Review Notes-- chapter 17
Thermodynamics
recommended problems: #23, 34, 40, 50, 62,
64, 66, 70, 88 (107)
Although the thermodynamics plays a role in a wide range of applications, the types of problems that might surface on the exam are quite limited.
* As with almost everything we have been doing this semester, one critical element is a
balanced chemical equation for the reaction being considered. Without the
equation (or with an incorrectly balanced equation) nothing works correctly.
* In this chapter, any temeprature used in a calculation must in in Kelvin
For the reaction you can compute DGo , DSo or DHo for the reaction
from the standard thermodynamc tables (and these would cetrainly be provided)
the superscript o means all species are in their standard state (1 atm, 1 M, or solid-liquid)
since the table is at 25oC, the calculations are at 25oC
Typical Problem--
compute DGo , DSo or DHo for a
reaction, or determine if a reaction
is exothermic or if it is
spontaneous (under standard conditions at 25 oC) Quite likely
to be given two of these (DGo , DSo or DHo) and asked to find the other or answer.
DHo is the heat of reaction
(negative if exothermic), typically
10-500 kJoule
DGo is the free energy of the
reaction (negative if reaction is spontaneous)
10-500 kJ
DSo is the entropy change (positive if disorder increases) typically a few Joule/deg
you can evaluate DGo from DGof values in the tables ..... or
you can evaluate DHo and DSo and compbine DGo=DHo - T DSo
if you do this, be careful to avoid mixing kJoule and Joule
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There's a very useful link between DGo (25oC and Kequil (25oC)
Kequil = exp (- DGo/RT) or
ln (Kequil) = - DGo / (RT)
so, from the thermodynamic tables, we can compute Kequil (first find DGo)
Typical problem:
Given value for DGo (or for DSo
and DHo)
find Kequil. Or if Kequil = xxx, what
is DGo for the reaction. Of course a spontaneous reaction has DGo <0 and so Kequil >1.
Note: the exp( )
function is the inverse of the ln ( ) function on your calculator
usually [second] + [ln] keys do this.
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We often want to know the value of DGo (or Kequil) at some other temperature.
We can
get a good estimate of DGo
(another T)
by computing DSo and DHo at 25oC
and combining with a T other than 298 K. That value of DGo can be used to find Keq (T)
or to determine if the reaction is spontaneous at this other temperature.
DGo= DHo - T DSo
Typical problem: if
a reactions has DHo= xxx and DSo
= yyy at 25oC
a. is the reaction spontaneous at 25oC? b. at 450oC ? c. at -140 oC ?
d. at what temperature will a
nonspontaneous reaction (25oC) beome spontaneous?
answer, when T = - DHo / DSo
e. There are four cases (DHo
, DSo are both
negative), (DHo , DSo
are both posative),
(DHo
negative but DSo postive) ,
and (DHo positive, DSo negative) . When
could equation above (d)
be used and why?
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The temperature dependence of Kequil was discussed in an earlier chapter
a plot of ln (Kequil) vs 1/T (K) shows data as a straight line
slope is
- DHo / RT (R=8.314, a value given on exam)
an exothermic graphic shows a + slope and Kequil decreases as T rises
an endothermic graphic shows a - slope and Kequil inccreases as T rises
(that's what le Chatlier said earlier, remember?)
Typical problem: a
table of Kequil is given at several temperatures
find Kequil
at another temeprature, find DHo for the reaction
(may be
easiest to calculate ln(K) and 1/T and plot the graph....
on exam you
may be given such data, already plotted)
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These rules / equations / ideas apply equally well to
chemical reactions
physical processes like evaporation (vapor pressure) or solubility (Ksp)
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Review Notes, Kinetics
there are really only a few major tasks (exam problems) in this chapter:
find the order of the reaction (from data)
find rate constant k (from data; you
must know the order)
how much remains (or is produced )
after __sec or how long until ___% has reacted ?
how to find and use half
life data ?
how will the rate of a reaction vary
with temperature
or, what is Ea for the
reaction?
Finding the order of a reaction
Initial rate (slope of [reactant] vs. time )
how does it vary, from run to run, when you change initial concentrations of each?
Rate when other species are in great excess (or are catalysts)
that is, when their concentration really doesn't vary during the run
Subsequent half lives
Fit data (conc of reactant vs. time ) to a graph
if conc vs time is straight line-- zeroth order
if ln (conc) vs time is straight line-- first order
if 1/[conc] vs time is straight line -- second order
Finding the rate constant
[rate of reaction] / {concentrations }
first order k = ln(2) / t1/2 second order k = 1 / ( [A]* t1/2 )
Finding out how much is left ... when
ln (A/Ao) = - k*t first
1/[A] - 1/[Ao] = kt second
[A] = [Ao] - kt zeroth
Energy of activation and temperature
plot of ln (k) vs 1/T is straight line
slope = -Ea / R