For a reaction that follows the general rate law, Rate = k[A][B]2, what will happen to the rate of reaction if the concentration

Question

Paladinen [302]

3 years ago

8

For a reaction that follows the general rate law, Rate = k[A][B]2, what will happen to the rate of reaction if the concentration

of B is increased by a factor of 4.00? The rate will

Chemistry

1 answer:

GuDViN [60] · Answer 1 · 2022-01-22T00:53:19+03:00

Answer : The rate is increased by 16 times.

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given rate law expression is:

$Rate=k[A][B]^2$

As per question, when the concentration of B increased by factor of 4.00 then the rate law expression will be:

$Rate=k[A][B]^2$

$Rate=k[A][4B]^2$

$Rate=16k[A][B]^2$

From this we conclude that, when the concentration of B increased by factor of 4.00 then the rate is increased by 16 times.

Thus, the rate is increased by 16 times.