Question

At 35^oC, K = 1.6 x 10^-5 for the following reaction

2NOCl(g) <----> 2NO(g) +    Cl₂(g)

Calculate the concentration of all species at equilibrium for each of the following original mixtures.

1) 3.0 moles of pure NOCl in a 2.0 L flask

2) 2.0 moles of NOCl and 2 moles of NO in a 3 L flask

3) 3.0 moles of NOCl 1 mole of Cl₂ in a 2.0 L flask

Show work please

Answer 1

2 NOCl 2NO + Cl2

Keq = 1.6*10-5

1. Concentration of NOCl = 3 moles/2 L = 1.5M

Now,

	[NOCl]	[NO]	[Cl2]
Initial	1.5	0	0
Change	-2x	+2x	+x
Equilibrium	1.5-2x	2x	x

During change -2x represnts the loss of concentration for 2 moles of reactant while +2x and +x represnts the increase in concentration of NO and Cl2 respectively.

Now,

Keq = [NO]2 [Cl2] / [NOCl]2

1.6*10-5 = 4x3 / (1.5-2x)2

Since the change in concentration of NOCl is negligible-

1.6*10-5 = 4x3 / (1.5)2

1.6*10-5 * (1.5-2x)2 = 4x3

On solving this we get x= 0.0208M

Thus, at equilibrium

[NOCl] = 1.4584M

[NO] = 0.0416 M

[Cl2] =0.0208 M

2. Concentration of NOCl = 2 moles/3 L = 0.66 M

Concentration of NO = 0.66M

Now,

	[NOCl]	[NO]	[Cl2]
Initial	0.66	0.66	0
Change	-2x	+2x	+x
Equilibrium	0.66-2x	0.66+2x	x

During change -2x represnts the loss of concentration for 2 moles of reactant while +2x and +x represnts the increase in concentration of NO and Cl2 respectively.

Now,

Keq = [NO]2 [Cl2] / [NOCl]2

1.6*10-5 = x* (0.66+2x)2 / (0.66-2x)2

1.6*10-5 = x* (0.66+2x)2 / (0.66)2

On solving this we get x= 0.000016 M

Thus, at equilibrium

[NOCl] = 0.66 M

[NO] = 0.660032 M

[Cl2] =0.000016 M

3. Concentration of NOCl = 3 moles/2 L = 1.5M

Concentration of Cl2 = 0.5M

Now,

	[NOCl]	[NO]	[Cl2]
Initial	1.5	0	0.5
Change	-2x	+2x	+x
Equilibrium	1.5-2x	2x	0.5+x

During change -2x represnts the loss of concentration for 2 moles of reactant while +2x and +x represnts the increase in concentration of NO and Cl2 respectively.

Now,

Keq = [NO]2 [Cl2] / [NOCl]2

1.6*10-5 = 4x2 * (0.5+x) / (1.5-2x)2

1.6*10-5 = 4x2 * (0.5+x) / (1.5)2

3.6*10-5 = 4x3 + 2x2

1.8*10-5 = 2x3 + x2

On solving this we get x= 0.0043 M

Thus, at equilibrium

[NOCl] = 1.49 M

[NO] = 0.0086 M

[Cl2] =0.5043 M