In a second-order reaction (one that is second order in one reactant), cutting in half the concentration of that reactant will have what effect on the reaction rate?

A) The reaction rate will remain the same.
B) The reaction rate will double.
C) The reaction rate will decrease by a factor of two.
D) The reaction rate will increase by a factor of four.
E) The reaction rate will decrease by a factor of four.

A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.

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kobenhavn kobenhavn · Answer 2 · 2019-05-06T13:59:38+02:00

Answer: D) The reaction rate will increase by a factor of four.

Explanation:

[tex]A\rightarrow products[/tex]

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

given order =2 , thus rate law is:

[tex]R=[A]^2[/tex] (1)

On cutting the concentration of A to half

[tex]R'=[\frac{A}{2}]^2[/tex] (2)

Dividing 2 by 1

[tex]\frac{R'}{R}=\frac{1}{4}[/tex]

[tex]R'=4R[/tex]