Question

I. $A+B\rightleftharpoons 2C+D$
II. $2C+D\rightleftharpoons A+B$
If the equilibrium constant for the reaction represented by equation I above is $4.0\times {10}^{-2}$. What is the value of the equilibrium constant for the reaction represented by equation II?

• A. $16\times {10}^{-4}$
• B. $4.0\times {10}^{-2}$
• C. $\frac{1}{4.0}$
• D. $\frac{1}{4.0\times {10}^{-2}}$
• E. $4.0\times {10}^2$

Answer 1

Answer: (D)

$\frac{1}{4.0\times {10}^{-2}}$

The given reaction is :-

$I$. $A+B\rightleftharpoons 2C+D$

$K_1=\frac{{\left[C\right]}^2\left[D\right]}{\left[A\right]\left[B\right]}=4\times {10}^{-2}\left(given\right)\to \left(I\right)$

Now,

$II$. $2C+D\rightleftharpoons A+B$

$K_2=\frac{\left[A\right]\left[B\right]}{{\left[C\right]}^2\left[D\right]}=\frac{1}{K_1}=\frac{1}{4\times {10}^{-2}}$