Question 6abf7

Nov 22, 2017

Explanation:

You know that for the following equilibrium

${\text{N"_ (2(g)) + 3"H"_ (2(g)) rightleftharpoons 2"NH}}_{3 \left(g\right)}$

you have

${K}_{c} = \left({\left[{\text{NH"_3]^2)/(["N"_2] * ["H}}_{2}\right]}^{3}\right) = 41$

Now, notice what happens to the equilibrium constant of the equilibrium if you divide all the coefficients by $2$.

$\frac{1}{2} {\text{N"_ (2(g)) + 3/2"H"_ (2(g)) rightleftharpoons "NH}}_{3 \left(g\right)}$

This time, you have

${K}_{{\text{c /2") = (["NH"_3])/(["N"_2]^(1/2) * ["H}}_{2}}^{\frac{3}{2}}$

This is, of course, equivalent to

K_ ("c /2") = (["NH"_3])/(sqrt(["N"_2]) * sqrt(["H"_2]^(3))

which, in turn, is what you get when you have

sqrt(K_ c) = sqrt( (["NH"_ 3]^2)/(["N"_ 2] * ["H"_ 2]^3)) = (["NH"_ 3])/(sqrt(["N"_ 2]) * sqrt(["H"_ 2]^(3))) = K_ ("c /2")#

You can thus say that

${K}_{\text{c /2}} = \sqrt{{K}_{c}} = \sqrt{41}$

For the reverse reaction, you have

${\text{NH"_ (3(g)) rightleftharpoons 1/2"N"_ (2(g)) + 3/2"H}}_{2 \left(g\right)}$

with

${K}_{\text{c /2")^' = (sqrt(["N"_ 2]) * sqrt(["H"_ 2]^(3)))/(["NH"_ 3]) = 1/((["NH"_ 3])/(sqrt(["N"_ 2]) * sqrt(["H"_ 2]^(3)))) = 1/K_ ("c /2}}$

Therefore, you can say that

${K}_{\text{c /2}}^{'} = \frac{1}{\sqrt{41}}$