Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...

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

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

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