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  • The graph between the log K versus 1/T is a straight line. The slope of the line is.Option: 1 <img alt="\mathrm{-\frac{2.303 R}{E_a}}" src="https://e

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

Option: 1

\mathrm{-\frac{2.303 R}{E_a}}


Option: 2

\mathrm{-\frac{E_a}{2.303 R} }


Option: 3

\mathrm{\frac{2.303 R}{E_a}}


Option: 4

\mathrm{\frac{E_a}{2.303 R}}


Answers (1)

best_answer

\mathrm{K}=\mathrm{A}_0 \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}
Taking logarithm of both sides,
2.303 \log \mathrm{K}=2.303 \log \mathrm{A}_0-\mathrm{Ea} / \mathrm{RT}

\log \mathrm{K}=\log \mathrm{A_o}-\frac{E a}{2.303 R} \times \frac{1}{T}
Clearly, it is a straight line (y=mx+c) graph between log K vs 1/T
\mathrm{ \text { Slope }=-\frac{E a}{2.303 R} }

Posted by

seema garhwal

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