NAME __________________
CHEMISTRY 162W15 Test#2
Show all calculations for full credit
Score __________/ 73
1. A – D – C – B – A
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2. B
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2
#3. 2
ways catalyst speeds up reaction.
a) _increase collisons_
b) _lowers Ea_
4. Hydration – H towards anions – ion/dipole interactions – O towards cations
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#5 Choose the best solvent that each solute would be most soluble in:
Solvents
Solultes
Solvent of Choice
EC: State which intermolecular forces are
CCl4
involved between the solute & solvent
HBr
_water_
__Dipole-Dipole_
water
KNO3
_water_
_Ion-Dipole_
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I2
CH3OH
C10H22
_CCl4_
_Dispersion_
_water_
_Hydrogen Bonding_
_CCl4_
_Dispersion_
E6
A 7a)EC:The change in FP or BP of a solution is dependent on the number of particles present in the solution
B 7b)EC: “Kf/b” values based on the pure solvent characteristics
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8.
rate law 1st order H2 & ICl, Rate = k[H2][ICl]: H2(g)+2 ICl(g) --2 HCl(g)+I2(g)
Which mechanisms consistent with observed rate law? b or d
PART II CALCULATION PROBLEMS ( pts)
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4 NH3(g)+3 O2(g) --2 N2(g)+ 6 H2O(g) N2(g) form at 0.68 mol·L-1·s-1. Use info, find:
1 [NH3 ]
1 [O 2 ]
1 [N2 ]
1 [H2O]


a) Rate  4 Δt
3 Δt
2 Δt
6 Δt
#1.
b) rate water formed
1 [ N 2 ] 1 [ H 2O]

2 t
6 t
6 0.68 mol N 2 mol H 2O
*
 2.04
2
Ls
mol N 2
c) rate NH3 react
d) rate O2 consume
1 [ N 2 ]
1 [ NH 3 ]
1 [ N 2 ]
1 [O2 ]
2 t
4 t
2 t
3 t
mol NH 3
mol O2
3 0.68 mol N 2
4 0.68 mol N 2
* - 1.06
* - 1.36
2
Ls
mol N 2
2
Ls
mol N 2
2
#2. solution = 12 g NaCl + 68 g H2O = 80.0 g solution mass % = (12 g/80.0 g) * 100 = 15%
#3) solubility Ar in water at 25 °C is 1.6  103 M ,when pressure of Ar above solution 1.0 atm.
Solubility Ar at 2.5 atm is __________ M.
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can set this problem up as a simple ration:
M Ar 25o C M Ar 25o C
0.0016 M
X



1 atm
2.5 atm
1 atm
2.5 atm
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X = (2.5*0.0016)/1 = 4.0*10-3 M
#4. data
for reaction diff temp and rate constants. Calculate the
a) label 2 columns and values correspond to 4 data points
3
b) determine activation energy, Ea
3
c) find constant, k, at 208oC
T, oC
k, s-1
1/T, K
ln k
190
2.52*10-5 0.00216 -10.589
199
5.25*10-5 0.00268
-9.855
-4
230
6.30*10
0.00199
-7.370
251
3.16*10-3 0.00191
-5.757
2
-5,000
0,00185
a) Ea determined from slope of line,
m = Y/X = -1.9*104 J
see bold inputs on chart
b) m = -Ea/R
Ea = -mR
Ea = -(-1.9*104 J)*(8.31) = 1.58*105 J
c) T2=208=481 K 1/T2 = 0.00208 k2 = ???
T1=1/T1 = 0.00216
Ln k1 = -10.589
0,0019
0,00195
0,002
0,00205
0,0021
0,00215
0,0022
-6,000
k 2  Ea  1 1 
 -  use any T data

k1
R  T2 T1 
Ln k 2
 1.58 *105  1
1 



- 10.589
8.31  481 463 
Ln
-7,000
-8,000
-9,000
-10,000
Ln k2 – -10.589 = -19013.24*(.00208 - .00216)
= -19013.24*(-.00008) = 1.521
-11,000
Ln k2 = 1.521 – 10.589 =-9.068
Ln k2 = e-9.068 = 1.15*10-4
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#5.
Complete and write the general rate equation: rate = k[reactant 1]p[reactant 2]q[reactant 3]r
Using the data in the table determine the values of k, p, q, & r in the rate equation for the reaction below.
BrO3-(aq) + 5 Br-(aq) + 6 H+(aq) ---- 3 Br2(aq) + 3 H2O(aq)
All values are initial concentrations (mols/L) & rate (mols/Ls)
[BrO3-]
0.10
0.20
0.10
0.20
[Br-]
0.10
0.10
0.30
0.10
[H+]
0.10
0.10
0.10
0.15
rate
1.2*10-3
2.4*10-3
3.5*10-3
5.4*10-3
solution
use rate2/rate1 to determine order for BrO3-
use rate3/rate1 to determine order for Br-
rate2 2.4 *10 2

2
rate1 1.2 *10 3
rate3 3.5 *10 3

 2.9
rate1 1.2 *10 3
k[0.20]p [0.10]q [0.10]r
2
 2p
p
p
r
k[0.10] [0.10] [0.10]
k[0.10] p [0.30]q [0.10]r
2.9 
 3q
p
q
r
k[0.10] [0.10] [0.10]
2  2p  1  p
3 .1  3 q 
use rate4/rate2 to determine order for H+
1q
use #1 to fine rate constant, k
3
rate4 5.4 *10

 2.25
rate2 2.4 *10 3
rate 1  1.2 *10 -3
k[0.20]p [0.10]q [0.15]r
2.25 
 1.5 r
p
q
r
k[0.20] [0.10] [0.10]
1.2 *10 3  k[0.10][0.1 0][0.10] 2
k  12 M-3s 1
 2.25  0.8
2.25  1.5 r Ln 
2

 1.5  0.4
2r
p=1
q=1
r=2
k = 12 M-3s-1
rate = 12 M-3s-1[BrO3-][ Br-][ H+]2
#6) What is the [He] in water when the partial pressure of helium above a sample of water is 494 mmHg.
PHe = KH*PoHe
KH = 3.70*10-4 M/atm
atm = 494 mmHg*(760 atm/mmHg) = 0.650 atm
PHe = (3.70*10-4)*(0.650 atm) = 2.41*10-4
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7) freezing point lowering: (freezing point)solv – [(# ions * m)*Kf)] = 8.60 – [(2*2.0)*0.78] = 8.60 – 3.12 = 5.48oC
3
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#8.
data decomposition of N2O5(g) at 45oC. a) graph find order b) Find value k c) Find half-life
d) initial amt 1.10 M, left after 4 half-lives.
time
[N2O5] ln [N2O5] 1/[N2O5]
5
a) 1st Order
(min)
0
6.67
13.33
20.00
26.67
33.33
1.40
1.10
0.87
0.68
0.53
0.42
0.336
0.095
-0.139
-0.386
-0.635
-0.868
0.714
0.909
1.149
1.471
1.887
2.381
b) from slope, use ln [ ] data
3
k
- 0.868 - 0.336
 3.61* 10- 2 /min
33.33  0.0
c) T0.5 = ln 2/k = 0.693/(3.61*10-2)
= 19.2 min
2
d) amt left = inital amt/2n = 1.10/24
= 0.069 mol
2
2nd Order [N2O5]-1 vs t
1st Order ln [N2O5] vs t
TAKE HOME TEST PART
25 PTS
#1) Soln made 27.1 g K2SO4 & 1000 g water. What vapor pressure of water above soln @ 100oC?
Assumptions: 1)pure water boils @ 100oC, 2)its vapor pressure is 1 atm (760 torr).
Vapor pressure above soln depends on mole fraction of water based on # moles solute in soln.
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Since 1 mol K2SO4 give 3 mols solute particles, then 27.1 g gives 3*(27.1/174.3) = 0.466 mol solute particles
OR,
K2SO4 = 174.3 g/mol
mols K2SO4 = 27.1 g*(1 mol/174.3 g) = 0.1555 mol*(3 ions) = 0.466 mol solute particles
mols H2O = 1000 g*(1 mol/18.0 g) = 55.56 mol H2O
XH2O = (mol fraction H2O)/(total mol soln) = (55.56 mol H2O)/(55.56+0.466) = 55.56/56.026 = 0.992
P = 1 atm*0.992 = 0.992 atm (754 torr)
#2) Benzene (C6H6) & toluene (C7H8) solution. 60.0oC VP benzene 0.507 atm & VP toluene 0.184 atm. What is VP
at 60.0oC of solution 6.5 g benzene & 23.0 g toluene. Which gas most
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info needed: Raoult’s Law: Psoln = X*Pgas benzene = 78.0 g/mol
toluene = 92.0 g/mol
st
1 find mols:
mol benzene = 6.5/78.0 = 0.0833 mol
mol toluene = 23.0/92.0 = 0.25 mol
2nd mole fraction: Xben = 0.0833/(0.0833+0.25) = 0.25
Xtolu = 0.25/(0.0833+0.25) = 0.75
3rd partial P:
Pben = 0.25*0.507 atm = 0.127 atm
Ptolu = 0.75*0.184 atm = 0.138 atm
4th find P:
Psoln = Pben + Ptolu = 0.127 + 0.138 = 0.265 atm
5th “most” & %:
% based on mole fraction: Toulene
0.138
*100  52.1%
0.138  0.127
Richer in Toulene (0.138 atm)
#3) Soln dissolve 1.27 g [mw: 310 g/mol] in 50 g water, f.p. soln is -0.762oC. Find amt solute particles formed
using: T = Kf*m*i need to solve for “i”
i = T/(Kf*m)
1st find mols: moles solute: 1.27 g*(1 mol/310 g) = 0.004097 mols (4.1*10-3)
2nd molality: m = moles solute/Kg solvent = (4.1*10-3 mol)/(0.050 Kg H2O) = 0.0819 m
3rd solve i:
i = T/(Kf*m) = (0.762 oC)/(1.86)(0.0819 m) = 0.762/0.152 = 5.0 particles
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