SATHEE: Chemical Kinetics - Result Question 11

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

(d) $y$ unit

Show Answer

Answer:

Correct Answer: 12. (a)

Solution:

The temperature dependence of rate of a chemical reaction is expressed by Arrhenius equation as, $k = A e^{- E_{α} / R T}$

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 = $y$ )

Chemical Kinetics - Result Question 11

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Chemical Kinetics - Result Question 11

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu