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+by=mx+b.

Order	Integrated Rate Law	Graph	Slope
0	[A]=−kt+[A]0[A]=−kt+[A]0	[A] vs. t[A] vs. t	−k−k
1	ln[A]=−kt+ln[A]0ln⁡[A]=−kt+ln⁡[A]0	ln[A] vs. tln[A] vs. t	−k−k
2	1[A]= kt+1[A]01[A]= kt+1[A]0	1[A] vs. t1[A] vs. t	kk

A.) The reactant concentration in a zero-order reaction was 0.100 MM after 165 ss and 4.00×10⁻² MM after 305 ss . 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:

k0thk0thk_0th =

4.29×10−4 MsMs

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

C.) The reactant concentration in a first-order reaction was 0.100 MM after 50.0 ss and 3.90×10⁻³ MM after 90.0 ss . 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.

D.) The reactant concentration in a second-order reaction was 0.600 MM after 235 ss and 2.70×10⁻² MM after 870 ss . 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

Ans.

a) 4.28×10^-3 mol.L^-1 s^-1

b) 2.93 × 10^-2 M

c) 0.081s^-1

d) 0.055 mol^-1.L.s^-1