Question

If a reaction is third order with respect to [A], doubling the concentration of [A] will:

a) double the rate of the reaction

b) result in a fourfold increase in the rate of reaction

c) result in an eightfold increase in the rate of reaction

d) not affect the rate of the reaction

Answer 1

Doubling the concentration of [A] in a third-order reaction concerning [A] results in an eightfold increase in the rate of the reaction, as the rate is proportional to the cube of the concentration of [A]. So, the correct option is c) result in an eightfold increase in the rate of reaction.

Step-by-step explanation:

If a reaction is third order concerning the concentration of a reactant [A], then the rate of the reaction depends on the cube of the concentration of [A]. In this scenario, if the concentration of [A] is doubled, the new rate of the reaction would be proportional to (2[A])3. This means that the new rate is 23, or eight times the original rate since 2 cubed equals 8.

The rate of a reaction and its dependence on reactant concentrations is defined by the reaction's rate law. For a third-order reaction to a single reactant A, the rate law is expressed as rate = k[A]3, where k is the rate constant. Therefore, by doubling the concentration of [A], the rate law becomes rate = k(2[A])3 = 8k[A]3, which indicates an eightfold increase in the rate of the reaction. So in answer to the student's question, doubling the concentration of [A] in a third-order reaction to [A] will result in an eightfold increase in the rate of the reaction. The mentioned correct option in the final part is (c) results in an eightfold increase in the rate of the reaction.