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  • Rate constant k varies with temperature as given by equation <img alt="\mathrm{\log k(min^{-1})=5-\frac{2000K}{T}}" src="/latex-image/?%5Cmathrm%7B%5Clog%20k%28min%5E%7B-1%7D%29%3D5-%5Cfr

Rate constant k varies with temperature as given by equation

\mathrm{\log k(min^{-1})=5-\frac{2000K}{T}}

Consider the following about this equation

(I) Pre-exponential factor is 105

(II) Ea is 9.212 kcal

(III) Variation of log k with (1/T) is linear

Option: 1

I,II,III


Option: 2

I,III


Option: 3

II,III


Option: 4

I,II


Answers (1)

best_answer

Arrhenius Equation -

\mathrm{k = A e^{-E_{a}/RT}}

k = Rate constant

Now,

\mathrm{\ln k=\ln A-\frac{E_{a}}{RT}}

\mathrm{\log k=\log A-\frac{ E_{a}}{2.303 R T}}

Given, equation :

\mathrm{\log k(min^{-1})=5-\frac{2000K}{T}}

Now, compare both equations

\mathrm{1. \log A = 5\Rightarrow A =10^5}

\\\mathrm{2.\ \frac{E_a}{2.303 R}=2000K}\\\\\mathrm{ \Rightarrow \mathrm{E}_{\mathrm{a}}=2.303 \times 2 \times 2000=9212 \mathrm{cal}=9.212 \mathrm{kcal} }

\\\mathrm{3.\ equation,\ \mathrm{\log k(min^{-1})=5-\frac{2000K}{T}}}\\\mathrm{\textup{it is form of y = mx +c, straight line or linear.}}

So, I, II, III all are correct.

Option 1 is correct.

Posted by

avinash.dongre

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