In an exothermic reaction at 10° C rise in temperature will:

Question

In an exothermic reaction at 10° C rise in temperature will: (A) decrease the value of equilibrium constant (B) double the value of Kc (C) do not produce any change in Kc  (D) produce some change in Kc

Tardigrade Admin · Accepted Answer

Equilibrium constant for a reversible reaction is given as:

K_{c} = \frac{k _{f}}{k _{b}}

In the exothermic reaction, the backward reaction increases more rapidly with increasing temperature than the forward reaction.
Therefore, the value of equilibrium constant is reduced.

Q. In an exothermic reaction at $1 0^{\circ} C$ rise in temperature will:

decrease the value of equilibrium constant

double the value of $K_{c}$

do not produce any change in $K_{c}$

produce some change in $K_{c}$

Equilibrium constant for a reversible reaction is given as: $K_{c} = \frac{k _{f}}{k _{b}}$
In the exothermic reaction, the backward reaction increases more rapidly with increasing temperature than the forward reaction.
Therefore, the value of equilibrium constant is reduced.

Q. In an exothermic reaction at 10∘C rise in temperature will:

decrease the value of equilibrium constant

double the value of Kc​

do not produce any change in Kc​

produce some change in Kc​

Equilibrium constant for a reversible reaction is given as: Kc​=kb​kf​​ In the exothermic reaction, the backward reaction increases more rapidly with increasing temperature than the forward reaction. Therefore, the value of equilibrium constant is reduced.

Q. In an exothermic reaction at $1 0^{\circ} C$ rise in temperature will:

double the value of $K_{c}$

do not produce any change in $K_{c}$

produce some change in $K_{c}$

Equilibrium constant for a reversible reaction is given as: $K_{c} = \frac{k _{f}}{k _{b}}$
In the exothermic reaction, the backward reaction increases more rapidly with increasing temperature than the forward reaction.
Therefore, the value of equilibrium constant is reduced.