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 8.00×10⁻²M after 200 s and 2.50×10⁻²Mafter 390 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 1.30×10⁻³Mafter 65.0 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 0.880 M after 160 s and 7.70×10⁻²Mafter 885 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.

Answer 1

Part A

For zero order reaction

[Ao] = [At] + kt

8.00 * 10^(-2) = 2.50 * 10^(-2) + k(390-200)

k = 5.50 * 10^(-2)/190 = 2.8947 * 10^(-4) Ms^(-1)

Part B

[Ao] = [At] + kt

[Ao] = 8.00 * 10^(-2) + 2.8947 * 10^(-4) * 200

[Ao] = 8.00 * 10^(-2) + 5.789 * 10^(-2) = 13.789 * 10^(-2) M

Part C

ln(Ao/At) = kt

ln(9.40 * 10^(-2)/1.30 * 10^(-3)) = k(65-50)

ln(72.307) = 15k

k = 0.2853 s^(-1)

Part D

1/At = 1/Ao + kt

1/7.70 * 10^(-2) = 1/0.880 + k(885-160)

k = 0.0163 M^(-1) s^(-1)