Question

The equilibrium constants for the reaction, $B{r}_2\rightleftharpoons 2Br$ 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 A endothermic

Chemical equilibrium constant for the reaction is $B{r}_2\rightleftharpoons 2Br$ is $K_c=\frac{[Br{]}^2}{\left[B{r}_2\right]}$

The value of $K_c$ at $500$ K is $1\times {10}^{-10}$. One increasing temperature $\left(700K\right)$. The value of $K_c$ is also increased, i.e., the concentration of the product is increased and obviously the increase in temperature will favour the forward reaction.

Thus, the forward reaction is endothermic.