Question #6abf7

1 Answer
Nov 22, 2017

Answer:

Yes, your answer is correct.

Explanation:

You know that for the following equilibrium

#"N"_ (2(g)) + 3"H"_ (2(g)) rightleftharpoons 2"NH"_ (3(g))#

you have

#K_c = (["NH"_3]^2)/(["N"_2] * ["H"_2]^3) = 41#

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

#1/2"N"_ (2(g)) + 3/2"H"_ (2(g)) rightleftharpoons "NH"_ (3(g))#

This time, you have

#K_ ("c /2") = (["NH"_3])/(["N"_2]^(1/2) * ["H"_2]^(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_ ("c /2") = sqrt(K_c)= sqrt(41)#

For the reverse reaction, you have

#"NH"_ (3(g)) rightleftharpoons 1/2"N"_ (2(g)) + 3/2"H"_ (2(g))#

with

#K_ ("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_ ("c /2")^' = 1/sqrt(41)#