Page 1 of 13
Memorial University of Newfoundland
Chemistry 1051
Final Examination
Fall 2017
NAME:
Time: 3 hours
MUN STUDENT #: _____________________
U
READ THE FOLLOWING CAREFULLY
1.
This examination paper has 14 pages. Ensure that this examination is complete.
2.
Failure to submit this paper in its entirety at the end of the examination may result in
disqualification.
3.
A Periodic Table and physical constants are provided, as are a sheet of useful equations on
the last two pages. These may be detached for use during the examination.
4.
Answer each question in the space provided. Should you require more space, use the back
of the page and indicate clearly where this has been done.
5.
When answering questions show all relevant equations, all calculations, and justify all
simplifying assumptions.
6.
Correct units should be maintained in all steps of a calculation, though you may show
cancellation of units in early steps to eliminate them in later steps, if possible.
7.
Numerical answers should be reported with the correct significant digits.
8.
Do not write in the space below.
QUESTION
TOPIC
VALUE
1, 2
Solutions
9
3, 4
Kinetics
9
5, 6
Equilibrium
16
7, 8
Acid-Base Equilibrium
17
9, 10
Aqueous Equilibrium
7
11, 12
Thermodynamics
11
13
Redox Balancing
4
15, 16, 17, 18
Short Answer
9
TOTAL
82
MARK
Page 2 of 13
[MARKS]
[5]
1.
An automobile antifreeze mixture is made by mixing 0.5000 L each of ethylene
glycol (density = 1.114 g mL-1 and molar mass = 62.07 g mol-1) and water (density
= 0.9982 g mL-1 and molar mass = 18.02 g mol-1). Express the amount of ethylene
glycol in the resultant solution in terms of the following concentration units:
a)
Volume percent
Since we have mixed equal volumes of the liquids and have no information about the total
volume of solution, we must assume it is additive and the resultant solution has twice the
volume of the original ethylene glycol. Therefore the volume percent is 50.0 %.
b)
Mass percent
We added together exactly 0.5000 L of each substance to create exactly 1.0000 L of
solution. The mass of each substance is the volume times the density:
massEG = d x V = 1.114 g mL-1 x 500.0 mL = 557.0 g
massWater = d x V = 0.9982 g mL-1 x 500.0 mL = 499.1 g
mass%EG = massEG / masssoln = 557.0 g/(557.0 g + 499.1 g) = 0.5274 = 52.74%
c)
Molarity
Based on the 1.000 L solution considered in part b), the number of moles of ethylene glycol
in the solution will be 557.0 g / 62.07 g mol-1 = 8.9737 mol.
Since the definition of molarity is number of moles of solute divided by the total volume of
solution, the molarity would be 8.974 mol L-1.
d)
Molality
The definition of molality in moles of solute divided by the mass of the solvent. From the
previous parts we see that for our 1.000 L solution we have 8.9737 mol of ethylene glycol
in 499.1 g = 0.4991 kg of the solvent water. Therefore the molality of ethylene glycol is
Molality = moles solute / mass solvent = 8.9737 mol / 0.4991 kg = 17.98 mol kg-1
e)
Mole fraction
From part c) we see the total amount of ethylene glycol is 8.9737 mol. From the mass of
water found in part b) we can calculate the number of moles of water as 499.1 g / 18.02 g
mol-1 = 27.697 mol and the mole fraction of ethylene glycol will be
EG = moles EG / total moles = 8.9737 mol / (8.9737 mol + 27.697 mol) = 0.2447
Page 3 of 13
[4]
2.
Calculate the molality and the van’t Hoff factor i for a 1.00% by mass aqueous
solution of sodium chloride NaCl where the freezing point of the solution has been
depressed by 0.593 C from the normal freezing point of water.
Kf for water = 1.86 C kg mol-1
The molality is by definition the number of moles of solute divided by the mass of
the solvent. Let’s assume our solution has a total mass of exactly 1.000 kg.
Therefore this solution is (by the definition of mass percent) 10.0 g sodium chloride
and 990.0 g = 0.990 kg water.
The number of moles of sodium chloride is 10.0 g / 58.44 g mol-1 = 0.1711 mol and
the molality will then be 0.1711 mol / 0.990 kg = 0.1728 mol kg-1
From the molality and the freezing point depression we can calculate the van’t Hoff
factor
T = im Kf
i = T/m Kf = 0.593 C/(0.1728 mol kg-1 · 1.86 C kg mol-1) = 1.845
[6]
3.
Consider the reaction between reactants A and B:
2 A (g) + B (g)  3 C (g)
The data obtained from several experiments are listed in the table:
Exp. [A], M [B], M Initial Rate of Disappearance of A, M s-1
1
2
3
4
1.72
3.44
2.91
1.59
2.44
2.44
0.100
1.22
0.680
5.44
0.135
0.269
Determine the rate law and the value of the rate constant.
In general, rate of disappearance of A is rate=k [A]m [B]n. By comparing experiments 1 and 2
m
n
m
rate2 k 2 [A]2 [B]2
[A]2


since [B] is constant in expt 1 and 2
m
rate1 k1 [A]1m [B]1n
[A]1
3.44 M  so 8.00 =( 2.00)m and m = 3
5.44 M s 1

1
0.680 M s
1.72 M m
Now by comparing experiments 2 and 3
3
n
rate 2 k 2 [A]2 [B]2

where here I' ve substitued for known m  3
rate3 k 3 [A]33 [B]3 n
m
5.44 M s 1 3.44 M  2.44 M 

0.135 M s 1 2.91 M 3 0.100 M n
3
n
so 40.30  1.652 24.4  or 24.4  
n
n
40.30
 24.4
1.652
so it can be seen that n = 1
rate of disappearance of A =k [A]3 [B]
Using data from experiment 1
k = rate/[A] 3 [B] = 0.680 M s-1/(1.72 M)3(2.44 M) = 0.680 Ms-1/12.42 M4
= 0.0548 M-3s-1
Page 4 of 13
[3]
4.
The gas phase decomposition reaction
SO2Cl2 (g)  SO2 (g) + Cl2 (g)
shows first order kinetics and has a half-life of 3.15 x 104 seconds. How long does
it take (in hours) for 80.0 % of a sample of SO2Cl2 (g) to decompose?
For a first order kinetic process k = 0.693 / t½ = 0.693 / 3.15 x 104 s = 2.20 x 10-5 s-1
 SO 2 Cl 2 t 
  kt for a 1st order reaction
ln 

 SO 2 Cl 2 o 
SO 2 Cl 2 t
here we want
to be 0.200 (the 20.0% remaining!)
SO 2 Cl 2 o


ln 0.200   2.20 x 10 -5 s -1 t
t
5.
- 2.302 6
 1 hr 
 7.31 x 10 4 s
  20.3 hr
-5 -1
 2.20 x 10 s
 3600 s 


Ammonium hydrogen sulfide has been detected in the atmosphere of Jupiter, and
decomposes to form ammonia and hydrogen sulfide gas:
NH4SH (s)  NH3 (g) + H2S (g)
Kp = 0.126 at 24.0 C
[2]
(a)
Suppose we have a sealed 2.50 L flask containing an equilibrium mixture of this
reaction system, and the partial pressure of H2S is found to be 0.355 bar. What is
the partial pressure of NH3?
𝐾𝑝 = (𝑃𝑁𝐻3 )(𝑃𝐻2 𝑆 ) = 0.126 𝑠𝑜 𝑃𝑁𝐻3 =
[2]
(b)
0.126 0.126
=
= 0.355 𝑏𝑎𝑟
𝑃𝐻2 𝑆
0.355
If we assume the equilibrium mixture from part a) resulted from the decomposition
some of the 10.00 g of ammonium hydrogen sulfide that were sealed in the flask
initially, how many grams of the solid remain in the equilibrium mixture.?
Since the PNH3 = PH2S = 0.355 bar, we can calculate the number of moles of either product gas
formed, which will equal the number of moles of solid that decomposed
n = PV/RT = [(0.355 bar)(2.50 L)]/[(0.083145 L bar K-1 mol-1)(297.15 K)] = 0.03592 mol
The molar mass of the solid is 51.1095 g mol-1 which means that 1.836 g of solid decomposed.
This means that 8.16 g of the solid remain (to the appropriate 2 decimal places).
[2]
(c)
What is Kc for the reaction at 24.0 C?
Kp = Kc (RT)ngas
0.126 = Kc [(0.0083145)(297.15)]2
Kc = 0.126/[(0.0083145)(297.15)]2
Kc = 2.06 x 10-4
Page 5 of 13
[10]
6.
Phosgene gas decomposes to give carbon monoxide and chlorine
Kp = 4.60 x 10-2 at 395 C.
COCl2 (g)  CO (g) + Cl2 (g)
A flask at 395 C starts with an initial reaction mixture of
0.200 bar of CO, 0.300 bar of Cl2, and 0.105 bar of COCl2.
Calculate the equilibrium pressures of each of the gases.
We can start with a Q calculation to determine the direction of reaction
PCO  PCl  0.2000.300
QP 
PCOCl   0.105  0.57
Since Qp > Kp, this system should react from products to reactants to establish equilibrium
2
2
COCl2 (g) 
0.105
+x
0.105 + x
(all in bar)
Initial Pressure
Pressure. Change
Equilibrium P
KP 
PCO  PCl
P
COCl 2

2
  4.6 x 10
CO (g)
0.200
-x
0.200 - x
2

+
Cl2 (g)
0.300
-x
0.300 - x
0.200 - x 0.300 - x 
0.105  x 
4.60 x 10 2 [0.105  x]  0.200 - x 0.300 - x 
4.83 x 10-3  4.6 x 10 2 x  0.0600  0.500 x  x 2
x 2  0.546 x  0.05517  0
This is a quadratic equation of the form ax2 + bx + c = 0, which has solutions given by
 b  b 2  4ac
x
2a
 (0.546)  (0.546) 2  4(1)(0.05517 )
so x 
2(1)
x
(0.546)  (0.546) 2  4(1)(0.05517 )
(0.546)  (0.546) 2  4(1)(0.05517 )
or x 
2(1)
2(1)
x
0.546  0.2981  0.22068
0.546  0.2981  0.22068
or x 
2(1)
2(1)
x
0.546  0.07744
0.546  0.07744
or x 
2
2
so
x
0.546  0.278
0.546  0.278
or x 
2
2
0.824
0.268
or x 
so x  0.412 bar or x  0.134 bar
2
2
If we were to choose the first value of x we would get an equilibrium pressure of [CO] = 0.200 - x
= 0.200 – (0.412) = -0.212 bar, which is not possible, so we can throw this answer out.
x
Our equilibrium pressures will be
[COCl2] = 0.105 + x = 0.105 + (0.134) = 0.239 bar
[CO] = 0.200 - x = 0.200 – 0.134 = 0.066 bar
[Cl2] = 0.300 - x = 0.300 – 0.134 = 0.166 bar
Page 6 of 13
[7]
7.
Calculate the pH, [H2C2O4], [HC2O4–] and [C2O42–] in a 0.20 M solution of the
diprotic acid, oxalic acid [H2C2O4] at 25 ºC.
Ka1 = 5.60 x 10–2 and Ka2 = 5.42 x 10–5
The first ionization is as given by the equation below. We will assume this reaction
contributes the significant portion of H3O+
H2C2O4 (aq) + H2O (l)  H3O+ (aq) + HC2O4- (aq)
Initial
0.20 M
0M
0M
Change
-x
+x
+x
Equilibrium
0.20 - x
+x
+x
HCO H O   x x 

H C O  0.20 - x

K a  5.60 x 10
-2

4
2
3

4
2

x 
2
0.20 - x
so x 2  0.0560 x - 0.0112  0
x
x
 b  b 2  4ac
2a
so x 
 (0.0560)  (0.0560) 2  4(1)(-0.0112)
2(1)
 (0.0560)  3.136 x 10 3  (0.0448)
2
or x 
 (0.0560)  3.136 x 10 3  (0.0448)
2
 (0.0560)  0.219
 (0.0560)  0.219
or x 
2
2
0.1629
 0.275
x
or x 
so x  0.0815 mol/L or x  -0.137 mol/L
2
2
x
Since x = [HC2O4-] can’t be negative then [HC2O4-] = x = 0.0815 M and
[H2C2O4] = 0.20 – x = 0.20 M – 0.0815 M = 0.118 M
For the second dissociation
HC2O4 (aq) + H2O (l)  H3O+ (aq) + C2O42- (aq)
0.0815 M
0.0815 M
0M
-x
+x
+x
0.0815M - x
0.0815 M + x +x
-
Initial
Change
Equilibrium
If we assume x << 0.0815 M, then
C O H O   x 0.0815
0.0815
HC O 
x  C O   5.4 x 10 M
2
K a  5.4 2 x 10 -5 
2
3

2
so

4
4
2
2
4
-5
2
Both assumptions are valid! (and check should be shown) and the pH should be
pH = - log [H3O+] = - log 0.0815 = 1.09
Page 7 of 13
[5]
8. (a) 25.00 mL of 0.800 M solution of the weak acid HNO2 (aq) is titrated with 1.25 M
NaOH (aq). Calculate the pH after 12.50 mL of titrant solution has been added to
the solution of HNO2.
Ka HNO2 = 4.5 x 10–4
After adding the NaOH then Vtotal = 25.00 mL + 12.50 mL = 37.50 mL.
Moles of HNO2 = CV = (0.800 M)(25.00 mL) = 20.00 mmol
Moles of OH- in the added NaOH is mol = CV = (1.25 M)(12.50 mL) = 15.63 mmol
(In mmol)
Initial
Change
Final
HNO2 (aq) + OH- (aq) 
20.00
15.63
-15.63
-15.63
3.37
0
NO2- (aq)
0
+15.63
15.63
+ H2O (l)
N/A
N/A
N/A
We now have a buffer solution. The buffer has concentrations of
[HNO2] = moles HNO2 / Vtot = 3.37 mmol / 37.50 mL = 0.0899 M
[NO2-] = moles NO2- / Vtot = 15.63 mmol / 37.50 mL = 0.4168 M
P
(In M) HNO2 (aq) + H2O (l) 
Initial
0.0899
N/A
Change
-x
N/A
Eq’m
0.0899-x
N/A
NO2- (aq)
0.4168
+x
0.4168+x
+ H3O+ (aq)
0
+x
+x
We will assume that x << 0.0899 M and we know Ka = 4.5 x 10-4
Ka 
NO H O   0.416

2

3
HNO 2 
8
 0.089 9
so x  
 0.416 8
x
0.089 9

K a   0.216  4.5 x 10  4  9.7 x 10 5 M  H 3 O 





Assumption check: 9.7 x 10-5 / 0.0899  0.1 % < 5 % and pH =- log [H3O+] = - log 9.7 x 10-5 = 4.02
[5]
(b)
The titration started in part a) is continued to the equivalence point. What volume
of strong base has been added, and what is the final pH?
At the equivalence point the 20.00 mmol of acid we start with has been reacted an equal amount of
strong base to give a solution with only 20.00 mmol NO2- in water. The total volume of this
solution is
Volume added base = moles / C = 20.00 mmol / 1.25 M = 16.0 mL
so the Vtot = Vbase + Vacid = 25.00 mL + 16.0 mL = 41.0 mL and
[NO2-] = moles NO2- / volume = 20.00 mmol / 41.0 mL = 0.4878 M
P
(In M)
Initial
Change
Eq’m
We now have a base dissociation reaction!
NO2- (aq) + H2O (l) → HNO2 (aq) + OH- (aq)
0.4878
N/A
0
0
-x
N/A
+x
+x
0.4878-x
N/A
+x
+x
We will assume that x << 0.4878 M and we know Kb for this reaction is
Kw / Ka HNO3 = 1.0 x 10-14 / 4.5 x 10-4 = 2.22 x 10-11
K b  2.2 2 x 10 -11 
HNO 2 OH -   x 2
NO 
2
0.487 8
so x 2  1.0 8 x 10 11 and x   3.2 9 x 10 -6 M
Assumption check: 3.3 x 10-6 / 0.488 < 5 % so the assumption is ok.
pH = 14.00 – pOH = 14.00 – ( - log [OH-]) =14.00 + log 3.3 x 10-6 = 8.52
Page 8 of 13
[4]
What is the mass solubility (in mg L-1) at 25 C of thallium (III) hydroxide
[Tl(OH)3]?
9.
Ksp of thallium (III) hydroxide at 25 C is 6.3 x 10-46
The molar mass of thallium (III) hydroxide is 255.4052 g mol-1
Tl(OH)3 (s) ∏ Tl3+ (aq) + 3 OH- (aq)
If the molar solubility of Tl(OH)3 is s, then we get [Tl3+] = s, and [OH-] = 3s
Ksp = [Tl+] [OH-]3 = 6.3 x 10-46
(s)·(3s)3 = 6.3 x 10-46
(s)·(27s3) =6.3 x 10-46
27·s4 = 6.3 x 10-46
s4 = 2.33 x 10-47
s = 2.2 x 10-12 M
The molar solubility of Tl(OH)3 is 2.2 x 10-12 M. The relationship of the molar solubility to the
solubility is
solubility = (molar solubility) (molar mass)
solubility = (2.2 x 10-12 M)( 255.4052 g mol-1)
solubility = 5.6 x 10-10 g L-1 (1000 mg / 1 g)
solubility = 5.6 x 10-7 mg L-1
[3]
10.
A 200.0 mL volume of 0.300 M aqueous lithium chloride [LiCl] is mixed with
250.0 mL of an aqueous solution of 0.0200 M sodium phosphate [Na3PO4] at 25
C. Will solid lithium phosphate precipitate when these solutions are mixed?
Ksp for lithium phosphate at 25 C is 3.2 x 10-9
After mixing the total volume of the solution is
200.0 mL + 250.0 mL = 450.0 mL = 0.4500 L.
Moles of lithium chloride is mol = CV = (0.300 M)(0.2000 L) = 0.0600 mol
Moles of sodium phosphate is mol = CV = (0.0200 M)(0.2500 L) = 0.00500 mol
Mixing will give concentrations of
[LiCl] = moles LiCl / volumetotal = 0.0600 mol / 0.4500 L = 0.133 M
[Na3PO4] = moles Na3PO4/ volumetotal = 0.00500 mol / 0.4500 L = 0.0111 M
The precipitation is the reverse of the dissolution reaction for which we define Ksp
Li3PO4 (s)  3 Li+ (aq) + PO43- (aq)
 



Qsp  Li PO34  0.0111 0.133  1.368 x 10-6 0.133  1.8 x 10-7  K sp
3
3
precipitation will occur!
Page 9 of 13
11.
The reaction for the formation of hydrazine from ammonia is:
2 NH3 (g)  N2H4 (l) + H2 (g)
We know the following data for the reaction species at 298.15 K:
ΔHf (NH3 )  -46.1 kJ mol -1
ΔHf (N2 H 4 )  50.6 kJ mol -1
S (NH3 )  192.3 J K 1 mol -1 S (H2 )  130.6 J K 1 mol -1 S (N2 H 4 )  121.2 J K 1 mol -1
[1]
a)
Why was S provided for H2 (g), but not Hf?
Since H2 (g) is the element in its standard reference state the enthalpy of formation is defined as
zero. The standard entropy, however, is absolute and its value is known.
[1]
b)
Calculate H for the reaction at 298.15 K.



ΔH rxn   ν prod ΔH f,prod   ν react ΔH f,react

 


ΔH rxn  1 50.6 kJ mol1 - 2  46.1 kJ mol1  142.8 kJ mol1
[1]
c)
Calculate S for the reaction at 298.15 K.


  1130.6 J K

ΔSrxn   ν prodSprod   ν reactS,react

 1 121.2 J K 1 mol-1
1



mol-1  2  192.3 J K 1 mol-1

 132.8 J K 1 mol-1
[1]
d)
Calculate G for the reaction at 298.15 K.

ΔG   ΔH   TΔ S


ΔG   142.8 kJ mol1  298 K   132.8 J K 1 mol-1

ΔG  142.8 kJ mol

[1]
e)
1
  39.6 kJ mol
1

 182.4 kJ mol1
Is the reaction spontaneous at standard conditions at 298.15 K?
No, since the standard Gibbs energy change is positive.
[3]
f)
What is G for the reaction at non-standard conditions when the pressures of the
gases in the system are NH3 = 0.8125 bar and H2 = 1.00 x 10-5 bar at 298.15 K.
G =G +RT ln Q = G + RT ln (PH2)/(PNH3)2
G = 182.4 kJ mol-1 + (8.3145 J K-1 mol-1)(298.15 K) ln (1.00 x 10-5)/(0.8125)2
G = 182.4 kJ mol-1 + (2479.0 J mol-1) ln (1.515 x 10-5)
G = 182.4 kJ mol-1 + (2479.0 J mol-1) (-11.097)
G = 182.4 kJ mol-1 - 27.5 kJ mol-1
G = 154.9 kJ mol-1
B
Page 10 of 13
[3]
12.
For the reaction
H2 (g) + Br2 (g)  2 HBr (g)
At what temperature would we expect the value of the equilibrium constant Kp to
be 50.0?
Kp = 2.1 x 106 and H = -107.7 kJ mol-1 @ 1000 K.
 K   H   1 1 
  
ln 2  
R  T2 T1 
 K1 
-1
 1
1 
 50.0   107700 J mol

 
ln
 
6 
-1
-1 
 2.1 x 10   8.3145 J K mol  T2 1000 K 


1

 10.645  12953 K   0.001000 K -1 
 T2

12953K   12.953
 10.645 
T2
2.308 
[4]
13.
12953 K 
T2
so T2 
12953 K
 5612 K
2.308
Balance the following unbalanced oxidation-reduction reaction in basic solution.
Mn2+ (aq) + BiO3- (aq)  Bi+ (aq) + MnO4- (aq)
Oxidation: Mn2+ (aq)  MnO4- (aq)
Reduction BiO3- (aq)
 Bi+ (aq)
Oxidation: Mn2+ (aq) + 4 H2O (l)  MnO4- (aq)
Reduction BiO3- (aq)  Bi+ (aq) + 3 H2O (l)
Oxidation: Mn2+ (aq) + 4 H2O (l)  MnO4- (aq) + 8 H+ (aq)
Reduction BiO3- (aq) + 6 H+ (aq)  Bi+ (aq) + 3 H2O (l)
Oxidation: Mn2+ (aq) + 4 H2O (l)  MnO4- (aq) + 8 H+ (aq) + 5 eReduction BiO3- (aq) + 6 H+ (aq) + 4 e-  Bi+ (aq) + 3 H2O (l)
Oxidation: 4 Mn2+ (aq) + 16 H2O (l)  4 MnO4- (aq) + 32 H+ (aq) + 20 eReduction 5 BiO3- (aq) + 30 H+ (aq) + 20 e-  5 Bi+ (aq) + 15 H2O (l)
Combined: 4 Mn2+(aq) + H2O(l) + 5 BiO3-(aq) 4 MnO4- (aq) + 2 H+ (aq) + 5 Bi+ (aq)
In basic: 4 Mn2+(aq) + 5 BiO3-(aq) + 2 OH- (aq) 4 MnO4- (aq) + H2O(l) + 5 Bi+ (aq)
Page 11 of 13
[2]
14.
CHOOSE ONE of the following questions to answer:
(i)
Explain why termolecular elementary reactions are very rare.
Termolecular reactions require a collision of three reactant species to occur at exactly the same
time. The probability of this occurring is quite low and so termolecular elementary reactions are
quite rare.
(ii)
HCl and HClO4 are both strong acids in water, and therefore it is hard to tell which
is the stronger acid. Describe how we could determine which acid is truly stronger.
Water is a sufficiently strong base that it can react with strong acids to completion. If we want to
differentiate which of two acids is stronger, we need to allow them to react with a weaker base
than water, such as diethyl ether, to see which acid is truly better at donating protons.
[2]
15.
CHOOSE ONE of the following questions to answer:
(i)
Explain why the reaction between NH3 (aq) and Cu2+ (aq) to form [Cu(NH3)4]2+
(aq) is considered to be a Lewis acid/Lewis base reaction.
NH3 molecules contain a lone pair which makes them an electron pair donor (a Lewis base), while
the metal ion is electron deficient and willing to accept electron pairs (A Lewis acid). The
resulting reaction is a Lewis acid/base reaction.
(ii)
Define the second and third laws of thermodynamics.
Second law: For any process to be spontaneous requires an increase in the entropy of the
universe.
Third law: The entropy of a pure, perfect crystal at 0K is zero.
Page 12 of 13
[3]
16.
CHOOSE ONE of the following questions to answer:
(i)
Of the three bases, CH3NH2, NH2OH and NH3, CH3NH2 is the strongest and
NH2OH is the weakest. Explain this order of strength.
In each case we have a nitrogen atom with a lone pair making it a BL base. Anything that
increases the availability of the lone pair increases the base strength. A methyl group can donate
electrons to the nitrogen relative to a hydrogen atom, while the hydroxide group withdraws
electrons relative to the nitrogen. Therefore the lone pair is enhanced in CH3NH2 making it the
strongest base and reduced in NH2OH making it the weakest base.
(ii)
Of the three acids, HClO2, HBrO2 and HClO3, HClO3 is the strongest and HBrO2 is
the weakest. Explain this order of strength.
Oxoacid strength depends on two factors: the oxidation number of the central atom (which
increases with increasing oxygen atoms bonded to it) and the electronegativity of the central atom.
Cl is more electronegative than Br, so HBrO2 is weakest as the bromine is less effective in drawing
electrons away from the protons. HClO3 is stronger than HClO2 since the extra oxygen enhances
the electron withdrawal from the proton.
[2]
17.
CHOOSE ONE of the following questions to answer:
(i)
The entropy of a sample of ideal gas molecules increases with temperature.
Explain why.
As temperature increases, more energy levels are available in terms of motion. Many more
microstates (ways of distributing the energy) to achieve the same state are available, increasing
the entropy of the sample.
(ii)
The entropy of a sample of ideal gas molecules at a constant temperature increases
as the gas expands into a vacuum. Explain why.
As the gas expands into the vacuum (a bigger volume) the allowed energy levels for translational
motion get closer together. Many more microstates (ways of distributing the energy) to achieve
the same state are available, increasing the entropy of the sample.
The End
Enjoy the Holidays!
Page 13 of 13