Solution W2009 NYB Final exam
Question 1
Consider one liter of solution with [H2SO4] = 3.75 mol/L. Then :
Mass of solution: 1000 mL x 1.23 g/mL = 1230 g solution
mass of H2SO4: 3.75 mol /L x 98.078 g/mol = 368 g
mass of water : 1230 g - 368 g = 862 g water
mole of water : 862g x (1 mol/18.02 g) = 47.9 mol water
a) mass percent H2SO4: 368 g / 1230 g x 100% = 29.9%
b) molality H2SO4: 3.75 mol / 0.862 kg solvent = 4.35 molal
c) mole fraction of H2SO4: 3.75 mol H2SO4/(3.75 mol + 47.9 mol) = 0.0727
d) number of mole of H2SO4 required: 1.5L x 0.10 mol/L = 0.15 mol H2SO4
volume of 3.75M corresponding to 0.15 mol: 0.15 mol x (1 L/3.75mol) = 0.040 L or 40. mL
Question 2
NaCl = 58.44 g/mol, osmotic pressure (∏) = 7.97 atm,
T = 37.0°C
Concentration of NaCl: (0.242 g)x(1 mol/58.44 g) / 0.0250 L = 0.166 mol/L
Since ∏ = i M R T then i =
i=
Π
MRT
7.97atm
(0.166 mol/L)(0.08206 L.atm/K.mol)(37.0 + 273)K
= 1.89
Question 3
E
From k = A exp(- a )
RT
we can get
ln
k1
E " 1
1 %
−
= a$
'
R # T2 T1 &
k2
since k is used to find the rate, therefore = M/t1 ∝ k and k1 can be replaced by M/t1.
Since M is the same in both cases the equation become:
ln
" 1
t2
1 % " Ea %
−
= $
' x$ '
t1
# T2 T1 & # R &
à
" 1
1 % " Ea %
−
ln t2 = ln t1 + $
' x$ '
# T2 T1 & # R &
1
1 % " 453x103 J/ mol%
"
ln t2 = ln 3.00 + $
−
' x$
'
# 368.6K 373.2K& $# 8.31 J/mol.K '&
t2 = 18.6 min
Question 4
a.
Reaction rate = k[ClO2]n[OH-]m
If [OH-] and k are kept constant between two experiments (example: exp 1 and exp 2) then
[ClO2 ]n
Rate2
=
Rate1
2.30x10 -1
2
[ClO2 ]n1
5.75x10 -2
& 0.100 #
= $
!
% 0.0500 "
n
4 = 2n
therefore n = 2 for [ClO2]
(the units cancelled out, therefore, they are not written here to make the numbers more apparent)
Since n = 2, then m can be found by using the combination of the experiments 2 and 3
2
- m
Rate2 [ClO2 ]2 [OH ]2
=
Rate3
2
- m
[ClO2 ]3 [OH ]3
2.30x10 -1
Finally the rate law is:
rate = k[ClO2]2[OH-]
2.30x10 -1
2
& 0.100 # & 0.100 #
= $
! $
!
% 0.200 " % 0.02500 "
m
4 = 4m therefore m = 1 for [OH-]
b. Any experiment can be used to determine the rate constant:
rate = k[ClO2]2[OH-] then
experiment 2:
k=
c.
k=
2.30x10 -1 mol.L−1 .s −1
rate
[ClO2 ]2 [OH-]
k = 230. L2mol-2s-1
[0.100 mol.L-1 ]2 [0.100 mol.L-1 ]1
The overall order of this reaction is n+m = 3
Question 5
a.
If 85% have reacted, then the NOBr concentration is: 15/100 x 0.080 M = 0.012 M
Since it is a second order reaction, then
1
1
−
[NOBr]
[NOBr]0
1
1
=kt +
the equation becomes: t =
[NOBr]0
k
[NOBr]
1
t=
[NOBr]
−
1
[NOBr]0
k
1
-1
=
[0.012 mol.L ]
−
1
[0.080 mol.L-1 ]0
0.80 mol.L-1 s -1
= 89 s.
b. The half-life for a second order reaction is:
t1/2 =
c.
1
k [NOBr]0
True: ii.
t1/2 =
1
-1 -1
0.80 M s [0.080 M]0
False: i, iii, iv, v, vi.
t1/2 = 16 s
Question 6
2
a.
Pressure of HI at equilibrium can be obtained from: Kp
(PHI )
=
PH2S
Since Kp uses values in atm, the initial pressure of H2S has to be converted:
10.1 kPa x 1 atm/101.3 kPa = 9.97x10-2 atm
H2S(g)
9.97x10-2 atm
-x
9.97x10-2 – x atm
I
C
E
+
I2(s)
-x
-
2HI(g)
0
+2x
+2x atm
+
S(s)
+x
-
(2x )2
Therefore: 1.34x10-5 =
where 2x is the pressure of HI at equilibrium
9.97x10 -2 − x
Since Kp is small, then we can assume that: 9.97x10-2 >> x. In this case, the equation becomes
1.34x10-5 x 9.97x10-2 = 4x2
x=
1.34x10 −6
4
= 5.79x10-4
(the approximation was right since
5.79x10-4/9.97x10-2x100% = 0.58%)
In this case, the HI pressure at equilibrium is: 2x = PHI = 1.16x10-3 atm
b. since
Kp =
c.
PV = nRT à
[MHI ]2 xRT
[M ]
H2S
P = MRT
therefore K =
Kp
RT
then
à K =
Kp =
(MHIRT )2
MH2SRT
1.34x10 -5
(0.08206 L.atm.K -1 mol −1 )x(273.15 + 60)K
the reaction is inverted and the amount of material is doubled. Therefore:
K 'p
& 1
= $
$ Kp
%
#
!
!
"
2
K 'p
à
1
&
#
= $
!
-5
% 1.34x10 "
2
= 5.57x109
Question 7
a.
True: i, ii, iii, iv,
b. Left: iii, iv.
v.
False: vi.
no change: i, ii, vi.
right: v.
= 4.90x10-7
Question 8
a) percent dissociation =
[dissociated] =
[dissociated salt or acid]
[initial concentration]
percent dissociation x [initial]
100%
X 100%
8.0% x [1.58]
=
= 0.13 M
100%
pH = -log[H+] à pH = - log (0.13) = 0.89
b)
H+(aq)
HClO2(aq)
I
C
E
Ka =
[H+ ][ClO-2 ]
[HClO2 ]
=
1.58
-x
01.58 – x
[0.13][0.13]
[1.58 - 0.13]
+
ClO2(aq)
0
+x
+x
≈0
+x
+x
= 1.1x10-2
c) Calculate first the concentration of H+ or ClO2- in solution:
Ka =
(x2 )
4.5 - x
if x << 4.5 then x =
-2
(4.5)x(1.1x10
The percent dissociation is
0.22
4.5
) = 0.22
x100% = 4.9% therefore the approximation was right
Question 9
a)
CH3NH2
ClO4-
H 2O
strongest base
weakest base
b)
NH4Cl
KF
KNO3
Most acidic
most basic
0.514 g x
c)
KCN Concentration =
+
6.31x10-2
-x
6.31x10-2 – x
I
C
E
(x 2 )
6.31x10
65.12 g
0.125 L
CN-(aq)
Kb =
1 mol
-2
- x
if x << 6x10-2 then x =
= 6.31x10-2 M
HOH(l)
-
Kb =
Kw
Ka
HCN(aq)
0
+x
x
=
10 -14
6.2x10-10
+
= 1.6x10-5
OH(aq)
≈0
+x
x
(6.31x10-2 )x(1.6x10-5 ) = 1.0x10-3 = [OH-].
-3
pH = 14 - (-log 1.0x10-3) = 11.00
-2
(Check: (1.0x10 /6.31x10 )x100% = 1.6% Therefore, the approximation was right)
Question 10
a) The system is already at equilibrium.
Since the pH = 4.00 then [H+] = 1.0x10-4. Therefore:
HNO2(aq)
H+(aq)
0.20 M
1.0x10-4
Eq.
Ka =
[H+ ][NO -2]
NO2- =
à
[HNO2 ]
Ka x [HNO2 ]
[H +]
=
+
4.6x10-4 x 0.20 M
1.0x10-4 M
NO2(aq)
x
= 0.92 M
The mass of KNO2 required is: (0.92 mol/L) x 0.500 L x 85.11 g/mol = 39 g
Question 11
a) After the addition of 15.0 mL Ba(OH)2, the concentration of the species are:
[HCOOH] =
(0.0200 L) x (0.10 M)
(0.0200 + 0.0150) L
2HCOOH(aq)
Initial
reaction
product
= 0.057 M
+
0.057 M
-2 x 0.021 M
0.015 M
[Ba(OH)2] =
(0.0150 L) x (5.0x10-2 M)
(0.0200 + 0.0150) L
Ba(OH)2(aq)
Ba(HCOO)2(aq)
0.021 M
-0.021 M
0
0
+0.021 M
+0.021 M
+
= 0.021 M
2H2O(l)
excess
+2 x 0.021 M
excess
After the reaction, the following equilibrium is present
H+(aq)
≈0
+x
x
HCOOH(aq)
0.015 M
-x
0.015 M - x
I
C
E
+
HCOO-(aq)
0.042 M
+x
0.042 M + x
If x << 0.015 M then:
[H+ ][HCOO- ]
Ka =
[HCOOH]
à [H+] =
Ka x [HCOOH]
-
[HCOO ]
=
1.8x10-4 x 0.015 M
0.042 M
= 6.4x10-5 M
pH = -log(6.0x10-5) = 4.19
[H+ ][HCOO- ]
check: Ka =
[HCOOH]
=
(6.4x10-5 )(0.042 + 6.4x10-5 )
-5
(0.015 - 6.4x10
therefore, the approximation was right.
)
= 1.8x10-4
Question 11 (Cont.)
b) At equivalence point, 2n mole acid = n mole base
n mole acid = (0.0200 L) x (0.10 mol/L) = 2.0x10-3 mole acid
2.0x10-3 mole acid x
1 base
2 acid
= 1.0 x10-3 mole of base required
The volume of Ba(OH)2 required is: 1.0x10-3 mole Ba(OH)2 x
1L
5.0x10-2 mole
= 0.020 L of 20. mL
c) At equivalence point, Ba2+ and HCOO- are in large amount. However, only HCOO- is responsible for the
pH (conjugated base of a weak acid). Therefore:
[HCOO-] =
2.0x10-3 mole
0.0400 L
= 5.0x10-2 M
HCOO-(aq)
+
(x 2 )
if x << 5.0x10-2
5.0x10-2 - x
[H+] =
Kw
Ka
10-14
=
-4
1.8x10
= 5.6x10-11
H-OH(l)
HCOOH(aq)
-x
-
0
+x
+x
5.0x10-2
-x
5.0x10-2 – x
I
C
E
Kb =
Kb =
then x =
-11
(5.6x10
+
OH(aq)
≈0
+x
+x
-2
)x(5.0x10 ) = 1.7x10-6 = [OH-].
The approximation was right since x is less than 5% of 5x10-2.
10-14
-6
1.7x10
= 6.0x10-9
pH = -log(6.0x10-9) = 8.23
Question 12
Two reactions are present
2 NaI(aq) + Pb(NO3)2(aq)
NaI(aq) + AgNO3(aq)
PbI2(s) + 2 NaNO3(aq)
Ksp = 1.4x10-8
Ksp = 1.5x10-16
PbI(s) + NaNO3(aq)
a)
1.4x10-8 = [Pb2+][I-]2
[I-] =
1.5x10-16 = [Ag+][I-]
[I-] =
1.4x10 -8
0.10
1.5x10-16
-4
= 3.7x10-4 M
= 7.5x10-13 M
2.0x10
Ag will be the first one to precipitate when [I-] = 7.5x10-13 M
+
b) The second species will begin to precipitate when [I-] = 3.7x10-4 M. Therefore, the concentration of
Ag+ in solution when Pb2+ will begin to precipitate will be:
Ksp AgI = [Ag+][I-]
à
1.5x10-16 = [Ag+] [3.7x10-4]
à [Ag+] = 4.0x10-13 M
Question 13
ΔS < 0
a) i.
ii.
ΔS > 0
A liquid that boils
Sugar that crystallized out from a supersaturated sugar solution
iii. Iron rusts (formation of Fe2O3 from pure Fe and O2)
iv. A-B(g) + C-D(s)
v.
N2O4(g)
vi. NaCl(s)
2NO2(g)
H O
2
b) ΔG = ΔH - TΔS.
-ΔGo
-ΔSo
A-B-C(g) + D(s)
=T
Na+(aq) + Cl-(aq)
ΔHsol = +4.0 kJ/mol
At T = boiling temperature, the system is at equilibrium. Therefore, ΔG° = 0
T=
-58.51x103 J.mol−1
= 629.7 K or 365.5°C
−92.92 J.K-1mol−1
Question 14
ΔS° = 2 mol x (240.5
ΔH° = 2 x (33.8
since :
kJ
mol
J
K ⋅ mol
) - 1 mol x 304
) - 1 mol x 9.67
kJ
mol
J
K ⋅ mol
= +177
J
K
or 0.177
kJ
K
= +57.9 kJ
ΔG° = ΔH° - TΔS°
therefore: ΔG° = +57.9 kJ - (T)K x 0.177
a) if T = 25.0°C, then T = (25.0 + 273.15)K
ΔG° = +57.9 kJ - (298.2K) x 0.177
kJ
K
kJ
K
K
T = 298.2K
à ΔG° = + 5.1 kJ ΔG°
b) if T = 60.0°C, then T = (60.0 + 273.15)K
ΔG° = +57.9 kJ - (333.2K) x 0.177
kJ
> 0 therefore not spontaneous
T = 333.2K
à ΔG° = - 1.1 kJ ΔG°
< 0 therefore spontaneous
Question 15
Data for the Unknown Solute/Cyclohexane Solution
Mass of empty test tube, stopper, beaker
g
185.2235
Mass of test tube, stopper, beaker, & cyclohexane
g
204.5736
Mass of test tube, stopper, beaker, & unknown solute/cyclohexane solution
g
204.9847
Mass of cyclohexane
g
19.3501
Mass of unknown solute
g
0.4111
Freezing Temperature of unknown solute/cyclohexane solution
Molar mass of unknown solute
4.27
°C
g⋅mol
−
1
Mass of cyclohexane: 204.5736 - 185.2235 = 19.3501 g
Mass of the unknown: 204.9847 - 204.5736 = 0.4111 g
ΔTf = Kf x m à m =
m=
mol solute
kg solvent
ΔTf
Kf
à
=
(6.55 - 4.27)°C
-1
20.2°C.mol
= 0.113
kg
mol solute = (m) x (kg solvent)
mol solute = (0.113
finally, the molar mass is:
mol
molar mass =
mass
n
mol
kg
=
) x (19.3501x10-3 kg) = 2.19x10-3 mole
0.4111 g
2.19x10 −3 mole
= 188 g/mol
188