Activation energy (Ea) and rate constant (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by -

Question

Activation energy (Ea) and rate constant (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by - (A) ln (k2/k1) = - (Ea/R) ( (1/T2) - (1/T1) )  (B) ln (k2/k1) = - (Ea/R) ( (1/T2) + (1/T1) )  (C) ln (k2/k1) = (Ea/R) ( (1/T1) - (1/T2) )  (D) ln (k2/k1) = - (Ea/R) ( (1/T1) - (1/T2) )

Tardigrade Admin · Accepted Answer

Ace. to Arrheneis equation ℓ n (K2/K1) = (Ea/R) [ (1/T1) - (1/T2) ] option (1) and (3) are same

Q. Activation energy $(E_{a})$ and rate constant $(k_{1}$ and $k_{2})$ of a chemical reaction at two different temperatures $(T_{1}$ and $T_{2})$ are related by -

$l n \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$

$l n \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} + \frac{1}{T _{1}})$

$l n \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$l n \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Ace. to Arrheneis equation
$ℓ n \frac{K _{2}}{K _{1}} = \frac{E _{a}}{R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$
option (1) and (3) are same

Q. Activation energy (Ea​) and rate constant (k1​ and k2​) of a chemical reaction at two different temperatures (T1​ and T2​) are related by -

lnk1​k2​​=−REa​​(T2​1​−T1​1​)

lnk1​k2​​=−REa​​(T2​1​+T1​1​)

lnk1​k2​​=REa​​(T1​1​−T2​1​)

lnk1​k2​​=−REa​​(T1​1​−T2​1​)

Ace. to Arrheneis equation ℓnK1​K2​​=REa​​[T1​1​−T2​1​] option (1) and (3) are same