On increasing temperature, the equilibrium constant of exothermic and endothermic reactions, respectively

Question

On increasing temperature, the equilibrium constant of exothermic and endothermic reactions, respectively (A) increases and decreases (B) decreases and increases (C) increases and increases (D) decreases and decreases

Tardigrade Admin · Accepted Answer

For For A+B <=>[Kf][Kb] C+D, K=(kf/kb) For exothermic reaction, kf decreases with increase of temperature so that K decreases. For endothermic reaction, kf increases with increase of temperature so that K increases.

Q. On increasing temperature, the equilibrium constant of exothermic and endothermic reactions, respectively

increases and decreases

decreases and increases

increases and increases

decreases and decreases

For For $A + B K_{f} K_{b} C + D, K = \frac{k _{f}}{k _{b}}$

For exothermic reaction, $k_{f}$ decreases with increase of temperature so that $K$ decreases. For endothermic reaction, $k_{f}$ increases with increase of temperature so that $K$ increases.

Q. On increasing temperature, the equilibrium constant of exothermic and endothermic reactions, respectively

increases and decreases

decreases and increases

increases and increases

decreases and decreases

For For A+BKf​Kb​​C+D,K=kb​kf​​ For exothermic reaction, kf​ decreases with increase of temperature so that K decreases. For endothermic reaction, kf​ increases with increase of temperature so that K increases.

For For $A + B K_{f} K_{b} C + D, K = \frac{k _{f}}{k _{b}}$

For exothermic reaction, $k_{f}$ decreases with increase of temperature so that $K$ decreases. For endothermic reaction, $k_{f}$ increases with increase of temperature so that $K$ increases.