Calculate the value of the equilibrium constant for the following reaction.

Question

Given the following reactions:

3C <=> D KC​=5.52×10⁻⁴

D <=> 3A+3B KC​=6.73×10⁻²

 

Calculate the value of the equilibrium constant for the following reaction.

A+B <=> C

Answer 1

Given: 3C <==> D ... Kc = 5.52x10^-4

Given: D <===> 3A + 3B ... Kc = 6.73x10^-2

Calculated: 3C + D <==> D + 3A + 3B which reduces to ...

3C <==> 3A + 3B and this is not only 3x the requested equation, but also the reverse of the requested equation. Thus, we find the Kc for this requested equation as follows:

We find the Kc for 3C <==> 3A + 3B as follows:

(5.52x10^-4)(6.73x10^-2) = 3.71x10^-5

Now we take the cube root of this to account for the requested eq. being 1/3 of the calculated eq.

cube root 3.71x10^-5 = 3.34x10^-2

And finally, we take the reciprocal of this to account for the fact that the requested eq. is in reverse of the calculated eq.

1 / 3.34x10^-2 = 29.9