The rate of a reaction A ---> B is expressed by the rate expression rate=k(A)2. If the molar concentration of A is tripled, the rate of the reaction should increase by a factor of some many?
32 = 9
To determine how the rate of the reaction changes when the molar concentration of A is tripled, we can use the rate expression given: rate = k(A)^2.
Let's consider the initial rate of the reaction as R1 when the concentration of A is given as [A]1, and the new rate of the reaction as R2 when the concentration of A is tripled and becomes [A]2.
According to the rate expression, we have:
R1 = k([A]1)^2
R2 = k([A]2)^2
We want to find the factor by which the rate changes, so we can divide R2 by R1:
R2 / R1 = (k([A]2)^2) / (k([A]1)^2)
Since k appears in both the numerator and denominator, it cancels out:
R2 / R1 = ([A]2)^2 / ([A]1)^2
Now let's substitute the values: when the molar concentration of A is tripled, [A]2 = 3[A]1. Plugging this in:
R2 / R1 = (3[A]1)^2 / ([A]1)^2
R2 / R1 = 9[A]1^2 / [A]1^2
Since [A]1^2 appears in both the numerator and denominator, it cancels out:
R2 / R1 = 9
Therefore, the rate of the reaction should increase by a factor of 9 when the molar concentration of A is tripled.