The activation energy for a certain reaction is $334.4 k J m o l^{- 1}$ . How many times larger is the rate constant at $610$ K than the rate constant at $600$ K?

Two times

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Three times

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Four times

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None of these

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Solution

The correct option is B Three times
Let $K_{1} a n d K_{2}$ be rate constants.
$log (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$log (\frac{K_{2}}{K_{1}}) = (\frac{334.4 \times 10^{3}}{2.303 (8.316)}) [\frac{610 - 600}{600 \times 610}]$

$\frac{K_{2}}{K_{1}} = 3$ $∴ K_{2}$ is .greater by three times

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The activation energy for a certain reaction is 334.4kJmol−1. How many times larger is the rate constant at 610 K than the rate constant at 600 K?

The correct option is B Three timesLet K1andK2 be rate constants.log(K2K1)=Ea2.303R[T2−T1T1T2]log(K2K1)=(334.4×1032.303(8.316))[610−600600×610]K2K1=3 ∴K2 is .greater by three times

The activation energy for a certain reaction is $334.4 k J m o l^{- 1}$ . How many times larger is the rate constant at $610$ K than the rate constant at $600$ K?