The activation energy for a hypothetical reaction reaction A $\to$ X is 12.49 kcal $m o l^{- 1}$ . If temperature is raised to 295 from 305 K, the reaction rate increased by 0.002 kcal $K^{- 1} m o l^{- 1}$ is almost equal to:

60 %

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50 %

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100 %

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Solution

The correct option is C $100 %$
$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g \frac{k_{1} + 0.002 k c a l / K / m o l}{k_{1}} = \frac{12490 c a l / m o l}{2.303 \times 1.99 c a l / m o l / K} (\frac{1}{295} - \frac{1}{305})$
$l o g \frac{k_{1} + 0.002 k c a l / K / m o l}{k_{1}} = 0.6941$
$\frac{k_{1} + 0.002 k c a l / K / m o l}{k_{1}} = 4.94$
$k_{1} + 0.002 k c a l / K / m o l = 4.94 k_{1}$
$0.002 k c a l / K / m o l = 3.94 k_{1}$
$k_{1} = 1972$
The increase in the reaction rate is 100% as the reaction rate is doubled.

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Similar questions

A ⟶

12.49 k c a l / m o l

295 K

305 K

350 K

400 K

[R = 8.314 J K^{- 1} m o l^{- 1}]

m o l^{- 1}

K^{- 1}

20^{0} C

35^{0} C

m o l^{- 1} K^{- 1}

5

350

400

= 8.314

K^{- 1} m o l^{- 1}

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