Question

The compound NOCl decomposes to nitric oxide and chlorine according to the following equation:

2 NOCl (g) → 2 NO (g) + Cl2 (g)

Suppose that 0.320 mol NOCl is placed in a 4.00-L flask at a given temperature. When equilibrium has been established, it is found that the concentration of NOCl is 0.0340 M. Calculate the equilibrium constant for this reaction.

Answer 1

Equilibrium constant is defined as the ratio of product of concentrations of products at equilibrium to product of concentrations of reactants at equilibrium raised to power their stoichiometric coefficients in the balanced chemical equation for the reaction.

Initial concentration of NOCl=number of moles/volume of flask

=0.320 mol/4.00 L=0.08 M

Equilibrium concentration of NOCl=0.0340 M

Change=0.0340 M -0.08 M=-0.046 M

(As per the balanced chemical equation given in question

For every 2 mol NOCl decomposing, 2 mol NO and 1 mol Cl₂ are formed

So for 0.046 M decrease in concentration of NOCl, 0.046 M NO and 0.046 M/2=0.023 M Cl₂ are formed)

Let us make an ICE chart for decomposition of NOCl to NO and Cl

Initial concentration (M) 0.08 0 0

Change -0.046 +0.046 +0.046/2

=+0.023

Equilibrium concentration(M)0.034 0.046 0.023

Equilibrium constant=[NO]²[Cl₂]/[NOCl]²

=(0.046 M)²(0.023 M)/(0.034 M)²

=4.21 x 10^-2

So the equilibrium constant for this reaction=4.21 x 10^-2