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*.

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^{−2}*M* after 135 s and
2.50×10^{−2}*M* after 315 s . What is the rate
constant for this reaction?

Express your answer with the appropriate units.

**Part B**

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

Express your answer with the appropriate units.

**Part C**

The reactant concentration in a first-order reaction was
9.40×10^{−2}*M* after 50.0 s and
1.50×10^{−3}*M* after 75.0 s . What is the rate
constant for this reaction?

Express your answer with the appropriate units.

**Part D**

The reactant concentration in a second-order reaction was 0.250
*M* after 170 s and 2.60×10^{−2}*M* after 835
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.t B**

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

Express your answer with the appropriate units.

Answer #1

Part A.

For a zero order reaction,

[A] = [A]o - kt

with,

[A] at 135 s = 7 x 10^-2 M

[A] at 315 s = 2.50 x 10^-2 M

So the rate constant k for the reaction becomes,

k = -[2.5 x 10^-2 - 7 x 10^-2)/(315 - 135)] = 2.5 x 10^-4 M/s

Part B.

Initial concentration [A]o

[A]o = 7 x 10^-2 + 2.5 x 10^-4 x 135 = 0.104 M

Part C

rate constant for a first order reaction,

k = -[(ln(1.5 x 10^-3) - ln(9.4 x 10^-2)/(75 - 50)] = 0.165 s-1

Part D

Rate constant for a second order reatcion,

k = (1/2.6 x 10^-2 - 1/0.25)/(835 - 170) = 0.052 M-1*s-1

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
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
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
5.00×10−2M after 200 s and
2.50×10−2M after 310 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
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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
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3.)The reactant concentration in a first-order reaction was
6.70×10−2
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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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The reactant concentration in a zero-order reaction was
6.00×10−2 mol L−1 after 140 s and 3.50×10−2
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This problem can easily be solved by calculating how far the car
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Part A
A certain first-order reaction (A→products) has a rate constant
of 7.20×10−3 s−1 at 45 ∘C. How many minutes does it take
for the concentration of the reactant, [A], to drop to 6.25% of the
original concentration?
Express your answer with the appropriate units.
Answer:
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Part B
A certain second-order reaction (B→products) has a rate constant
of 1.35×10−3M−1⋅s−1 at 27 ∘Cand an initial
half-life of 236 s . What is the concentration of the reactant B
after...

For a first-order reaction, the half-life is constant. It
depends only on the rate constant k and not on the reactant
concentration. It is expressed as t1/2=0.693k For a second-order
reaction, the half-life depends on the rate constant and the
concentration of the reactant and so is expressed as
t1/2=1k[A]0
Part A
A certain first-order reaction (A→products) has a rate constant
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for the concentration of the reactant, [A],...

For a first-order reaction, the half-life is constant. It
depends only on the rate constant k and not on
the reactant concentration. It is expressed as
t1/2=0.693k
For a second-order reaction, the half-life depends on the rate
constant and the concentration of the reactant and so is
expressed as
t1/2=1k[A]0
Part A
A certain first-order reaction (A→products ) has a rate constant
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Part A
The rate constant for a certain reaction is k = 3.70×10−3 s−1 .
If the initial reactant concentration was 0.200 M, what will the
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Part B
A zero-order reaction has a constant rate of
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Express your answer with the appropriate units.

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