Question

slope of the plot lnk versus 1/T for Arrhenius equation is
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Answer 1

Answer:

-Ea/R

Explanation:

K= Ae^(-Ea/RT)      ,where Ea: activation energy

                                           R: universal gas const.

so, lnK= lnA-Ea/RT

so, Slope = -Ea/R

Answer 2

Answer: slope of plot $ln k$ versus $\frac{1}{T}$ for Arrhenius equation is $-\frac{E_{a} }{R}$.

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Explanation:

The relationship between the rate constant of a reaction and temperature is given by the Arrhenius equation as:

$k = A e^{-E_{a}/RT }$    .....$(1)$

where $A$ is pre-exponential factor, $E_{a}$ is Activation energy and $R$ is gas constant.

Taking natural log of equation $(1)$

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

slope of plot $ln k$ versus $\frac{1}{T}$

on comparing this equation with straight line equation;

 $y = c + mx$

where $c$ is intercept, $m$ is slope of line.

$c = ln A$, $m = -\frac{E_{a} }{R}$

Therefore slope of plot $ln k$ versus $\frac{1}{T}$ is $-\frac{E_{a} }{R}$.

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