Question

The $K_c$ for the reaction $A+B\leftrightharpoons C$ is 4 and $K_c$ for the reaction $2A+D\leftrightharpoons C$ is 6, then

the value of $K_c$ for the reaction, $C+D\leftrightharpoons 2B$ is :

• A. 0.675
• B. 0.375
• C. 2.7
• D. 1.5

Answer 1

Answer: (B)

0.375

Following three reactions are given,

$A+B\leftrightharpoons C$ $K_1$ ......(1)

$2A+D\leftrightharpoons C$ $K_2$......(2)

$C+D\leftrightharpoons 2B$ $K_c$......(3)

To obtain reaction (3), reaction (1) is multiplied by 2 and reversed. It is then added to the reaction (2).

Hence, the relationship between the equilibrium constant values is $K_c=\frac{K_2}{K_1^2}$.

Substitute values in the above expression.

$K_c=\frac{6}{4^2}=0.375$.

Option B is the correct answer