For a given exothermic reaction K_p and K'_p are the equilibrium constants at temperatures T_1 and T_2 respectively (T2.>T1 ). Assuming that heat of reaction is constant in temperatures range between T_1 and T_2, it is readily observed that:

Correct answer: (A) K_p>K'_p. log⁡K_2K_1=(∆H^o)/2.303R1T_1-1T_2 T_2>T_1 So K_p> K'_p (exothermic reaction) (Since T_2>T_1 )

Q66 / 180 · NEET 2014 · Chemistry

For a given exothermic reaction $K_{p}$ and ${K'}_{p}$ are the equilibrium constants at temperatures $T_{1}$ and $T_{2}$ respectively (T₂.>T₁ ). Assuming that heat of reaction is constant in temperatures range between $T_{1}$ and $T_{2},$ it is readily observed that:

Correct answer
A
$K_{p} > {K'}_{p}$
B
$K_{p} < {K'}_{p}$
C
$K_{p} = {K'}_{p}$
D
$K_{p} = \frac{1}{{K'}_{p}}$

Solution

$\log \frac{K_{2}}{K_{1}} = \frac{∆ H^{o}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$T_{2} > T_{1}$ So $K_{p} > {K'}_{p}$ (exothermic reaction)
(Since $T_{2} > T_{1}$ )

Hence, the correct answer is option: 'A'.

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

Solution

More Chemical Equilibrium Questions

For a given exothermic reaction $K_{p}$ and ${K'}_{p}$ are the equilibrium constants at temperatures $T_{1}$ and $T_{2}$ respectively (T₂.>T₁ ). Assuming that heat of reaction is constant in temperatures range between $T_{1}$ and $T_{2},$ it is readily observed that: