# Activation energy (Ea) and rate constants (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2 ) related by : Option 1) Option 2) Option 3) Option 4)

-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)

This is correct option

Option 2)

This is incorrect option

Option 3)

This is correct option

Option 4)

This is incorrect option

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