Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.

Order	Integrated Rate Law	Graph	Slope
0	[A]=−kt+[A]0	[A] vs. t	−k
1	ln[A]=−kt+ln[A]0	ln[A] vs. t	−k
2	1[A]= kt+1[A]0	1[A] vs. t	k

Part A

The reactant concentration in a zero-order reaction was 5.00×10⁻²M after 200 s and 2.50×10⁻²M after 310 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.

k0th =

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.

Part C

The reactant concentration in a first-order reaction was 9.30×10⁻²M after 10.0 s and 1.50×10⁻³M after 100 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.

Part D

The reactant concentration in a second-order reaction was 0.580 M after 275 s and 8.00×10⁻²M after 840 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.

Answer 1

Part a

For zero order reaction

[At] = -kt + [Ao]

Compare it with y = mx + b

For y = [At], x = t

Slope of the equation m = - k

Rate constant k = - ([A2] - [A1]) / (t2 - t1)

= - (2.5x10^-2M - 5x10^-2M) / (310s - 200s)

= 2.27 x 10^-4 M/s

Part b

For zero order reaction

[At] = -kt + [Ao]

5 x 10^-2 M = - 2.27 x 10^-4 M/s x 200s + [A0]

5 x 10^-2 M = - 0.0454 M + [A0]

[A0] = 0.0954 = 9.54 x 10^-2 M

Part C

For first order reaction

ln[A]= −kt + ln[A]0

Rate constant k = - (ln[A2] - ln[A1]) / (t2 - t1)

= - (ln[9.30×10^−2] - ln[1.50×10−3]) / (10 - 100)

= - (-2.375 + 6.502) / (-90)

= 0.4585 s-1

Part d

For second order reaction

1/[A]= kt + 1/[A0]

Rate constant k = (1/[A2] - 1/[A1]) / (t2 - t1)

k = (1/[0.580] - 1/[8 x 10^-2]) / (275 - 840)

= (1.7241 - 12.5) / (-565)

= 0.01907 M-1s-1