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  • For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2, r

For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2, respectively. Assuming that heat of the reaction is constant in the temperature range between T1 and T2, it is readily observed that :

Option: 1

Kp > K'p


Option: 2

Kp < K'p


Option: 3

Kp =K'p


Option: 4

Kp=1/K'p


Answers (1)

best_answer

Let's assume that T_{2}> T_{1}

We know that log \frac{K^{1}_{p}}{K_{p}} =\frac{\Delta H}{2.303R} \left [ \frac{1}{T_{1}}-\frac{1}{T_{2}} \right ]

 

For a given exothermic reaction, KP? and KP′? are the equilibrium constants at temperatures T1? and T2? respectively. Assuming that heat of reaction is constant in temperatures range between T1? and T2?, it is readily observed that Kp? > Kp′?

\log \frac{K_{2}}{K_{1}}=\frac{\Delta H^{o}}{2.303 R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)

T2? > T1?      So, KP ?> KP′? (exothermic reaction)

(assuming T2? > T1?, although it is not mentioned, which temperature is higher. If T1 ?> T2? then Kp ?< Kp′?).
Thus for an exothermic reaction, when the temperature is increased, the equilibrium will shift in the reverse direction and the value of the equilibrium constant will decrease. This is in accordance with the Le-Chatalier principle.

The Correct answer is option 1

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