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Standard XII

Chemistry

Activation Energy

The activatio...

Question

The activation energy of one of the reaction in a biochemical process is $532611 J m o l^{- 1}$ . When the temperature falls from $310 K$ to $300 K$ , the change in rate constant observed is $k_{300} = x \times 10^{- 3} k_{310}$ . The value of $x$ is
[Given: $I n 10 = 2.3, R = 8.3 J K^{- 1} {mol}^{- 1}$ ]

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Solution

$l o g (\frac{K_{310}}{K_{300}}) = \frac{532611}{8.3} (\frac{1}{300} - \frac{1}{310})$
$\frac{K_{310}}{K_{300}} = 10^{3} \Rightarrow K_{300} = 1 \times 10^{- 3} \times K_{310}$

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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 particular reaction doubles when the temperature changes from $300 K$ to $310 K$ . Calculate the energy of activation of the reaction. [Given $R = 8.314 J K^{- 1} {mol}^{- 1}$ ]

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The activation energy of one of the reaction in a biochemical process is 532611 J mol−1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300=x×10−3 k310. The value of x is [Given: In 10=2.3, R=8.3 JK−1 mol−1]

log(K310K300)=5326118.3(1300−1310) K310K300=103⇒K300=1×10−3×K310

$l o g (\frac{K_{310}}{K_{300}}) = \frac{532611}{8.3} (\frac{1}{300} - \frac{1}{310})$
$\frac{K_{310}}{K_{300}} = 10^{3} \Rightarrow K_{300} = 1 \times 10^{- 3} \times K_{310}$