For a concentration of is initially 0.150 M before it proceeds to equilibrium, the equilibrium concentration of B is mathematically given as
x=3.0*10^{-7}m
What is the equilibrium concentration of B?
Generally, the equation for the change in Gibbs free energy is mathematically given as
dG=-RTInK
Therefore
31.5*10*^{3}=-(8.314)(298)(ink)
K=3*10^(-6)
Eqution
A--->B
K=x/0.1-x
x=3.0*10^{-7}m
In conclusion, the equilibrium concentration
x=3.0*10^{-7}m
Read more about Concentration
https://brainly.com/question/16979235
