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 7.00×10⁻²M after 135 s and 2.50×10⁻²M after 315 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

Part B

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

Express your answer with the appropriate units.

Part C

The reactant concentration in a first-order reaction was 9.40×10⁻²M after 50.0 s and 1.50×10⁻³M after 75.0 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

Part D

The reactant concentration in a second-order reaction was 0.250 M after 170 s and 2.60×10⁻²M after 835 s . What is the rate constant for this reaction?

Express your answer with the appropriate units. Include an asterisk to indicate a compound unit with mulitplication, for example write a Newton-meter as N*m.t B

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

Express your answer with the appropriate units.

Answer 1

Part A.

For a zero order reaction,

[A] = [A]o - kt

with,

[A] at 135 s = 7 x 10^-2 M

[A] at 315 s = 2.50 x 10^-2 M

So the rate constant k for the reaction becomes,

k = -[2.5 x 10^-2 - 7 x 10^-2)/(315 - 135)] = 2.5 x 10^-4 M/s

Part B.

Initial concentration [A]o

[A]o = 7 x 10^-2 + 2.5 x 10^-4 x 135 = 0.104 M

Part C

rate constant for a first order reaction,

k = -[(ln(1.5 x 10^-3) - ln(9.4 x 10^-2)/(75 - 50)] = 0.165 s-1

Part D

Rate constant for a second order reatcion,

k = (1/2.6 x 10^-2 - 1/0.25)/(835 - 170) = 0.052 M-1*s-1