2. (a) False. The so called equilibrium constant, K, is really an equilibrium coefficient and is dependent on T. If you think of the equilibrium consant as comming from a ratio of rate constants when the rate of the forward reaction becomes equal to the rate of the reverse reaction and recall that the rate constants are temperature dependent as given by the Arhenius equation then you have to conclude that K is a function of T.
(b) True.
We can think of it this way: If there is a solid in excess then the partial pressure of that solid will be a constant, the vapor pressure at the temperature of the solid, and this constant can be combined in one equilibrium constant.
There will be more 
in the chloride case because, given that both compounds form the same
number of ions, the 
Let us compute the concentrations. We are given 
and when the solid is placed in
a beaker of water initially there is nothing in solution. Then a certain amount of
dissolves, lets say x dissolves. There is still excess around. For every x of lead there
are 2x of chloride according to the stoichiometry. We can write the concentrations
= x, 
= 2x. and 
.
x = []
= 0.0075M. Doing the same thing with the fluoride get x = [] = 0.0021M.
In a tabular way:
6. Calcium carbonate decomposition is endothermic 
. Treating heat as a reagent and
using LeChatlier's principle raising the temperature will cause more of the carbonate
to dissolve. If more calcium carbonate is added [assuming
that all of it had vaporized, if not then adding more to an excess of CaCO3(s)
would not change the equilibrium.] more carbon dioxide will form according
to LeChatlier's principle. If the partial pressure of carbon dioxide is raised above
the equilibrium value then more calcium carbonate will form.
We can think globally. The earth and its atmosphere is a closed system (forgetting about sunlight, protons and other particles and the occasional meteroite) created by the gravational field of its mass. If carbon dioxide is released into the atmosphere by our mining and burning fossil fuel why is it not true that, by LeChatlier's principle, the cabon dioxide dissolves in the oceans and becomes part of carbonates?
8.
(a) 
(b) 
(c) 
(d) 
(e) 
10.
The second chemical equilibrium is obtained from the first by multiplying by 3. This
is the same as adding three equilibria together and multiplying their equilibrium
constants. The answer is that the final K = (K')^3.
14.
(a) The same basic problem as 10. above. Multiplying the reaction by 1/2 means that
the equilibrium constant is the square root of the reference reaction equilibrium
constant. K = 2.2x10^5.
(b) The equilibrium constant for the reaction written in reverse is the reciprocal of K or 4.6 x 10^-6.
16.
.
More iodine, I_2, must dissociate building up [I] untill the equilibrium constant
is reached.
18.
.
Note: I dissagree with the value of the concentration equilibrium constant given
in the book-lets work the problem with what I think is the correct value.
and 
so that
The system concentrations adjust to decrease Q to the value of Kc = 165/M at equilibrium.
That means 
decreases and 
increases.
22.
Note: I think the temperature should be 562C for the equilibrium system with the
concentrations below.
,
and 
in the presence of excess C(s). (Note : We started with 0.4 mole carbon and the carbon
in carbon monoxide and carbon dioxide corresponds to 0.3 moles so we still have excess
carbon in the system and therefore can have an equilibrium according to the reaction.)
(a) 
(b) Assuming C(s) is in excess, which it is, the equilibrium constant will not change
on addition of further C(s). This is because the concentration (or partial pressure)
of gas phase carbon is a constant at a given temperature when there is excess carbon-just
the vapor pressure of carbon.
26.
Initally,[CO]o =0.0102M, [Cl2]o = 0.0069M. At equilibrium [Cl2] = 0.00301M.
(a) What are the concentrations at equilibrium and
(b) what is the equilibrium constant?
(a) Any COCl2 comes from the dissociation of Cl2 and this is [COCl] = [Cl2]o - [Cl2] = 0.00609 - 0.00301M = 0.00308M. [CO] = 0.0102 - 0.00308M.
(b) The equilibrium constant is therefore, 
30.
and x = methly cyclopentane = 0.0017M; cyclohexane = 0.0144M
36.
40.
Predict the effect of the changes listed in the table.
42.
For the equilibrium: 
(a) What are the concentrations of 
and 
?
From the chemical equation we know = .
Solving, 
(b) At the start have an additional 0.020M of . What concentrations at the new
equilibruim?
50.
1% bromine gas dissociated means 0.99 left and the formation of 0.02 bromine atoms.
Kc = (0.02)^2/0.99 = 0.0004M.
52.
,
.
78.7% 
is consumed. Therefore, 
is left and 
.
.
54.
.
What is Ptotal? Each mole of ammonia formed means a mole of hydrogen sulfide formed.
and Ptotal = 0.663 atm.
56.
K = 2.5 at 298K.
(a) If at the start [butane]_o = 1.75 mole and [iso-butane]_o = 1.25 mole (Note:
units do not matter since the equilibrium constant is dimensionless.) Is there equilibrium?
(b) [iso-butane] must increase and [butane] must decrease untill Q -> K.
(c) 
and [butane] = 1.75 - 0.893 = 0.857 mol.
[iso-butane] = 1.25 + 0.893 = 2.143 mol.
58.
Kc = ?
60.
the system is not at equilibrium and
(b) with Q > Kc concentrations will shift towards reactants.
(c) 
The sign of x will take care of itself.
.
[NO] = [O2] = 0.0042 + 2(-0.00155) = 0.0011M
73.
Mw(NO) = 30 g/mol; Mw(NOBr) = 109.9 g/mol; Mw(Br2) = 159.8 g/mol
(a) 3.5 g 
= 3.5g/30 g/mol = 0.1196 mol
+ 9.67 g 
= 9.67g/159.8 g/mol = 0.0605 mol. This is a slight excess of bromine so the most
NOBr would be 0.1196 mol, or 13.1 g NOBr.
However, there is an equilibrium dependent on the magnetude of an equilibrium constant. We need to know this in order to compute the amount of NOBr formed. If there were a large excess of Bromine then we could make the statement that the [NOBr] = 0.1196 mole. In part (c) we compute an equilibrium constant. Let us add a part (d) in which we repeat (a) and determine the equilibrium concentrations of NO, NOBr, and Br2.
(b) 
(c) (i) 
(ii) bent 
(iii) a polar molecule because of the electronegativity of Br and O and the fact
that the molecule is bent.
(c) 34 percent NOBr dissociates. That means 1 - 0.34 is left and 0.34 NO produced and 1/2*).34 bromine produced. The total amount of gas present is the sum or (1 - 0.34) + 0.34 + 0.17 = 1.17. The mole fraction of each is therefore
mole fraction NO = 0.34/1.17 = 0.29
mole fraction NOBr = 0.66/1.17 = 0.564
mole fraction Br2 = 0.17/1.17 = 0.145
The partial pressure is given by the mole fraction times the total pressure. The pressures of each component are:
P(NOBr) = 0.564*190 mmHg = 107.2 torr
P(NO) = 0.291*190 mmHg = 55.3 torr
P(Br2) = 0.17*190 mmHg = 27.5 torr
From this Kp = 2.71 mmHg^1/2
A question is: can we use the tabular method to do this calculation? Yes, but not without basically doing this partial pressure thing! We may use Kp and determine Kc using it in the tabular method below.
(d) Using moles of NO, NOBr, Br2 from (a), and assuming 1 L since volume is not
given, the table becomes
The equilibrium constant expression: 
give us an equation in x = [NOBr]. Square
this expression and solve the resulting cubic equation (use Maple, or Mathematica)
to find [NOBr] = 0.0330 M. [NO] = 0.0866M, and [Br2] = 0.0440, Kc = 0.55.
This is quite a bit less than 0.1196 mole from (a)