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.

1.) The reactant concentration in a zero-order reaction was 6.00×10⁻²M after 165 s and 3.50×10⁻²Mafter 385 s . What is the rate constant for this reaction?

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

3.)The reactant concentration in a first-order reaction was 6.70×10⁻²

M after 40.0 s and 2.50×10⁻³Mafter 95.0 s . What is the rate constant for this reaction?

4.)The reactant concentration in a second-order reaction was 0.850

M after 255 s and 3.40×10⁻²M after 860 s . What is the rate constant for this reaction?

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

Answer 1

According to the equations on your table, we can actually calculate all that.

For a zero order reaction, the equation to use would be:

A = Ao - kt

We already know the concentrations after some time, so the expression for a linear equation to get the slope is:

k = A2 - A1 / t2 - t1

We have both concentrations to a given time so:

k = 0.035 - 0.06 / 385 - 165

k = -1.136x10^-4 M/s

Now that we have the constant, we can calculate the innitial concentration:

A = Ao - kt

Ao = A + kt

Ao = 0.06 + (-1.136x10^-4)*(165)

Ao = 0.04125 M

As for a first order reaction the procedure is exactly the same but using the equation for a first order:

lnA = lnAo - kt

k = lnA2 - lnA1 / t2 - t1

k = ln(0.0025) - ln(0.067) / 95-40

k = -0.0598 s^-1

For the last question use the second order reaction:

k = 1/A2 - 1/A1 / t2 - t1

Solve for k, replacing the values.

Hope this helps