Question

Consider a simple reaction in which a reactant AA forms products: A→productsA→products What is the rate law if the reaction is zero order with respect to AA? First order? Second order? For each case, explain how a doubling of the concentration of AA would affect the rate of reaction.

Answer 1

Answer :

The rate law expression for zero order reaction will be:

$Rate=k[A]^0$

The rate law expression for first order reaction will be:

$Rate=k[A]^1$

The rate law expression for second order reaction will be:

$Rate=k[A]^2$

Zero order reaction : There is no affect on the rate law.

First order reaction : The rate law becomes doubled.

Second order reaction : The rate law becomes quadrupled.

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 reaction is:

$A\rightarrow Products$

The rate law expression for zero order reaction will be:

$Rate=k[A]^0$

The rate law expression for first order reaction will be:

$Rate=k[A]^1$

The rate law expression for second order reaction will be:

$Rate=k[A]^2$

Now we have to determine that if doubling of the concentration of A then the rate of reaction will be:

As we know that the zero order reaction does not depend on the concentration of reactant. So, there is no affect on the rate law.

As we know that the first order reaction depend on the concentration of reactant. So, the rate law becomes doubled.

As we know that the second order reaction depend on the concentration of reactant. So, the rate law becomes quadrupled.