Question

If a reaction follows the Arrhenius equation, the plot \[\text{lnK}\]vs\[\dfrac{1}{\text{(RT)}}\] gives a straight line with a gradient \[\left( \text{-y} \right)\] unit. The energy required to activate the reactant is:
A) \[\text{y unit}\]
B) \[\text{-y unit}\]
C) \[\text{yR unit}\]
D)$\dfrac{\text{y}}{\text{R}}\text{unit}$

Answer 1

Hint: the Arrhenius equation is the relation of rate constant k with activation energy\[{{\text{E}}_{\text{a}}}\], gas constant R, and absolute temperature T.The logarithmic form of the equation \[\text{lnK= }\dfrac{\text{-}{{\text{E}}_{\text{a}}}}{\text{RT}}\text{+ln(A)}\]resembles the line equation$\text{y=mx+c}$. Rearrange the equation such that the plot of \[\text{lnK}\]against the \[\dfrac{1}{\text{RT}}\] gives \[\text{-}{{\text{E}}_{\text{a}}}\]as the slope.

Complete step by step answer:
Arrhenius relation proposed the empirical relation between the rate of constant K, activation energy \[{{\text{E}}_{\text{a}}}\] and absolute temperature T.The relation used for the calculating energy of activation:
 $\text{K=A}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}$
Here, \[{{\text{E}}_{\text{a}}}\] is called the Arrhenius activation energy, and A is called the Arrhenius pre-exponential factor. Since the exponential factor is dimensionless, the pre-exponential factor A has the same units as that of the rate constant K.
We know the Arrhenius equation. Let’s take a natural log of the equation (1), we have
$\text{lnK=ln(A}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}\text{)}$
Or $\text{lnK=ln(A)+ln(}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}\text{)}$
Using logarithmic rules.
Since we know that,$\text{ln(}{{\text{e}}^{\text{x}}}\text{)=x}$ we have
 \[\text{lnK= }\dfrac{\text{-}{{\text{E}}_{\text{a}}}}{\text{RT}}\text{+ln(A)}\]
\[\text{lnK= (}\dfrac{\text{-}{{\text{E}}_{\text{a}}}}{\text{R}}\text{)(}\dfrac{1}{\text{T}}\text{)+ln(A)}\]
This equation resembles with the equation of the line,
$\text{y=mx+c}$
To understand the conditions given in the problem let’s consider
Y equals to the \[\text{lnK}\]
X is equal to the \[\dfrac{1}{\text{RT}}\]
The slope m of the line is represented by \[\text{-}{{\text{E}}_{\text{a}}}\] and the constant ‘c’ is\[\text{ln(A)}\].
Thus it is evident that the plot \[\text{lnK}\] versus the reciprocal of the product of a gas constant and absolute temperature that is \[\dfrac{1}{\text{RT}}\] , gives a straight line with the slope equals to \[\text{-}{{\text{E}}_{\text{a}}}\]and y-intercept \[\text{ln(A)}\].
The plot of \[\text{lnK}\]versus the \[\dfrac{1}{\text{RT}}\]is as shown below:



We are given the gradient of the plot \[\text{lnK}\]against the \[\dfrac{1}{\text{RT}}\] as the $\text{(-y)}$ units.
Since from the general plot, we know that the gradient for the lot is\[\text{-}{{\text{E}}_{\text{a}}}\].
Therefore, for the given condition the activation energy can be calculated by equating the s gradient of the, we have
Activation energy, \[\text{-}{{\text{E}}_{\text{a}}}\text{=-y}\] units
Cancel out the negative signs from the left and right-hand side. We get,
\[{{\text{E}}_{\text{a}}}\text{= y}\]
Thus the activation energy required to activate the reactant which is \[{{\text{E}}_{\text{a}}}\]is equal to y units.

Hence, (A) is the correct option.

Note: The gradient of the line is calculated as,
The gradient of the line $\text{=}\dfrac{\text{Change in y-coordinates}}{\text{Change in x-coordinates}}\text{=}\dfrac{{{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{1}}}}{{{\text{x}}_{\text{2}}}\text{-}{{\text{x}}_{\text{1}}}}$
The gradient is also called the slope.