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Arrhenius Equation

On increasing...

Question

On increasing temperature from $200 K$ to $220 K$ rate of reaction $A$ increases by $3$ times and rate of reaction $B$ increases by $9$ times then correct relationship between activation energy of $A$ and $B$ is :-

A

E_{A}

=

3 E_{B}

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B

3 E_{A}

=

E_{B}

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C

E_{B}

=

2 E_{A}

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D

E_{A}

=

2 E_{B}

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Solution

The correct option is C $E_{B}$ = $2 E_{A}$
Reaction A
$I n (\frac{K_{2}}{K_{1}}) = - \frac{E_{1}}{R} [\frac{1}{220} - \frac{1}{200}]$
$\Rightarrow I n 3 = - \frac{E_{1}}{R} [\frac{1}{220} - \frac{1}{200}]$
$\Rightarrow E_{1} = - \frac{R I n 3}{\frac{1}{220} - \frac{1}{200}} ⟶ (1)$
Reaction B
$I n (\frac{K_{2}}{K_{1}}) = - \frac{E_{2}}{R} [\frac{1}{220} - \frac{1}{200}]$
$\Rightarrow I n 9 = - \frac{E_{2}}{R} [\frac{1}{220} - \frac{1}{200}]$
$\Rightarrow 2 I n 3 = - \frac{E_{2}}{R} [\frac{1}{220} - \frac{1}{200}]$
$\Rightarrow E_{2} = - \frac{2 R I n 3}{\frac{1}{220} - \frac{1}{200}} ⟶ (2)$
Considering equation :- $(1)$ and $(2)$ , $E_{2} = 2 E_{1}$

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