If the kpseudo of the reaction of crystal violet with a substantially larger amount of naoh (0.1 m) is 0.5 s-1 what is k for the overall reaction? (show inverse units as unit^-1)

5 s-1

Explanation:

In order to explain the meaning of pseudo rate constant, consider the following;

Let's say we have a rate law, Rate = k[A][B]. k is a second order rate constant. If B is in a large excess over A, then it logically follows that [B] will not change significantly during the reaction compared to A so it can be approximated as constant. Then, [B] can be absorbed into the rate constant, so we write Rate = k'[A] where k'=k[B] and k' is now a pseudo-first-order rate constant.

The rate law is now first-order since it looks like only one reactant [A], but pseudo- since we had to make an approximation and rely on having excess B.

Let us apply what we just learnt to the question at hand. We can write

k'= k[OH^-]

From the question

k'= 0.5 s-1

[OH^-] = 0.1 M

Where ;

k'= pseudo rate constant

k= overall rate constant

[OH^-]= hydroxide ion concentration

Then;

0.5= k(0.1)

k= 0.5/0.1

k= 5 s-1

answer: your mm

explanation: