# Key solution for assignment 01 (1) ```CHE3702/201/0/2022
Tutorial letter 201/0/2022
Physical Chemistry 3 (Theory)
CHE3702
Year module
Department of Chemistry
CONTENTS
1
KEY TO ASSIGNMENT 01........................................................................................................... 3
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CHE3702/201/0
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KEY TO ASSIGNMENT 01
QUESTION 1
The thermodynamic equilibrium constant values were provided in the question and the relevant
temperature values.
Using the question:
ΔrGΘ = - RT ln K,
we compute the following values of ΔrGθ at the specified temperatures:
T/K
K
ΔrGθ / kJ mol-1
300
4.0 x 1024
- 141
500
2.5 x 1010
- 100
700
3.0 x 104
- 60
It was assumed that ΔrHθ to be constant over this temperature range, then by using any two pairs of
values of K and T in the above table, we compute the value of Δ rHθ using the following equation:
3
Solving for ΔrHθ :
= 8.3145 J. K-1 mol-1 In (3.0 x 104 / 4.0 x 1024) / (700-1 – 300-1)
= 202 kJ mol-1
And using the following equation to find the vaue for ΔrSθ:
= 203 J K-1 mol-1.
= 0.203 kJ K-1 mol-1.
For all the the specified temperatures, using the ΔrHθ value obtained above and any ΔrGθ and the
relevant temperature value.
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QUESTION 2
In order to determine the concentration of H+ for the reaction, we have to know the details of the
stoichiometry of the reaction:
H2 (g) → 2 H+(aq) + 2e- Eθ = 0.0 V
2 AgCl (s) + 2e- → 2 Ag (s) + 2 Cl- (aq) Eθ = 0.22 V
.....................................................................................................................................
2 AgCl (s) + H2 (g) → 2 Ag (s) + 2 H+ (aq) + 2 Cl- (aq) Eθ = 0.22 V
The Nernst equation for the reaction is
where the number of electrons involved is v = 2
for which we also make the following assumptions:
(i) p(H2) = pθ
(ii) a(H+) = a(Cl-)
so that the equation becomes
5
from which we obtain
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QUESTION 3
The details of the stoichiometry of the reaction are:
R:
Bi3+ (aq) + 3e− → Bi (s)
L:
Bi2S3 (s) + 6e− → 2Bi (s) + 3S2- (aq)
Rearranging:
Bi3+ (aq) + 3e− → Bi (s)
2Bi (s) + 3S2- (aq) → Bi2S3 (s) + 6e−
Then Multiply the first equation by 2, the overall reaction is:
2Bi3+ (aq) + 3S2- (aq) → Bi2S3 (s),
where the number of electrons involved is v = 6
at equilibrium, Ecell = 0 and Q = K, then the Nernst equation gives
Rearranges into:
= {6(96485 J. V-1. mol-1) x (0.96 V)} / (8.31447 J K-1 mol-1) x 298K
= 224
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K = e224 = 1.91 x 1097
So, the equilibrium expression, is
K = 1 / [Bi3+]2[ S2-]3 = 1.91 x 1097
Where K is the equilibrium constant.
Remember from your first year general chemistry, the definition of a “Heterogeneous equilibrium
reaction ” when considering equilibrium expression.
However, the solubility equilibrium expression is written as the reverse of the cell reaction.
Remember that again from your first-year general chemistry,
Bi2S3 (s) → 2Bi3+ (aq) + 3S2- (aq)
Ks = [Bi3+]2[ S2-]3 = 1/K
Ks = 1/K = 1/ 1.91 x 1097
= 5.2 x 10-98
Where Ks = 5.2 x 10-98, is the solubility constant for Bi2S3.
For the solubility of Bi2S3 is done as follows:
Bi2S3 (s)  2Bi3+ (aq) + 3S2- (aq)
Let s = molar solubility of Bi2S3 (s)
s mol/L of Bi2S3 (s) dissolves to reach equilibrium.
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
Bi2S3 (s)
Start
∆
solid
-s
Equilibrium
2Bi3+ (aq)
+
3S2- (aq)
-
-
+2s
+3s
2s
3s
Remember that solids are not considered in the equilibrium calculations.
Thus:
Ks = 5.2 x 10-98 = [Bi3+] [ S2-] = (2s)2 x (3s)3 = 108s5
s5 = (5.2 x 10-98) x (9.26 x 10-3)
s = (5.2 x 10-98)1/5 x (9.26 x 10-3)1/5 ……………….where = 9.26 x 10-3 = 1/108
= (3.49 x 10-20) x (3.92 x 10-1)
= 1.4 x 10-20 mol.L-1
Then:
s = molar solubility = 1.4 x 10-20 mol.L-1
So, Bi2S3 (s) has a solubility of = 1.4 x 10-20 mol.L-1
From above, can you calculate the equilibrium concentrations of both ions, [Bi3+]and [ S2-]?
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QUESTION 4
The fundamental measurement used to study the motion of ions is that of the electrical resistance,
R, of the solution. The conductance (measured in Siemens units, S), G, of a solution is the inverse
of its resistance R: G = 1/R.
Alternatively: The resistance R (measured in per Siemens units, S-1 or Ohms, Ω) of a uniform
conductor is directly proportional to its length, ℓ, and inversely proportional to its cross-sectional
area, A:
R = (1/κ)( ℓ /A), κ is the conductivity (ohm-1 cm-1 or S cm-1.)
The conductance of a sample decreases with its length ℓ and increases with its cross-sectional area
A. We therefore write:
where κ is the conductivity
The expression for conductance, in units of S (Siemens), is
Conductance, G, = 1/R = κ(A/ℓ)
The conductivity of a solution depends on the number of ions present, and it is normal to introduce
the molar conductivity, Λm, which is defined as
In units of S cm2 mol-1
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And c = concentration of the added electrolyte in mol.cm-3 units,
The essential requirements for measuring molar conductivities are a suitable cell containing metal
electrodes, between which the current passes through the solution, and a means of measuring the
resistance of the cell and solution. This is achieved by measuring or determination of the cell
constant
The cell constant, defined by: C = (ℓ /A) cm-1
So cell constant gives, C = κR = κ / conductance so that κ = C / R
And Λm = C / (cR),
……… C = cell constant, c = concentration, R = resistance
The concentration dependence of molar conductivities indicates that there are two classes of
electrolyte. The characteristic of a strong electrolyte (like KCl) is that its molar conductivity
depends only slightly on the molar concentration (and in general decreases slightly as the
concentration is increased,). The characteristic of a weak electrolyte (like CH3COOH) is that its
molar conductivity is normal at concentrations close to zero but falls sharply to low values as the
concentration increases. See figure below.
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Fig. 4.1 The concentration dependence of the molar conductivities of (a) a typical strong electrolyte
(aqueous potassium chloride) and (b) a typical weak electrolyte (aqueous acetic acid)
Strong electrolytes:
Strong electrolytes are fully ionized in solution, e.g, NaCl, KCl. Strong electrolytes dissociate in
aqueous solution; especially at low electrolyte concentration, the conductivity depends on the
square root of the concentration (see the Kohlrausch equation below). The molar conductivity is the
conductivity of one mole of electrolyte in a litre of solution; the limiting molar conductivity is obtained
by extrapolating conductivity measurements as function of concentration to zero concentration (i.e.
the intercept of a Λm versus c&frac12; plot).
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Kohlrausch’s law gives: Λm =Λ&deg;m − K c1/2, The constant Λ&deg;m is the limiting molar conductivity,
the molar conductivity in the limit of zero concentration.
Plotting the value of Λm against c1/2 yields a straight line at low concentrations. This line can be
extrapolated to c = 0 to yield Λ&deg;m, the molar conductivity at infinite dilution. Weak electrolytes do
not give a straight line; the line exhibits a characteristic curvature.
If the limiting molar conductivity of the cations is denoted λ+ and that of the anions λ−, then his law
of the independent migration of ions states that
Where ν+ and ν− are the numbers of cations and anions per formula unit of electrolyte (for example,
ν+ = ν− = 1 for HCl, NaCl, and CuSO4, but ν+ = 1, ν− = 2 for MgCl2).
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QUESTION 5
(a) If the limiting molar conductivity of the cations is denoted λ+ and that of the anions λ−, then the
law of the independent migration of ions states that
Where ν+ and ν− are the numbers of cations and anions per formula unit of electrolyte
From table 21.5 Limiting ionic conductivities in water at 298 K, λ /(mS m 2 mol−1) of Cl- is 7.635 mS
m2 mol−1 = 76.35 S cm2 mol-1
Given that, Λom (AgCl) = 138.3 S cm2 mol-1 =
it follows that Λom (AgCl) = ν+ (Ag+) + ν- (Cl-) = 1 x (Ag+) + 1 x (76.35 S cm2 mol-1)
thus, the the limiting ionic conductivity λ, of Ag+ = Λom (AgCl) – λ (Cl-)
= 138.3 S cm2 mol-1 - 76.35 S cm2 mol-1
= 61.95 S cm2 mol-1
(b) Using the equation below, and that the resulting conductivity of 0.1888 m S m-1 = 1.887 x 10-6 S
cm-1
Which rearranges into:
c = κ / Λm
And assuming the concentration to be very small,
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c = κ / Λm ≈ k / Λ&deg;m
= 1.887 x 10-6 S cm-1 /138.3 S cm2 mol-1
= 1.36 x 10-8 mol cm-3
= 1.36 x 10-5 mol L-1
the solubility of AgCl.
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QUESTION 6
Experimental methods used to monitor chemical reactions.
In order to monitor chemical reactions, we use chemical kinetics which is concerned with the rates of
chemical reactions. Chemical kinetics deals with how rapidly reactants are consumed and products
formed, how reaction rates respond to changes in the conditions or the presence of a catalyst, and the
identification of the steps by which a reaction takes place.
Few reaction properties that, when measured over time, can be useful as reaction rate indicators:
a)
Changes in pressure (indicates gas consumption or formation).
b)
Changes in pH (indicates acid/base formation or consumption).
c)
Changes in volume.
d)
Changes in concentration of one or more of the molecules in a solution. This is the most
widely used method, as concentrations of many molecules can be determined by analytical
techniques.
In general, the following, methods are used to monitor chemical reactions.
isolation method:
The determination of a rate law is simplified by the isolation method, in which all the reactants
except one are present in large excess (or constant). We can find the dependence of the rate on
each of the reactants by isolating each of them in turn—by having all the other substances present
in large excess. Consider the reaction: A + 2 B →3 C + D, If a reactant B is in large excess, for
example, it is a good approximation to take its concentration as constant throughout the reaction,
while A is isolated at the same time.
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Although the true / ideal rate law might be: v (rate law) = kr[A][B], we can approximate [B] by its initial
value [B]0 (from which it hardly changes (remember is constant) in the course of the reaction) and
write: v = kr′[A] and kr′ = kr[B]0
Because the true rate law has been forced into first order form by assuming a constant B
concentration, the effective rate law is classified as pseudo first-order and kr′ , is called the
effective rate constant for a given fixed concentration of B.
If, instead, the concentration of A were in large excess (constant), while B is isolated this time, , then
the rate law would simplify to: v = kr′[B] and kr′ = kr[A]0
method of initial rates:
In the method of initial rates, which is often used in conjunction with the isolation method, the
instantaneous rate is measured at the beginning of the reaction for several different initial
concentrations of reactants. For example, suppose the rate law for a reaction with A isolated (while
B is constant) is
v = k′r[A]a;
Then the initial rate of the reaction, v0, is given by the initial concentration of A:
v0 = k’r[A]0a
Taking logarithms gives:
log v0 = log k′r + a log [A]0
This equation has the form of the equation for a straight line:
So that, for a series of initial concentrations, a plot of the logarithms of the initial rates against
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the logarithms of the initial concentrations of A should be a straight line with slope a.
similarly, suppose the rate law for a reaction with B isolated (while A is constant) is
v0 = k’r[B]a0
then, log v0 = log k′r + a log [B]0
Integrated rate laws:
An integrated rate law is an expression that gives the concentration of a species as a function of
the time.
Integrated rate laws have two principal uses. One is to predict the concentration of a species at any
time after the start of the reaction. Another is to help find the rate constant and order of the reaction.
Because rate laws as discussed above are differential equations, we must integrate them if we want
to find the concentrations as a function of time.
The method of initial rates might not reveal the full rate law, for the products may participate in the
reaction and affect the rate. To avoid this, the rate law should be fitted to the data throughout the
reaction. The fitting may be done, in simple cases at least, by using a proposed rate law to predict
the concentration of any component at any time and comparing it with the data. Because rate laws
are differential equations, we must integrate them if we want to find the concentrations as a function
of time. In order to determine the rate law, one plots the right-hand side of the integrated rate laws
against time in order to see which of them results in a straight line through the origin. The one that
does, is the correct rate law.
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QUESTION 7
The following table is constructed: in which a plot of logarithms of the initial rates (log vo) vs the
logarithms of the initial concentrations (log [J]o) should give a straight line. A plot is shown below.
[J]o / mmol L-1 vo / mol L-1 s-1
log [J]o
log vo
5.0
3.6 x 10−7
0.699
-6.44
8.2
9.6 x 10−7
0.914
-6.02
17
41 x 10−7
1.23
-5.39
30
130 x 10−7
1.48
-4.89
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A straight line with slope a = 1.9862 ≈ 2, which the is order of the reaction with respect to X (I,e,
the reaction is 2nd order). and the intercept given as log k′r = -7.8315. therefore, the rate constant k
= 1.44 &times; 10−2 L mol−1 s−1.
Or by using any pairs of values (log vo) vs (log [J]o), we can calculate the rate constant k. using
the 1st pair,
v0 = k’r[J]0a
3.6 x 10−7 mol L-1 s-1 = k x (5.0 x 10-3 mol L-1)2
k = 3.6 x 10−7 mol L-1 s-1 / 2.5 x 10-5 mol2 L-2
= 1.44 &times; 10−2 L mol−1 s−1.
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QUESTION 8
(i)
When data of time and concentration for a reaction are given, we use the method of integrated rate laws:
the rate law of a reaction is integrated, and an equation containing time and a function of concentration
is obtained. The function of the concentration depends on the order of the reaction, and it is plotted
against time to yield a straight line.
Alternatively, we can calculate an expression for the half-life of the reaction (the time necessary for a
concentration to decrease to half its value); if this equation is obeyed by the given data, we can determine
the order of the reaction and the rate constant.
(ii)
By inspecting the given data, it is observed that the half-lives obtained at various starting concentrations
are not constant; meaning that this is NOT a characteristic for a first order half-lives (make sure that
you know the equation); i,e, the reaction appear not to be a first order.
The integrated rate law is then used, and for a second-order reaction the equation is:
Or 1 / [A] = 1 / [A]o + kt, where [A]0 is the initial concentration of A (at t = 0).
So that a plot of plot 1/[A] against t should give a straight line, to test for a second-order reaction.
The slope of the graph is k.
We therefore plot 1/[A] against time, to test for a second-order reaction
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Time / s
0
120
240
360
480
600
[A] / mol dm-3
1.3720
0.7231
0.4910
0.3716
0.2990
0.2501
1/ [A] / dm3 mol-1
0,729
1,383
2,037
2,691
3,344
3,998
We obtain a straight line (confirmation that we have a second-order reaction) with a slope of
0.0054 dm3 mol-1 sec-1 as obtained directly from the plot. Therefore, we have for the rate constant:
k = 0.0054 dm3 mol-1 sec-1
= 0.324 dm3 mol-1 min-1
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QUESTION 9
(i)
For a first-order reaction, the half-life of a reactant is independent of its initial concentration.
Therefore, if the concentration of A at some arbitrary stage of the reaction is [A], then it will have
fallen to 1/2 [A]
At a half-life of a reactant A, whose initial concentration is [A]o, we have:
[A] = &frac12; [A]0
Where [A]0 is the initial concentration of A (at t = 0).
For first-order reactions, the equation is given by
Hence by integrating the above equation, gives
&frac12; [A]0
[A]0
𝑑[𝐴]
= −
[𝐴]
𝑡1/2
𝑘𝑑𝑡
𝑜
The time (from t = 0 to t&frac12;) for [A] to decrease from [A]0 to &frac12; [A]0 in a first-order reaction is given by
equation:
23
So that:
Accordingly, no concentration term appears in the above equation, meaning that the half-life of a
reactant is independent of its initial concentration’.
(ii)
The integrated rate law for the reaction
A→P
where the order n ≥ 2, is
kt =
1 
1
1 


n−1
n - 1  ([A] 0 - x )
[A] n-01 
Let [A]0 be the initial initial concentration of A (at t = 0).
Now at t&frac12;, x = &frac12; [A]0
So that,
k t&frac12; =
=
=
1
[
𝑛−1
1
𝑛−1
1
𝑛−1
[
[
1
1
2
([𝐴]0 − [𝐴]0 )
𝑛−1
−
1
[𝐴]𝑛−1
0
1
1 𝑛−1
[𝐴0 ]𝑛−1 −( )
[𝐴]𝑛−1
0
2
1
1 𝑛−1
[𝐴]𝑛−1
( )
0
2
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−
1
[𝐴]𝑛−1
0
−
]
]
1
[𝐴]𝑛−1
0
]
CHE3702/201/0
=
=
1
𝑛−1
1
𝑛−1
2𝑛−1
1
0
[𝐴]𝑛−1
0
[[𝐴]𝑛−1 −
]
2𝑛−1 −1
[ [𝐴]𝑛−1 ]
0
Therefore, an expression for the half-life of the reaction is:
t&frac12; = [
2𝑛−1 −1
(𝑛−1)𝒌[𝐴]𝑛−1
0
]
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QUESTION 10
Note that the first column in the given table in the question shows that:
10-7 x k / L mol-1 s-1 = 1.44
So, k = 1.44 (L mol-1 s-1) / 10-7
= 1.44 x 107 L mol-1 s-1
The rates of most chemical reactions increase as the temperature is raised. For many reactions
that a plot of ln k against 1/T gives a straight line. This is described by Arrhenius equation:
The parameter A, which corresponds to the intercept of the line at 1/T = 0 (at infinite temperature),
is called the pre-exponential factor or the ‘frequency factor’. The parameter Ea, which is obtained
from the slope of the line (−Ea /R), is called the activation energy. Collectively the two quantities
are called the Arrhenius parameters.
A plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by the
Arrhenius equation. The slope gives −Ea/R and the intercept at 1/T = 0 gives ln A.
The given table is now constructed as follows:
k ( L mol-1 s-1)
T (K)
Ink
1/T (K-1)
1.44 x 107
300.3
16.483
0.003330
3.03 x 107
341.2
17.227
0.002931
6.90 x 107
392.2
18.050
0.002550
A plot of ln k against 1/T is provided below with a slope gives −Ea/R = -2007.8 K and the intercept
at 1/T = 0 gives ln A = 23.15.
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Therefore, the pre-exponential factor, A, is:
In A = intercept
= 23.15
A = 1.13 x 1010 L mol-1 s-1
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and the activation energy of the reaction:
− Ea/R = - 2007.8 K
Ea = 2007.8 K x R
= -2007.8 K x 8.3145 J K-1 mol-1
= 16 693 J mol-1
= 16.7 kJ mol-1
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```