Question

for reaction 2A + B---------2C ,k=x

Answer 1

Answer : (3) $\frac{1}{\sqrt{x}}$

Solution : Given reactions are,

1st reaction : $2A+B\rightleftharpoons 2C$                 $K_1=x$

2nd reaction : $C\rightleftharpoons A+\frac{1}{2}B$      $K_2=?$

The equilibrium constant expression for 1st reaction is,

$k_1=\frac{[C]^2}{[A]^2[B]}=x$         ..............(1)

By rearranging the equation(1), the value of [B] is

$[B]=\frac{[C]^2}{[A]^2(x)}$

The equilibrium constant expression for 2nd reaction is,

$K_2=\frac{[B]^{1/2}[A]}{[C]}$            ...............(2)

Now, Put the value of [B] in equation(2), we get

$K_2=\frac{[B]^{1/2}[A]}{[C]}=\frac{\left [\frac{[C]^2}{[A]^2(x)} \right ]^{1/2}[A]} {[C]}=\frac{1}{\sqrt{x}}$

Therefore, the equilibrium constant will be $\frac{1}{\sqrt{x}}$.

Answer 2

Answer:

1/√x

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