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

Engineering

Chemistry

Order of Reaction and Law of Mass Action

Methods of Determination of Order of Reaction

Question

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

rate of Reaction = K[M]^x

$\frac{r_{1}}{r_{2}} = \frac{K {[M]}^{x}}{K {[2 M]}^{x}}$

$\frac{1}{8} = \frac{1}{2^{x}}$

x = 3

Note : Here either this reaction is not elementary or if it is elementary then not balanced.

For an elementary reaction M → N, the rate of disappearance of M increases by a factor of 8 when the concentration of M is doubled. We need to find the order of the reaction with respect to M.

The rate law for the reaction is given by:

r = k [M]^{n}

where r is the rate, k is the rate constant, [M] is the concentration of M, and n is the order of the reaction with respect to M.

Step 1: Let the initial concentration be [M]. The initial rate is:

r_{1} = k {[M]}^{n}

Step 2: When the concentration is doubled, new concentration is 2[M]. The new rate is:

r_{2} = k {(2 [M])}^{n} = k 2^{n} {[M]}^{n}

Step 3: Given that the rate increases by a factor of 8, so:

\frac{r_{2}}{r_{1}} = 8

Substitute the expressions:

\frac{k 2^{n} {[M]}^{n}}{k {[M]}^{n}} = 2^{n} = 8

Step 4: Solve for n:

2^{n} = 8 = 2^{3}

Therefore, n = 3.

Final answer: The order of the reaction with respect to M is 3.

[A] (mol L^–1)	[B] (mol L^–1)	Initial Rate (mol L^–1S^–1)
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

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