Question

Given the information below:

4M(g) <--->3Z(g) KC=6.68×10−3

3Z(g)<--->4Q(g)+8X(g) KC=2.79×10−2

Calculate the value at Kc for the hypothetical reaction: Q(g)+2X(g)<--->M(g)

Answer 1

To get the value of kc for this hypothetical reaction . We have to perform some operation with the given another equations.

Adding :

(Reverse equation (I) x 1/4) + (Reverse equation (II) x 1/4)

when we reverse any reaction and multiplied by any factor , new equilibrium constant for that reaction is given by

K= (1/Kc)^n  

Where '''n'' is any multiplied factor

So, K for first reaction

K1= [ 1/0.00668]^1/4 = [149.70]^0.25 = 3.497

and K for another reaction

K2 = [1/0.0279]^0.25 = [35.84]^0.25 = 2.446

And when these two equation added to get our required equation

For added reaction Kc is given by

Kc = K1 x K2

so, Kc for hypothetical reaction is

Kc = 3.497 x 2.446 = 8.553

​​​​​