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The half-life of a first-order reaction is 12 min. if the initial concentration of reactant is 0.13 m it takes _______min for it to decrease to 0.080 m.

User POPI
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Final answer:

The half-life of a first-order reaction can be used to determine the time it takes for the reactant concentration to decrease to a certain value. In this case, it takes approximately 7.60 min for the reactant concentration to decrease from 0.13 M to 0.080 M.

Step-by-step explanation:

The half-life of a first-order reaction is related to the rate constant for the reaction and is given by the equation t₁/2 = 0.693/k. In this case, the half-life of the reaction is 12 min. To determine how long it takes for the reactant concentration to decrease to 0.080 M from an initial concentration of 0.13 M, we can use the half-life equation. Since the reactant concentration is decreasing, the reaction is first order. We can set up the equation as follows:

0.080 M = 0.13 M * (1/2)^n, where n is the number of half-lives.

Solving for n, we get:

n = log1/2(0.080/0.13)

n ≈ 2.303 * log1/2(0.080/0.13)

n ≈ 2.303 * -0.275

n ≈ -0.633

Since n represents the number of half-lives, it cannot be negative. So, the answer is that it takes approximately 0.633 half-lives for the reactant concentration to decrease to 0.080 M. Since each half-life is 12 min, we can calculate the time as:

Time = 0.633 * 12 min

Time ≈ 7.60 min

User MZB
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