The graph between the log K versus 1T is a straight line- The slope of the line is-

The correct option is A E_a2.303R K=A_oe^ E_aRT #10; #10;Taking log on both sides #10; #10; 2.303 log K= 2.303 log A_o E_aRT ...

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Introduction

The graph bet...

Question

The graph between the $l o g K$ versus $\frac{1}{T}$ is a straight line. The slope of the line is:

- \frac{2.303 R}{E_{a}}

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- \frac{E_{a}}{2.303 R}

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\frac{2.303 R}{E_{a}}

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\frac{E_{a}}{2.303 R}

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Solution

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Similar questions

log k

\frac{1}{T}

O X = 5, tan θ = (\frac{1}{2.303})

E_{a}

K_{c}

K_{p} .

K_{p}

\frac{1}{T}

O X = 5, tan Θ = (\frac{1}{2.303})

E_{a}

Column I	Column II
(A) Zero order	(i) $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
(B) First order	(ii) $Δ H > 0$
(C) Endothermic reaction	(iii) $k = \frac{2.303}{t} log \frac{[A]_{0}}{[A]}$
(D) Activation energy	(iv) $k = \frac{1}{t} ([A]_{0} - [A])$

E_{A}

E_{B}

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