Question

For a given exothermic reaction, $K_p$ and $K_p'$ are the equilibrium constants at temperatures $T_1$ and $T_2$, respectively. Assuming that heat of reaction is constant in temperature range between $T_1$ and $T_2$, it is readily observed that :-

• A. $K_p > K'_p$
• B. $K_p < K'_p$
• C. $K_p = K'_p$
• D. $K_p = \frac{1}{ K'_p}$

Answer 1

Answer: (A)

For a given exothermic reaction, $K _{ P }$ and $K _{ P }^{\prime}$ are the equilibrium constants at temperatures $T _{1}$ and $T _{2}$ respectively. Assuming that heat of reaction is constant in temperatures range between $T _{1}$ and $T _{2}$, it is readily observation that $K _{ P }> K _{ P }^{\prime}$.

$\log \frac{ K _{2}}{ K _{1}}=\frac{\Delta H ^{\circ}}{ 2 . 3 0 3 R }\left(\frac{1}{ T _{1}}-\frac{ 1 }{ T _{2}}\right)$

$T _{2}> T _{1} \quad$ So, $K _{ P }> K _{ P }^{\prime}$ (exothermic reaction)

(assuming $T _{2}> T _{1}$, although it is not mentioned, which temperature is higher. If $T _{1}> T _{2}$

then $K _{ P }< K _{ P }^{\prime}$ then answer should be (2)).



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 Le-Chatalier principle.