Activation energy is given by the formula

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Question

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.Taking the logarithm on both sides at two different temperatures $T_1$ and $T_2$ with rate c

Vedclass · Accepted Answer

$\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$

Vedclass · Answer

(A) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.Taking the logarithm on both sides at two different temperatures $T_1$ and $T_2$ with rate constants $K_1$ and $K_2$ respectively,we get:$\log K_1 = \log A - \frac{E_a}{2.303RT_1}$$\log K_2 = \log A - \frac{E_a}{2.303RT_2}$Subtracting the two equations:$\log K_2 - \log K_1 = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$$\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Thus,option $A$ is the correct formula.

$Step$	$Rate \ constant$	$Activation \ energy$
$Step-1$	$k_1$	$E_{a1} = 180 \ kJ/mol$
$Step-2$	$k_2$	$E_{a2} = 80 \ kJ/mol$
$Step-3$	$k_3$	$E_{a3} = 50 \ kJ/mol$

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