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For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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Final answer:

Using the formula for the fraction remaining in a first-order reaction after n half-lives, we find that reaching 99% completion takes twice as many half-lives as reaching 90% completion, demonstrating that the time required is twice as long.

Step-by-step explanation:

To show that the time required for 99% completion is twice the time required for the completion of 90% of a first-order reaction, we rely on the concept of half-lives.

For a first-order reaction, the time it takes for a substance to decay to half its initial amount is constant and is known as the half-life (t1/2).

In a first-order reaction, the relationship between the percent completion and the number of half-lives passed can be described by the formula (1/2)^n, where n is the number of half-lives. After n half-lives, the fraction remaining of the original is (1/2)^n.

For a reaction to be 90% complete, 10% of the reactant remains, which means the reaction has gone through approximately 3.32 half-lives, since (1/2)^3.32 equals approximately 0.1. To reach 99% completion, 1% of the reactant remains, which corresponds to approximately 6.64 half-lives, because (1/2)^6.64 equals approximately 0.01.

Therefore, 6.64 half-lives is twice 3.32 half-lives, indicating that the time required for 99% completion is indeed twice the time required for 90% completion.

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