Question

If the equilibrium constant for the reaction A 2B C 5/2 D has a value of 4.0, what is the value of the equilibrium constant for the reaction 2C 5D 2A 4B at the same temperature

Answer 1

Answer:

$K'=\frac{1}{16}$

Explanation:

Hello!

In this case, since the given reaction is:

$A+ 2B \rightleftharpoons C+ \frac{5}{2} D$

Whereas the equilibrium constant is:

$K=\frac{[C][D]^{5/2}}{[A][B]^2} =4.0$

However, the new target reaction reverses and doubles the initial reaction to obtain:

$2C+5D \rightleftharpoons 2A+4B$

Whereas the equilibrium constant is:

$K'=\frac{[A]^2[B]^4}{[C]^2[D]^5}$

Which suggest the following relationship between the equilibrium constants:

$K'=\frac{1}{K^2}$

So we plug in to obtain:

$K'=\frac{1}{4.0^2}\\\\K'=\frac{1}{16}$

Best regards!