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

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

Answer 2

Answer: For the given chemical reaction, value of
`k_(eq)` is
`1.616* 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)=([NO]^2[Cl_2])/([NOCl]^2)`

Given:

`[NO]=1.2* 10^(-3)M`

`[Cl_2]=2.2* 10^(-3)M`

`[NOCl]=1.4* 10^(-2)M`

Putting values in above equation we get:

`k_(eq)=((1.2* 10^(-3))^2(2.2* 10^(-3)))/((1.4* 10^(-2))^2)`

`k_(eq)=1.616* 10^(-5)`