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

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Answer:

Rate law if;

a. [A] is zero order

Rate law = k [products]

b. [A] is first order

Rate law = k [A] [Products]

c. [A] is second order

Rats law = k[A]^2 [products]

Doubling the concentration of each case will results to:

A. In reaction 1, when the concentration of A is doubled, it has no effect on the reaction. This is so because the reaction is not dependent on the concentration of the reactant A.

B. In reaction 2, when concentration of A is doubled, rate of reaction doubles. There is an increase in the rate of the reaction by a product of 2. This increases the reaction and more products formed.

C. In reaction 3, when the concentration of A is doubled, the rate of thsii reaction is quadrupled that is increases by four.

Order of reaction is K , K[A] , K[A]²

Order based reaction:

Zero order reaction - Doubling the conc - Does nothing to the reaction rate - Rate = [A⁰] = K

Frist order reaction - Doubling the conc - Doubles the reaction rate - Rate = [A¹] = K[A]

Second order reaction - Doubling the conc - Quadruples the reaction rate -

Rate = Rate = K[A]²

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