In collision theory according to arrhenius plot, intercept is equal to

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$\log_{10} K$

$In A$

$In K$

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

answer is B.

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Detailed Solution

Explaining the Arrhenius equation first. The formula is:
The formula is:

$k = {Ae}^{- \frac{E_{a}}{RT}}$
'k' represents the reaction's rate constant in the Arrhenius equation. The following terms are used to describe a chemical reaction: A is regarded as the pre-exponential component, T is temperature, R is the universal gas constant, and is the activation energy.

We need to locate the equation's intercept in this case. Now that we have the logarithm on both sides, we can write it as follows:

$lnK = \ln ({Ae}^{- \frac{E_{a}}{RT}})$

If we solve it further then it can be written as:

$lnK = lnA - \frac{E_{a}}{RT}$
$Now, we will compare this equation with another equation, i.e. y = mx + cIn this equation m represents the slope, and c represents the intercept of the plot.Thus, after the comparison of the equations, we get slope = -E a R, and the intercept is equal to ln A.$

We will now contrast this equation with another one, $y = mx + c$ .
In this equation, m stands for the slope and c for the plot's intercept.
The slope is therefore equal to $\frac{- Ea}{R}$ and the intercept is equal to $\ln A$ after comparing the two equations.