Question

The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached.The integrated rate law for a first-order reaction is:[A]=[A]0e−ktNow say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute [A]02 for [A] and rearrange the equation to:t1/2=0.693k This equation calculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life.Half-life equation for first-order reactions:t1/2=0.693k where t1/2 is the half-life in seconds (s), and kis the rate constant in inverse seconds (s−1).Part AWhat is the half-life of a first-order reaction with a rate constant of 5.40×10−4 s−1?Express your answer with the appropriate units.SubmitHintsMy AnswersGive UpReview PartPart BWhat is the rate constant of a first-order reaction that takes 349 seconds for the reactant concentration to drop to half of its initial value?Express your answer with the appropriate units.SubmitHintsMy AnswersGive UpReview PartPart CA certain first-order reaction has a rate constant of 8.70×10−3 s−1. How long will it take for the reactant concentration to drop to18 of its initial value?Express your answer with the appropriate units.

Answer 1

Answer:

A) Half-life = 1,283 s = 21.38 minutes

B) Rate constant = (1.99 × 10⁻³) s⁻¹

C) The time it will take for this reactant concentration to drop to (1/8) of its initial value

= 239 s = 3.98 minutes

Explanation:

The question already explained how the half-life of a first order reaction (T½)

is related to its rate constant (k)

T½ = (0.693/k)

Part A

What is the half-life of a first-order reaction with a rate constant of (5.40×10⁻⁴) s⁻¹?

T½ = (0.693/k)

T½ = ?

k = (5.40×10⁻⁴) s⁻¹

T½ = (0.693) ÷ (5.40×10⁻⁴) = 1,283.33 s = 1283 s

Part B

What is the rate constant of a first-order reaction that takes 349 seconds for the reactant concentration to drop to half of its initial value?

T½ = (0.693/k)

T½ = 349 s

k = ?

349 = (0.693/k)

k = (0.693/349) = 0.0019856734

= (1.99 × 10⁻³) s⁻¹

Part C

A certain first-order reaction has a rate constant of 8.70×10−3 s−1. How long will it take for the reactant concentration to drop to (1/8) of its initial value

T½ = (0.693/k)

T½ = ?

k = (8.70×10⁻³) s⁻¹

T½ = (0.693) ÷ (8.70×10⁻³) = 79.655 s

- The half life is the time it takes the concentration of the reactant to reach half of its initial value.

- In two Half-Lives, the concentration of the reactant drops to one-quarter (1/4) of its initial value.

- In three Half-Lives, the concentration of the reactant drops to one-eight (1/8) of its initial value.

Hence, the time it will take for this reactant concentration to drop to (1/8) of its initial value

= 3 × T½ = 3 × 79.655 = 238.97 s = 239 s

Hope this Helps!!!!