The reactant concentration in a zero-order reaction was 9.00×102 M after 120 s a

Question

The reactant concentration in a zero-order reaction was 9.00×102 M after 120 s and 1.50×102 M after 300 s . What is the rate constant for this reaction?

Part B

What was the initial reactant concentration for the reaction described in Part A?

Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

The reactant concentration in a second-order reaction was 0.430 M after 285 s and 6.60×102M after 720 s . What is the rate constant for this reaction?

Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

Explanation / Answer

Part A

Given-

Reactant concentration [A]1 = 9.00 ×102 M

Reactant concentration [A]2 = 1.50 ×102 M

Time t1 = 120 s

Time t2 = 300 s

[A] is initial concentration

Integrated zero order rate law

[A] = [A] - kt

We have concentration at two different times, i.e.

[A] = [A] - kt

[A] = [A] - kt

Eliminate unknown by taking the difference

[A] - [A] = -kt + kt

[A] - [A] = k(t - t)

Hence:

k = ([A] - [A]) / (t - t)

= (9.00 ×10²M - 1.5×10²M) / (300s - 120s) = 4.166 ×10Ms¹

The rate constant for this reaction is 4.17 ×10M.s¹

Part B- Calculate initial concentration

[A] = [A] - kt

[A] = [A] + kt

       = 9.00 ×10²M + 4.17×10Ms¹120s

        =9.00 ×10²M + 5.00 x10²M

[A] = 13.00×10²M

PartC-

Given- Reactant concentration [A]1 = 0.430 M

Reactant concentration [A]2 = 6.60×102 M

Time t1 = 285 s

Time t2 = 720 s

Integrated second order rate law-

1/[A] = 1/[A] - kt

We have concentration at two different times, i.e.

1/[A] = 1/[A] + kt

1/[A] = 1/[A] + kt

1/[A] - 1/[A] = k(t - t)

k = (1/[A] - 1/[A] )/(t - t)

= ln( 1/6.60 x 10²M - 1/0.430M)/(720s - 285s)

= ln(12.82 )/435

= 2.55/435

= 5.86×103 M¹s¹

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