Question

the rate constant k1 of a reaction is found to be double that of rate constant k2 of another reaction. How can be the relationship between corresponding activation energy of two reactions at same temperature can be represented?​

Answer 1

Answer:  $E_a_2=E_a_1+ln2\times RT$.

Explanation:

The effect of temperature on rate constant is given by Arrhenius equation:

$k=Ae^{\frac{-Ea}{RT}$

k= rate constant

$Ea$= activation energy

R= gas constant

T= temperature

where, A= pre exponential factor  

for Ist reaction : $k_1=Ae^{\frac{-Ea_1}{RT}$     (1)

for 2nd reaction : $k_2=Ae^{\frac{-Ea_2}{RT}$      (2)

Given $k_1=2k_2$

Dividing 1 by 2

$\frac{2k_2}{k_2}=\frac{Ae^{\frac{-Ea_1}{RT}}}{Ae^{\frac{-Ea_2}{RT}}}$  

$2={e^\frac{Ea_2-Ea_1}{RT}}$      

$ln2=\frac{Ea_2-Ea_1}{RT}$

$E_a_2=Ea_1+ln2\times RT$

Thus activation energy of two reactions at same temperature can be represented by $E_a_2=E_a_1+ln2\times RT$.

Answer 2

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