Question

For a given exothermic reaction $Kp$ and $Kp'$ are equilibrium constants at temperatures ${T_1}$ and ${T_2}$ respectively. Assuming that the heat of the reaction is constant in temperature range between ${T_1}$ and ${T_2}$ it is readily observed that:
A.$Kp = Kp'$
B.$Kp = \dfrac{1}{{Kp'}}$
C.$Kp > Kp'$
D.$Kp < Kp'$

Answer 1

Hint: An exothermic reaction is a reaction in which energy is released in the form of heat or light and the total energy of the products is less than the total energy of the reactants in the reaction. For an exothermic reaction, the change is enthalpy is negative. Equilibrium constant only depends on the change in temperature and according to the Le Chatalier principle, when temperature is increased for a reaction, the reaction shifts in the reverse direction and the value of equilibrium constant decreases.
Formula used:
$\log \dfrac{{Kp}}{{Kp'}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$


Complete step by step answer:
Exothermic reactions are reactions which release energy when they take place in the form of heat or light. In this reaction the total energy of the products is less than the total energy of the reactants. Some examples of an exothermic reaction are rusting of iron, combustion reaction, neutralization reaction and others. The equilibrium constant of a reaction is the ratio of the reactants to the products in a reaction at a given temperature in the case of dynamic chemical equilibrium. The rate of forward reaction becomes equal to the rate of reverse reaction.
It is given in the above question that the reaction is exothermic
$\therefore \Delta H = - ve$
$Kp$ is the equilibrium constant at temperature ${T_1}$
$Kp'$ Is the equilibrium constant at temperature ${T_2}$
According to the formula,
$\log \dfrac{{Kp}}{{Kp'}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$
We know that for an exothermic reaction heat is evolved and therefore,
We assume that ${T_2} > {T_1}$
$\therefore \log \dfrac{{Kp'}}{{Kp}} = - ve$
$ \Rightarrow \log Kp' - \log Kp = - ve$
According to the above equation we can conclude that $Kp > Kp'$
Also, when in an exothermic reaction temperature is increased, the reaction shifts in the reverse direction and the value of equilibrium constant decreases.

So, the correct option is (C) $Kp > Kp'$.

Note:
 The equilibrium constant of a reaction is useful in determining whether the concentration of the reactant is more than the concentration of the product or not at equilibrium. Equilibrium constants are independent of the change in the concentration of the reactant or the product in a chemical reaction; it only depends on the change in temperature.