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
5.00×10^{−2}*M* after 110 s and
4.00×10^{−2}*M* after 375 s . What is the rate
constant for this reaction?

----------

Part B

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

-------------

Part C

The reactant concentration in a first-order reaction was
9.60×10^{−2}*M* after 30.0 s and
5.20×10^{−3}*M* after 60.0 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 0.650
*M* after 175 s and 5.60×10^{−2}*M* after 815
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 multiplication, for example write a Newton-meter as N*m.

Answer #1

A. rate constant, k = - slope

= - d[A]/dt

= - [0.04-0.05] M /[375-110]s

= 3.77*10^{-5} M.s^{-1}

B. Initial reactant concentration = [A]_{0}

= intercept in y axis

look at the graph(the graph is qualitative). Draw your own graph in a graph paper to determine the initial concentration.

C. rate constant, k = - slope

= - dln[A]/dt

= - [ln 0.0052 M - ln 0.096 M]/(60-30)s

= - ln (0.0052/0.096)/30 s

= 0.097 s^{-1}

D. rate constant, k = slope

= d(1/[A])/dt

= (1/0.056 - 1/0.65) M^{-1} /(815-175)s

= 0.0255 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
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
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.
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 ....

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...

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
Part A
The reactant concentration in a zero-order reaction was
6.00×10−2 mol L−1 after 140 s and 3.50×10−2
mol L−1 after 400 s . What is the rate constant for...

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marker numbers. What mile marker will the car reach after 2 hours?
This problem can easily be solved by calculating how far the car
travels and subtracting that distance from the starting marker of
145. 55 mi/hr×2 hr=110 miles traveled milemarker 145−110
miles=milemarker 35 If we...

Part A The rate constant for a certain reaction is k = 2.40×10−3
s−1 . If the initial reactant concentration was 0.650 M, what will
the concentration be after 18.0 minutes? Express your answer
with the appropriate units.
Part B
A zero-order reaction has a constant rate of
4.50×10−4M/s. If after 40.0 seconds the
concentration has dropped to 4.50×10−2M, what
was the initial concentration?
Express your answer with the appropriate units.

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:
6.42 min
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
of 4.20×10−3 s−1 at 45 ∘C. How many minutes does it take
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
of 5.10×10−3 s−1 at 45 ∘C . How many minutes does it
take for the concentration of the...

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