Question

The radioisotope studies today decays following either zero order or first order, or second order kinetics. The graphs produced by the integrated rate law plot different variables on the x and y axis depending on the order of the reaction. Look up in your textbook for the straight line plots of these three types of rate laws. Draw the plots in your notebook and label it. Give an equation that can be use to determine the half life of the radioisotope.

This is a Chemistry 1046 question..

Answer #1

Shown below are the straight line plots,

zero-order reaction : [concentration] vs time

first-order reaction : ln[concentration] vs time

second order reaction : 1/[concentration] vs time

[concentration] represents the amount of substrate remaining at time t

k = rate constant

Half life for,

zero-order reaction : t1/2 = initial [concentration] of substrate/2k

first order reaction : t1/2 = 0.693/k

second order reaction : t1/2 = 1/k x iniitial [concentration] of substrate

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 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
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 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
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...

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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