SATHEE: Chemical Kinetics - Result Question 19

Chemical Kinetics - Result Question 19

####22. For the elementary reaction, $M ⟶ N$ , the rate of disappearance of $M$ increases by a factor of 8 upon doubling the concentration of $M$ . The order of the reaction with respect to $M$ is

(a) 4

(b) 3

(d) 1

(2014 Adv.)

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Answer:

Correct Answer: 22. (a)

Solution:

For the elementary reaction, $M ⟶ N$

Rate law can be written as

$\begin{aligned} Rate \propto [M]^{n} & Rate = k [M]^{n} \end{aligned}$

When we double the concentration of $[M]$ , rate becomes 8 times, hence new rate law can be written as

$\begin{aligned} 8 \times Rate & = k [2 M]^{n} \frac{Rate}{8 \times Rate} & = \frac{k [M]^{n}}{k [2 M]^{n}} \Rightarrow \frac{1}{8} = \frac{1}{[2]^{n}} \Rightarrow [2]^{n} & = 8 = [2]^{3} \Rightarrow n = 3 \end{aligned}$

Chemical Kinetics - Result Question 19

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Chemical Kinetics - Result Question 19

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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