The integrated rate law for a first-order reaction is: [A]=[A]0e k t Now say we

Question

The integrated rate law for a first-order reaction is:
[A]=[A]0ekt

Now 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 (s1).

Part A

What is the half-life of a first-order reaction with a rate constant of 5.40×104  s1?

Express your answer with the appropriate units.

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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?

Express your answer with the appropriate units.

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Part C

A certain first-order reaction has a rate constant of 8.70×103 s1. How long will it take for the reactant concentration to drop to18 of its initial value?

Express your answer with the appropriate units.

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]0ekt

Now 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 (s1).

Part A

What is the half-life of a first-order reaction with a rate constant of 5.40×104  s1?

Express your answer with the appropriate units.

SubmitHintsMy AnswersGive UpReview Part

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?

Express your answer with the appropriate units.

SubmitHintsMy AnswersGive UpReview Part

Part C

A certain first-order reaction has a rate constant of 8.70×103 s1. How long will it take for the reactant concentration to drop to18 of its initial value?

Express your answer with the appropriate units.

Explanation / Answer

Part A :

What is the half-life of a first-order reaction with a rate constant of 5.40×104  s1?

Half-life, t = 0.693 / k

              = 0.693 / (5.40×104 )

              = 1283.3 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?

Given the time taken for the reactant concentration to drop to half of its initial value is 349 s

Thatmenas half-life is = 349 s

So rate constant , k = 0.693 / half-life

                             = 0.693 / 349 s-1

                             = 1.986x10-3 s-1

Part C

A certain first-order reaction has a rate constant of 8.70×103 s1. How long will it take for the reactant concentration to drop to18 of its initial value?

For a first order reaction rate constant , k = ( 2.303 /t )x log ( Mo / M)

Where

k = rate constant = 8.70x10-3 s-1

Mo = initial concentration

M= concentration left after time t = (1/18) Mo

t = time = ?

Plug the values we get

t = ( 2.303 /k )x log ( Mo / M)

= ( 2.303/(8.70x10-3)) x log 18

= 332.3 s

Therefore the time taken is 332.3 s

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