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) $\frac{y}{R}$ unit

(b) $-y$ unit

(c) $y R$ unit

(d) $y$ unit

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Answer:

Correct Answer: 12. (a)

Solution:

  1. The temperature dependence of rate of a chemical reaction is expressed by Arrhenius equation as, $k=A e^{-E _{\alpha} / R T}$

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 $\ln k=\ln A-\frac{E _a}{R T}$

On comparing with equation of straight line $(y=m x+c)$, the nature of the plot of $\ln k v s \frac{1}{R T}$ 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 = $y$ )



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