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The activation energy of one of the reactions in a biochemical process is $532611 mathrm{~J} mathrm{~mol}^{-1} ext {. }$ When the temperature falls from $310 mathrm{~K} ext { to } 300 mathrm{~K} ext {, }$ the change in rate constant observed is $mathrm{k}_{300}=x imes 10^{-3} mathrm{k}_{310}$ . The value of x is ___________.

$mathrm{ ext { [Given: } quad ln 10=2.3}$

$mathrm{left.mathrm{R}=8.3 mathrm{~J} mathrm{~K}^{-1} mathrm{~mol}^{-1} ight]}$

Option: 1
1

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

As we have learnt,

$\mathrm{K =A \exp \left(-\frac{E a}{R T}\right)} \\$

$\mathrm{\therefore K_{300} =A \exp \left(-\frac{E a}{300 R}\right) }\\$

$\mathrm{K_{310} =A \exp \left(\frac{-E a}{310 R}\right)}$

$\mathrm{\therefore \frac{K_{300}}{K_{310}} =\exp \left\{\frac{E a}{R}\left(\frac{1}{310}-\frac{1}{300}\right)\right\}} \\$

$\mathrm{=\exp \left\{\frac{-532611}{8.3} \times \frac{10}{310 \times 300}\right\}}$

$\mathrm{=\exp (-6.9)} \\$

$\mathrm{=e^{-6.9}} \\$

$\mathrm{=\left(e^{-2.3}\right)^{3}} \\$

$\mathrm{=10^{-3}}$

$\mathrm{\therefore K_{300}=10^{-3} \times K_{310}}$

$\mathrm{\therefore x=1}$

Hence answer is 1

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Pankaj

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