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

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$E_{A} = 3 E_{B}$

$3 E_{A} = E_{B}$

$E_{B} = 2 E_{A}$

$E_{A} = 2 E_{B}$

answer is C.

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Detailed Solution

For a reaction A $\to$ P

$\ln \frac{k_{2}}{k_{1}} = \frac{E_{A}}{R} [\frac{1}{200} - \frac{1}{220}] = \ln 3$

$\ln \frac{k_{2}}{k_{1}} = \frac{E_{B}}{R} [\frac{1}{200} - \frac{1}{220}] = \ln 9$

$\frac{\ln 3}{\ln 3^{2}} = \frac{E_{A}}{E_{B}} \Rightarrow 2 E_{A} = E_{B}$

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