Let’s assume I want to double the rate of a reaction. The constraint is that I can only increase the temperature by 10 K. Calculate the minimum activation energy required if the temperature is 300K.

$54 k J m o l^{- 1}$

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$64 k J m o l^{- 1}$

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$94 k J m o l^{- 1}$

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$84 k J m o l^{- 1}$

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Solution

The correct option is A
$54 k J m o l^{- 1}$

Now you must be thinking that the rate constant is not given.
How do we proceed?

Let’s take some assumptions.

When $T_{1} = 300 K, k_{1} = k$
When $T_{2} = 300 + 10 = 310 K, k_{2} = 2 k$

Substituting these values in the equation

$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$

We get, log $\frac{2 k}{k} = \frac{E_{a}}{2.303 \times 8.314 \times 10^{- 3}} (\frac{310 - 300}{300 \times 310}) l o g 2 = \frac{E_{a} \times 10^{3}}{2.303 \times 8.314} \times (\frac{10}{300 \times 310}) E_{a} = 53.6 k J m o l^{- 1}$

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Similar questions

Let’s assume I want to double the rate of a reaction. The constraint is that I can only increase the temperature by 10 K. Calculate the minimum activation energy required if the temperature is 300K.

Q. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

Q. Which equation is true to calculate the energy of activation, if the rate of reaction is doubled by increasing temperature from

T_{1} K t o T_{2} K ?

The rate of a reaction doubles when its temperature changes form 300 K to 310 K. Activation energy of such a reaction will be $(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

$(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

(IIT-JEE-2013)

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