Question

The equilibrium constant for the reaction
${Br}_2\rightleftharpoons 2Br$ at 500 K and 700 K are $1\times {10}^{-10}$ and $1\times {10}^{-5}$ respectively. The reaction is:-
(A) Endothermic (B) Exothermic (C) Fast (D) Slow

Answer 1

The correct option is C

We have,

$A_2\rightleftharpoons 2A$

$K_c=\frac{[A{]}^2}{A_2}$

$K_c$ at $500k$ and $700k$ are $1\times {10}^{-10}$and $1\times {10}^{-5}$.

This shows tha the formation of the A is favored with an increase in temperature.

So, reaction is not a slow reaction because with increase in temperature decomposition is favored