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.

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

explanation:

i think option no d is the correct answer