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In the following graph.
The slope of line $A B$ gives the information of the
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The slope of line $A B$ gives the information of the
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Verified Answer
The correct answer is:
value of $-\frac{E_a}{2.303 R}$
$\because \operatorname{In} K=\ln A-\frac{E_a}{R T}$
$\operatorname{In} K=\operatorname{In} A-\frac{E_a}{R} \cdot \frac{1}{T}$
Thus, $\log _{10} K=\log A-\frac{E_a}{2.303 R T}$
$\log K=\log A-\frac{E_a}{2.303} \cdot \frac{1}{T}$
$\operatorname{In} K=\operatorname{In} A-\frac{E_a}{R} \cdot \frac{1}{T}$
Thus, $\log _{10} K=\log A-\frac{E_a}{2.303 R T}$
$\log K=\log A-\frac{E_a}{2.303} \cdot \frac{1}{T}$
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