Question

Item 4

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]t=−kt+[A]0	[A]t vs. t	−k
1	ln[A]t=−kt+ln[A]0	ln[A]t vs. t	−k
2	1[A]t= kt+1[A]0	1[A]t vs. t	k

Part A

The reactant concentration in a zero-order reaction was 6.00×10⁻² mol L−1 after 140 s and 3.50×10⁻² mol L−1 after 400 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

k0th =

9.62×10−5 mol⋅L−1⋅s−1

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Part B

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

Express your answer with the appropriate units.

[A]0 =

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Part C

The reactant concentration in a first-order reaction was 8.50×10⁻² mol L−1 after 35.0 s and 4.50×10⁻³mol L−1 after 80.0 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

k1st =

0.0141s−1

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Part D

The reactant concentration in a second-order reaction was 0.500 mol L−1 after 215 s and 7.70×10⁻² mol L−1 after 745 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.

k2nd =

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Answer 1