Chemical Kinetics - Result Question 11
12. If a reaction follows the Arrhenius equation, the plot $\ln k v s$ $1 /(R T)$ gives straight line with a gradient $(-y)$ unit. The energy required to activate the reactant is
(2019 Main, 11 Jan I)
(a) $\dfrac{y}{R}$ unit
(b) $-y$ unit
(c) $y R$ unit
(d) $y$ unit
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Answer:
Correct Answer: 12. (d)
Solution:
- The temperature dependence of rate of a chemical reaction is expressed by Arrhenius equation as, $k=A e^{-E _{\alpha} / R T}$$\quad$ …….. (i)
where, $A=$ Arrhenius factor or frequency factor or pre-exponential factor
$R=$ Gas constant,$E _a=$ Activation energy
Taking $\log$ on both sides of the Eq. (i), the equation becomes $\ln k=\ln A-\dfrac{E _a}{R T}$
On comparing with equation of straight line $(y=m x+c)$, the nature of the plot of $ln \hspace {1mm} k \hspace {1mm} v s \dfrac{1}{R T}$ will be:
(i) Intercept $=C=\ln A$
(ii) Slope/gradient $=m=-E _a=-y \Rightarrow E _a=y$
So, the energy required to activate the reactant, (activation energy of the reaction, $E _a$ is = $y$ )
BETA
