 + cl2

now

usinng ICE table

at equilibrium

[NOCl ] = 2 - 2x

[N0] = 2x

[Cl2] = x

now

Kc = [Cl2] [N0]^2 / [NOCl]^2

so

Kc = [x] [2x]^2 / [2-2x]^2

Kc = 4x^3 / [2-2x]^2

Kc = 4x^3 / 4(1-x)^2

Kc = x3/ (1-x)^2

1.6 * 10-5 = x3 / (1-x)^2

solving

we get

x = 0.02478

so

at equilibrium

[NOCl] = 2 - 2x = 1.95044

[Cl2] = x = 0.02478 M

[NO] = 2x = 0.04956 M