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

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.