રાસાયણિક પ્રકિયા ની સક્રિયકરણ ર્જા $(E_a)$ અને અચળ વેગ $(k_1$ અને $ k_2)$ બે જુદા જુદા તાપમાને પ્રક્રિયા $(T_1$ અને $T_2)$ કોના દ્વારા સંબંધિત છે

Question

A$\ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right)$
B$\ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)$
C$\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right)$
D $(b)$ અને $(c)$ બંને

AIPMT 2012, Medium

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Solution

d
Rate constant of a reaction is given by arrhenius equation Arrhenius equation

$K=A e^{-E a / R T}$

Where $K \rightarrow$ rate contant.

$\text { A } \rightarrow \text { Collision factor. }$

$E_{a} \rightarrow \text { Activation energy }$

$T \rightarrow$ Temperature on kelvin scale

Let $K_{1} \;and\; K_{2}$ be rate constant of a reaction of temperatures $T_{1}\;and\; T_{2}$

Applying Arrhenius equation.

$K_{1}=A e^{-E a / R T_{1}} \longrightarrow(1)$

$K_{2}=A e^{-E a / R T_{2}} \quad \longrightarrow(2)$

$\Rightarrow$ Taking natural log on both sides,

$\ln K_{1}=\ln A-\frac{E_{a}}{R T_{1}} \quad \longrightarrow \dots (1)$

$i n K_{2}=\ln A-\frac{E_{a}}{R T_{2}} \quad \longrightarrow \dots (2)$

Now substracting $(1)$ from $(2)$

$i n K_{2}-i n K_{1}=\frac{-E_{a}}{R T_{2}}+\frac{E_{a}}{R T_{1}}$

$\Rightarrow \ln \left(\frac{K_{2}}{K_{1}}\right)=\frac{E_{a}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]$

OR

$\Rightarrow \ln \left(\frac{K_{2}}{K_{1}}\right)=\frac{-E_{a}}{R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right]$

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