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*=*m**x*+*b*.

1.) The reactant concentration in a zero-order reaction was
6.00×10^{−2}*M* after 165 s and
3.50×10^{−2}*M*after 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^{−3}*M*after
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^{−2}*M* 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

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
5.00×10−2M after 110 s and
4.00×10−2M after 375 s . What is the rate
constant for this reaction?
----------
Part B...

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
7.00×10−2M after 135 s and
2.50×10−2M after 315 s . What is the rate
constant for this reaction?
Express your answer with...

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
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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...

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
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2.50×10−2M after 310 s . What is the rate
constant for this reaction?
Express your answer with...

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
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A zero-order reaction has a constant rate of
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