1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constants for the reaction Br2 leftharpoons 2Br at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is:

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Q. The equilibrium constants 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:

Equilibrium

Solution:

Chemical equilibrium constant for the reaction is $Br _{2} \rightleftharpoons 2 Br$ is $K _{ c }=\frac{[ Br ]^{2}}{\left[ Br _{2}\right]}$
The value of $K _{ c }$ at $500\, K$ is $1 \times 10^{-10}$. One increasing temperature $(700\, K ) .$ 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.