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

  • Option 1)

    \ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = - \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_2 }} - \frac{1} {{\text{T}_1 }}} \right)

  • Option 2)

    \ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = - \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_2 }} + \frac{1} {{\text{T}_1 }}} \right)

  • Option 3)

    \ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_1 }} - \frac{1} {{\text{T}_2 }}} \right)

  • Option 4)

    \ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = - \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_1 }} - \frac{1} {{\text{T}_2 }}} \right)

 

Answers (1)

-Ea/RT1

K1=A.e

and K2=A.e-Ea/RT2

lnK_{1}=lnA-\frac{E_{a}}{RT_{1}}\rightarrow 1\\ and\:lnK_{2}=lnA-\frac{Ea}{RT_{2}}\rightarrow 2\\ From equation 1 and 2\\ lnK_{2}-lnK_{1}=\frac{Ea}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})

ln(\frac{K_{2}}{K_{1}})=\frac{Ea}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})\\ ln(\frac{K_{2}}{K_{1}})=-\frac{Ea}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})

Both 1 and 3 are correct option 


Option 1)

\ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = - \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_2 }} - \frac{1} {{\text{T}_1 }}} \right)

This is correct option 

Option 2)

\ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = - \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_2 }} + \frac{1} {{\text{T}_1 }}} \right)

This is incorrect option 

Option 3)

\ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_1 }} - \frac{1} {{\text{T}_2 }}} \right)

This is correct option 

Option 4)

\ln \frac{{\text{k}_2 }} {{\text{k}_1 }} = - \frac{{\text{E}_\text{a} }} {\text{R}}\left( {\frac{1} {{\text{T}_1 }} - \frac{1} {{\text{T}_2 }}} \right)

This is incorrect option 

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