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 temperature range between T1 and T2 , it is readily observed that (T2 T1)

Question

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 temperature range between T1 and T2 , it is readily observed that (T2 > T1) (A) KP=(1/KP') (B) KP < KP' (C) KP=KP' (D) KP>KP'

Tardigrade Admin · Accepted Answer

log (kp/k p') = (Δ H/2.303 R) ((1/T1) - (1/T2)) for exothermic reaction Δ H=-ve T2>T1 In exothermic reactions on increasing temperature value of equilibrium decreases. So, Kp>Kp'

Q. 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
$(T_{2} > T_{1})$

$K_{P} = \frac{1}{K _{P}^{^{'}}}$

$K_{P} < K_{P}^{^{'}}$

$K_{P} = K_{P}^{^{'}}$

$K_{P} > K_{P}^{^{'}}$

$l o g \frac{k _{p}}{k p ^{^{'}}} = \frac{Δ H}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$
for exothermic reaction
$Δ H = - v e$
$T_{2} > T_{1}$
In exothermic reactions on increasing temperature value of equilibrium decreases.
So, $K_{p} > K_{p}^{^{'}}$

Q. 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 temperature range between T1​ and T2​ , it is readily observed that (T2​>T1​)

KP​=KP′​1​

KP​<KP′​

KP​=KP′​

KP​>KP′​