Question

Two reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.37 2 C − ⇀ ↽ − D K 2 = 0.170 Calculate the value of the equilibrium constant for the reaction D − ⇀ ↽ − A + 2 B .

Answer 1

Answer: The value of $K_c$ for the net reaction is 13.94

Explanation:

The given chemical equations follows:

Equation 1:  $A+2B\xrightarrow[]{K_1} 2C$

Equation 2:  $2C\xrightarrow[]{K_2} D$

The net equation follows:

$D\xrightarrow[]{K_c} A+2B$

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times \frac{1}{K_2}$

We are given:

$K_1=2.37$

$K_2=0.170$

Putting values in above equation, we get:

$K_c=2.37\times \frac{1}{0.170}=13.94$

Hence, the value of $K_c$ for the net reaction is 13.94