Question

Consider the reaction. At equilibrium, the concentrations are as follows. [NOCl] = 1.4 ´ 10–2 M [NO] = 1.2 ´ 10–3 M [Cl2] = 2.2 ´ 10–3 M What is the value of Keq for the reaction expressed in scientific notation?

Answer 1

Answer: For the given chemical reaction, value of $k_{eq}$ is $1.616\times 10^{-5}$

Explanation: The chemical reaction according to the question is:

$2NOCl\rightarrow 2NO+Cl_2$

Equilibrium constant, $k_{eq}$ for the following reaction is given by:

$k_{eq}=\frac{[NO]^2[Cl_2]}{[NOCl]^2}$

Given:

$[NO]=1.2\times 10^{-3}M$

$[Cl_2]=2.2\times 10^{-3}M$

$[NOCl]=1.4\times 10^{-2}M$

Putting values in above equation we get:

$k_{eq}=\frac{(1.2\times 10^{-3})^2(2.2\times 10^{-3})}{(1.4\times 10^{-2})^2}$

$k_{eq}=1.616\times 10^{-5}$

Answer 2

2NOCl ⇄ 2NO + Cl₂

K = [NO]²[Cl₂]/[NOCl]
K = [0,0012]²[0,0022]/[0,014]²
K = 0,000000003/0,000196
K = 0,000016163