Consider this second-order reaction: (A B C and rate = k[A]2). What will happen to the reaction rate if the concentration of A i

Question

Basile [38]

3 years ago

12

Consider this second-order reaction: (A B C and rate = k[A]2). What will happen to the reaction rate if the concentration of A i

s doubled?

Chemistry

2 answers:

hammer [34] · Answer 1 · 2021-12-28T03:26:47+03:00

hammer [34]3 years ago

5 0

It will increase by a factor of four.

Usimov [2.4K] · Answer 2 · 2021-12-28T05:05:24+03:00

Answer:

The reaction rate will be multiplied by four.

Explanation:

The reaction rate is the measure of how fast a reaction is happening, and it can be calculated by how fast the reactants are disappearing, or how fast the products are being formed.

For a generic reaction:

A → B + C

The rate (r) is:

-r = k*[A]ⁿ

The minus signal refers to the disappearing of the reactant, k is the velocity constant of the reaction, and n is the reaction order. So, for a second-order reaction:

-r = k*[A]²

If the concentration of A is doubled: [A]' = 2[A]

-r' = k*(2[A])²

-r' = k*4*[A]

-r' = 4*k*[A]

-r' = 4*(-r)