# Balance the following reaction: Cr(s) + CrO 4 2- (aq)

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```					    Balance the following reaction:
Cr(s) + CrO42-(aq)           Cr(OH)3 (s)

•   Cr                  Cr(OH)3
•   Balance as if it is in acidic solution
•   Cr + 3H2O                     Cr(OH)3 + 3H+
•   Balance the H+ by adding equal number of OH- to each side
•   Cr + 3H2O +3OH-                  Cr(OH)3 + 3H+ + 3OH-
•   Balance electrons
•   Cr + 3H2O +3OH-                 Cr(OH)3 + 3H+ + 3OH- + 3e-1
•   Make every H+ and OH- as H2O on each side and sum up
•   Cr + 3OH-               Cr(OH)3 + 3e-1 (Eq 1)
•   CrO42-(aq)                          Cr(OH)3 (s)
•   Balance as if it is in acidic medium
•   CrO42-(aq) + 5H+                             Cr(OH)3 (s) + H2O
•   Balance the H+ by adding equal number of OH- to each side
•   CrO42-(aq) + 5H+ + 5OH-                     Cr(OH)3 (s) + H2O + 5OH-
•   Balance the electrons
•   CrO42-(aq)+5H++5OH- + 3e-1                 Cr(OH)3 (s)+H2O +5OH-
•   Make every H+ and OH- as H2O on each side and sum up
•   CrO42-(aq)+4 H2O + 3e-1                 Cr(OH)3 (s) +5OH- (Eq 2)
•   Add Eq 1 to Eq 2
•   Cr + 3OH-                Cr(OH)3 + 3e-1        (Eq 1)
•   CrO42-(aq)+4 H2O + 3e-1                 Cr(OH)3 (s) +5OH- (Eq 2)

• Cr(s) + CrO42-(aq) + 4H2O               2Cr(OH)3 (s) + 2OH-
Chapter 5

Gases
Topics
 Pressure
 Gas laws: Boyle, Charles and Avogadro
 The ideal gas law
 Gas stoichometry
 Dalton’s law of partial pressures
 The kinetic molecular theory of gases
 Effusion and diffusion
 Real gases
5.1 Pressure

Properties of a gas
-   Uniformly fills any container.
-   Mixes completely with any other gas
-   Exerts pressure on its surroundings.
-   Compressible
-   Gas pressure varies with altitudes and
storms
Measuring atmospheric pressure
Torricellian barometer

   Torricelli (1608-1647) studied the problem using
mercury rather than H2O.
   Mercury is denser than water, so the column wasn’t
quite so high.
Pascal (SI units)
   Gas Pressure
Force (N)
P (Pa) =
Force per unit area            Area (m2)
Pascal and Torricelli

Blaise Pascal    Evangelista Torricelli
(1623-1662)      (1608-1647)
Torricellian Barometer

760 mmHg
P = d·g·h

d - density         h
g - acc. of
atmospheric
gravity             atmospheric
pressure                            pressure
Units of Pressure

One atmosphere (1 atm)
is the average pressure of the atmosphere
at sea level
is the standard atmospheric pressure
Standard Atmospheric Pressure:
1 atm = 76 cm Hg = 760 mm Hg = 760 Torr =
101,325 Pa     Very small unit, thus it is not
commonly used
Example
A. What is 475 mm Hg expressed in atm?
485 mm Hg    x   1 atm   = 0.625 atm
760 mm Hg

B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg
14.7 psi 1.00 atm
Manometer
Device for Measuring the Pressure of a confined gas
5.2 The Gas Laws of Boyle, Charles

 Boyle’s Law:       PV = const
 at constant n, T

 Charles’ Law:      V/T = const
 at constant n, P

 Avogadro’s Law:    V/n = const
 at constant P, T
Boyle’s Law                               1
V
Slope= 1/k
Boyle's Law
p
0.012

0.01
Boyle's Law                   PV =k
1/Pressure (1/Torr)

800
0.008
700
(at constant
0.006
600                                              T and n)
Pressure (Torr)

500
0.004
P1V1 = P2V2
400
0.002
300
0
200
0    50          100       150   200
100                     Volume (L)

0
0       50          100        150   200
Volume (L)
A Plot of PV Versus P for Several Gases at
Pressure Below 1 atm
Boyle’s holds
Only at very
Low pressures

A gas strictly
obeys Boyle’s
law is called
Ideal gas
Charles’s Law             •V/T = b
•V = bT
(constant P & n)
•V1/T1 = V2/T2
Charles' Law
35

30

25
Volume (L)

20

15

10

5

0
0   100      200      300     400   500
Temperature (K)
Plots of V Versus T(Celsius) for Several Gases

Volume of a gas
Changes by 1
273
When the temp.
Changes by
1oC.
I.e., at -273oC ,
V=0 ???
All gases will solidify or liquefy before reaching zero volume.
• Vn
Avogadro’s Law                 • V = an
(constant P& T)

V V
 1       2

120

100

80
Volume (L)

60

40

20

0
0     1        2 moles 3   4   5
5.3 The Ideal Gas Law
k
V         Boyle’s law
p

V  bT    Charles's law
Universal gas
constant

Tn
V  R( )         PV  nRT
P
The Ideal Gas Law

PV = nRT
R = 0.0821 atm L mol-1 K-1
V nR P nR                    V RT
                           
T   P T V                    n   P
The Ideal Gas Law can be used to derive
the gas laws as needed!
Molar Volume
At STP

4.0 g He      16.0 g CH4    44.0 g CO2
1 mole          1 mole        1mole
(STP)           (STP)        (STP)

V = 22.4 L     V = 22.4 L   V = 22.4 L
The value of R
 What is R for 1.00 mol of an ideal gas at
STP (0 oC and 1.00 atm)?Given that
V of 1 mol of gas at STP= 22.4L

PV
PV  nRT R 
nT
1.00 atm22.4 L
atm. L

R 0.0821
R 1.00 mol(273K )
mol.K
Example
A reaction produces enough CO2(g)
to fill a 500 mL flask to a pressure of
1.45 atm at a temperature of 22 oC.
How many moles of CO2(g) are
produced?
PV
PV = nRT         n
RT

mol K      0.500 L
1.45 atm              
0.0821 atm L 295 K
The Ideal Gas Law: Final and initial state problems

PV
 nR  Constant
T
PV   PV
1
     1   2       2
(Ideal gas equation)
T    1
T          2
Ideal Gas Law

 The  ideal gas law is an equation of
state.
 Independent of how you end up
where you are at. Does not depend
on the path.
 The state of the gas is described by:
P, V, T and n
Example

A sample of helium gas has a volume of
0.180 L, a pressure of 0.800 atm and a
temperature of 29°C. What is the new
temperature(°C) of the gas at a volume of
90.0 mL and a pressure of 3.20 atm?
Data Table

Set up Data Table

P1 = 0.800 atm      V1 = 0.180 L   T1 = 302 K

P2 = 3.20 atm       V2= 90.0 mL         ??
T2 = ??
Solution

Solve for T2
Enter data

T2 = 302 K x        atm x    mL =   K
atm      mL

T2 =           K - 273 =     °C
Calculation

Solve for T2

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K
0.800 atm 180.0 mL

T2 = 604 K - 273 = 331 °C
Example

A gas has a volume of 675 mL at 35°C and
0.850 atm pressure. What is the
temperature in °C when the gas has a
volume of 0.315 L and a pressure of 802
mm Hg?
Solution
T1 = 308 K         T2 = ?
V1 = 675 mL        V2 = 0.315 L = 315 mL
P1 = 0.850 atm     P2 = 802 mm Hg
= 646 mm Hg

T2 = 308 K x 802 mm Hg x        315 mL
646 mm Hg          675 mL
= 178 K - 273 = - 95°C
5.4 Gas Stoichiometry

 At  Standard Temperature and Pressure
(STP, 0ºC and 1 atm) 1 mole of gas
occupies 22.4 L.
 If not at STP, use the ideal gas law to
calculate moles of reactant or volume
of product.
Example

A.What is the volume at STP of 4.00 g of CH4?
4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH4    1 mole CH4

B. How many grams of He are present in 8.0 L of gas
at STP?
8.00 L x   1 mole He x    4.00 g He = 1.43 g He
22.4 He        1 mole He
Example
A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC.
Determine the pressure in the cylinder.
P = nRT
PV = nRT
V
= (3.74 mol)(0.082L•atm)(297.5K)
P= ?                                    (12.25 L)     mol•K
V = 12.25 L
75.5 g   mol       = 3.74 mol           = 1.009 atm
n=
20.18 g                        = 5670 torr
R = 0.082 L•atm
mol•K
T = 24.5 + 273 = 297.5 K
Example

 30.2mL of 1.00 M HCl are reacted
with excess FeS. What volume of
gas is generated at STP?

2 HCl + FeS  FeCl2 + H2S(g)

# moles HCl = Vol (L) X M
2 HCl + FeS  FeCl2 + H2S
             1.00 mol HCl mol H 2S  0.0821atm L 
 0.0302 L 1..00 mol HCl  mol H 22S  0.0821atm L  273.15 K 
             1 00 mol HCl mol H 2S 
0..0302 L 1 .00 mol HCl  2 mol HCl           

 mol K  273.15 K 
0 .0302 L

 0 0302 L           L                             
                    L
L
L        2 mol HCl  mol K
2mol HCl            

1.00 atm

 0.339 L
PV = nRT
V = nRT/P
Example
The decomposition of sodium azide, NaN3, at
high temperatures produces N2(g). What
volume of N2(g), measured at 735 mm Hg and
26°C, is produced when 70.0 g NaN3 is
decomposed.

2 NaN3(s) → 2 Na(l) + 3 N2(g)
2 NaN3(s) → 2 Na(l) + 3 N2(g)
Determine moles of N2:
1 mol NaN3       3 mol N2
nN2 = 70 g NaN3 X                   X              = 1.62 mol N2
65.01 g N3/mol N3 2 mol NaN3

Determine volume of N2:
nRT   (1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)
V=     =
P                        1.00 atm
(735 mm Hg)
760 mm Hg

= 41.1 L
Molar mass of a gas
PxV=       nxRxT
m
n
P x   V=  mxRxT                M
M
 m = mass, in grams
 M = molar mass, in g/mol
 Molar mass = m R T
PV
Density
 Density(d) is mass divided by volume
P x V =   mxRxT
M
P   =       mxRxT
V xM
d   =       m
V                PM
P      =   dxRxT
     d
M              RT
Example
A glass vessel weighs 40.1305 g when clean, dry and
evacuated; it weighs 138.2410 when filled with water
at 25°C (d=0.9970 g cm-3) and 40.2959 g when filled
with propylene gas at 740.3 mm Hg and 24.0°C.
What is the molar mass of polypropylene?

PV  nRT                       PV 
m
M
RT

mRT
M                      Volume of the vessel
PV
m   H 2O
=
138.2410 g – 40.1305 g
-3

d                  (0.9970 g cm )
H 2O

= 98.41   cm3 = 0.09841 L

mgas = mfilled - mempty = (40.2959 g – 40.1305 g)

= 0.1654 g
m                  m RT
PV = nRT       PV =   RT            M=
M                   PV

(0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K)
M=
(0.9741 atm)(0.09841 L)

M = 42.08 g/mol
Example
Calculate the density in g/L of O2 gas at STP.

From STP, we know the P and T.

P = 1.00 atm         T = 273 K
Rearrange the ideal gas equation for moles/L

d=      PXM
RxT
PXM
d
RXT
g
(1.00 atm ) X (32.0    )
d                      mol    1.43 g / L
L.atm
(0.0821         )X (273K)
mol.K

The density of O2 gas at STP is
1.43 grams per liter
Example
 2.00 g sample of SX6(g) has a volume of
329.5 Cm3 at 1.00 atm and 20oC. Identify the
element X. Name the compound
 P= 1.00 atm
1L
V  392.5 Cm3X          0.3295 L
1000Cm3
T   = 273+20 = 293K
m RT
M=
PV
L.atm
(2.00 g )(0.0821       )(293 K )
MM                   mol.K
(1.00atm)(0.3295 L)
= 146 g SX6 /mol
f
Molar mass of (X6 )= 146- 32 = 114 g/mol

Molar mass of X = (114 g/mol X6) /6 = 19

X = with a molar mass of 19 = F

The compound is SF6
5.5 Dalton’s Law of Partial Pressures

 For a mixture of gases in a container,
the total pressure is the sum of the
pressure each gas would exert if it were
alone in the container.
 The total pressure is the sum of the
partial pressures.
 PTotal = P1 + P2 + P3 + P4 + P5 ...

nRT
 For   each gas:   P
V
•Partial pressure
–Each component of a gas mixture exerts a pressure
that it would exert if it were in the container alone
 PTotal   = n1RT + n2RT + n3RT +...
V     V       V
 Inthe same container R, T and V are
the same.
 PTotal   = (n1+ n2 + n3+...)RT
V
RT
Thus,      PTotal
n ( )
Total
V
A 250.0 mL flask contains 1.00 mg of He and 2.00 mg
of H2 at 25.0oC. Calculate the total gas pressure in the
The total pressure is due to the partial pressures of each
of these gases.
 RT 
so:   Ptotal  PHe  PH 2    (n He  n H 2 )    
 V 
For He:
1.00 x 10-3 g He mol = 2.50 x 10-4 mol He
_____________________
4.00 g
For H2:
2.00 x 10-3 g H2   mol
______________________ = 9.92 x 10-4 mol H2
2.016 g
A 250.0 mL flask contains 1.00 mg of He and and 2.00
mg of H2 at 25.0oC. Calculate the total gas pressure in
so: Ptotal  PHe  PH  (n He  n H )
 RT 

 V 
2               2

For He:      1.00 x 10-3 g He mol = 2.50 x 10-4 mol He
_____________________
4.00 g

For H2:       2.00 x 10-3 g H2   mol
______________________ = 9.92 x 10-4 mol H2
2.016 g

And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V)

= (0.001242 mol)(0.0821 L•atm)(25 + 273)K
mol•K (0.2500 L)
Ptotal= 0.1216 atm
A 250.0 mL flask contains 1.00 mg of He and and 2.00
mg of H2 at 25.0oC. Calculate the total gas pressure in
so: Ptotal  PHe  PH  (n He  n H )
 RT 

 V 
2              2

For He:    1.00 x 10-3 g He   mol
_____________________ = 2.50 x 10-4 mol He
4.00 g

For H2:      2.00 x 10-3 g H2   mol
______________________ = 9.92 x 10-4 mol H2
2.016 g
Calculate the pressure due just to He (???):

n He RT = 0.0245 atm
PHe 
V
and Phydrogen= ?   0.1216 - 0.0245 = 0.0971 atm
Magnesium is an active metal that replaces hydrogen
from an acid by the following reaction:

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

How many g of Mg are needed to produce 5.0 L of H2 at a
temperature of 25 oC and a pressure of 745 mmHg?

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
?g                             5.0 L
Hint: find moles of H2 using PV = nRT then work as a
stoichiometry problem.
mol•K
n = PV =____________________________________
745 mmHg 5.0 L
RT                   62.4 L•mmHg 298 K
n = 0.20 mol
The mole fraction

 Mole fraction: number of moles of one
component in a mixture relative to the
total number of moles in the mixture

 symbol     is Greek letter chi   c

n                     n
c           1
=           1

n
1

Total
n  n  n  ....
1   2        3
Mole fraction expressed in pressures
n          n
c                             1                                  1

n     n  n  n  ....    Total           1            2           3

V            V
n  P ( ); n  P ( ); ..........
1               1                               2            2
RT           RT
V
n
P(  1
)                                       V
c            1                                                 RT                                       P( )    1
1
n       V       V       V                                                             c                RT
P ( )  P ( )  P ( )  .....
Total

RT      RT      RT
1                   2                    3                1
V
( )( P  P  P  ....)
1           2   3

P            P                                                                          RT
c                                              1                                  1

( P  P  P  ....) P
1
n    P
1               2           3                        Total
c                 1                 1

P cP
1
n    P     Total                Total
1                               1           Total
Example
   A 1.00 L sample of dry air at 786 Torr and
25 oC contains 0.925 g N2 plus other
gasses (such as O2, Ar and CO2.) a) What
is the partial pressure of N2? b) What is
the mole fraction of N2?
mol                 PV = nRT
0.925 g           0.0330 mol P = nRT/V
28 .0 g
0.0330 mol 0.0821atm L mol 1 K 1 298 K   0.807 atm
1.00 L
PN  c N PTotal
760Torr                   613 Torr
0.807 atm           613Torr X N 2            0.780
atm                     786 Torr
Collecting gas over water
An insoluble gas is passed into a container of water, the
gas rises because its density is much less than that of
water and the water must be displaced
O2 gas
KClO3
Collection of Gases over Water

Assuming the gas is saturated with water vapor, the
partial pressure of the water vapor is the vapor
pressure of the water.

Ptotal = Pgas + PH2O(g)

Pgas = Ptotal – PH2O(g)
Example
Oxygen gas generated was collected over water. If the
volume of the gas is 245 mL and the barometric pressure
is 758 torr at 25oC, what is the volume of the “dry” oxygen
gas at STP? (Pwater = 23.8 torr at 25oC)

PO2 = PTotal - Pwater = (758 - 23.8) torr = 734 torr

P1V1 P2V2           P1V1T2
P1= PO2 = 734 torr;
P2= SP = 760. torr
        V2  P T
T1   T2              2 1
V1= 245mL
T1= 298K;
T2= 273K;      V2 = (245mL)(734torr)(273K)
V2= ?                 (298K)(760.torr)
=217 mL
5.6 The Kinetic Molecular Theory of Gases
   It explains why ideal gases behave the way they do.
 Postulates of the kinetic Theory:
 Gas particles (atoms or molecules) are
so small compared with distances
between them, thus we can ignore their
volume.
 The particles are in constant motion
and their collisions with walls cause
pressure exerted by the gas
Kinetic Molecular Theory
 The particles do not affect each other,
neither attracting or repelling
 The average kinetic energy (KE ) is
proportional to the Kelvin temperature.
 KE = 1/2 mu2
 m=mass of gas particle
 v=average velocity of particles
KMT explains ideal gas laws
• P&V: P = (nRT) . (1/V)  P 1/V
– When V decreases # collisions increases
• P & T: P = (nR/V).T  P T
– When T increases, hits with walls become stronger and more frequent
• V & T: V=(nR/P).T  V  T
– When T increases hits with walls become stronger and more frequent.
To keep P constant, V must increase to compensate for particles speeds
• V & n: V= (RT/P). N  V  n
– When n increase P would increase if the volume to be kept constant. V
must increase to return P to its original value
• Dalton’s law: Individual particles are independent of each
other and their volumes are negligible. Thus identities of gas
particles do not matter and would be treated as if they belong
to one gas
Driving the ideal gas law from KMT
• The following expression was derived for pressure of an
ideal gas:
2 nN A (1 / 2mu 2 )
P [                  ]
3        V

A

m  mass of each particle
u  average of square velocities of particles
2

(KE)  N (1/2mu )
avg         A
2
2  n(KE) 
P  [       ]
avg

3  V     
PV 2
 ( KE )  T
avg
n  3
PV
T
n
PV
RT
n
The meaning of temperature

PV                    2
RT            =     ( KE)   avg
n                    3

( KE)            3
avg   =   RT
2
• Thus, Kelvin temperature is an index of the random
motion of the gas particles. That is, with higher
temperature there is a greater motion.
• The same is applicable for liquids and solids
Root mean square velocity
Combine   these two equations

 (KE)avg = NA(1/2 mu 2 )
 (KE)avg = 3/2 RT

 Where M is the molar mass in kg/mole,
and R has the units 8.3145 J/Kmol.
 The velocity will be in m/s
Molecular speed for same gas at two different temperatures

3RT
u (  T1
)          1    1/ 2

M
3RT
u (
T2
)       2       1/ 2

M
u   T
( )    T1       1      1/ 2

u   T     T2       2
Molecular speed for two different gases at two different
temperatures

3RT
u (
T1
)         1           1/ 2

M         1

3RT
u (
T2
)            2        1/ 2

M            2

u    TM
(  T1
)  1               2      1/ 2

u    TM
T2   2               1
Range of velocities

 The  average distance a molecule
travels before colliding with another is
called the mean free path and is small
(near 10-7)
 Many collisions among gas molecules
produce large number of velocities
 The actual distribution of molecular
velcities is shown in the next slids
Effects of Molar Mass on urms

At constant T, 273 K, the
most probable speed for
O2 > CH4 > H2

urms  M–1/2
so smaller molar masses
result in higher molecular
speeds
EOS
Effects of Temperature on urms

At constant mass the most
probable speeds for O2
increase with temperature

urms  T1/2 so higher
temperatures result in
higher molecular speeds
Summary of Behaviors
EOS
number of particles   273 K

Molecular Velocity
number of particles   273 K

1273 K

Molecular Velocity
number of particles   273 K

1273 K

1273 K

Molecular Velocity
5.7 Effusion and Diffusion
 Effusion   is the passage of gas through
a small hole, into a vacuum.
 The effusion rate measures how fast
this happens.
 Graham’s Law : the rate of effusion is
inversely proportional to the square
root of the mass of its particles.
Rate of effusion for gas 1   M2

Rate of effusion for gas 2   M1
Deriving Graham’s law

 The   rate of effusion should be
proportional to urms
   Effusion Rate 1             urms for gas 1
=
Effusion Rate 2              urms for gas 2
3RT
effusion rate 1     u rms 1      M1       M2
                   
effusion rate 2     u rms 2      3RT      M1
M2
Diffusion

 The  spreading of a gas through a room.
 Slow considering molecules move at
100’s of meters per second.
 Collisions with other molecules slow
down diffusions.
 Best estimate is Graham’s Law.

Distance traveled by gas A rms for gas A   MB
               
Diatance traveled by gas B rms for gas B   MA
5.8 Real Gases

 Real  molecules do take up space and
they do interact with each other
(especially polar molecules).
 Correction factors are needed to be
added to the ideal gas law to account
for these factors.
Volume Correction

of molecules is less than the ideal volume
because of particle size.
 More molecules will have more effect.
 Actual volume
V   V - nb
 b is a constant that differs for each gas.
 The ideal gas equation is modified as follows:

nRT                  Volume of gas particles

P                        Has been taken into

V  nb
account
Pressure correction

 Because   the molecules are attracted to
each other, the pressure on the container
will be less than ideal
 This effect will depend on the
concentration of gas molecules (n/V)
 since two molecules interact, the effect
must be squared: (n/V)2
Pressure correction
   Thus the pressure corrected for
attraction of molecules is given by

n 2
Pobserved  P  a( )
V
a is a proportionality constant
Corrected Pressure

nRT        n 2
Pobs          a( )
V - nb     V
Volume correction   Pressure correction
Volume of container
Van der Waals equation

n 2
[ Pobs  a( ) ]  (V - nb)  nRT
V
Corrected pressure   Corrected volume

Pideal            Pideal
   a and b are determined by experiment.
   Different for each gas.
   Bigger molecules have larger b.
   a depends on both size and polarity.
Example

 Calculate the pressure exerted by
0.5000 mol Cl2 in a 1.000 L container at
25.0ºC
 Using the ideal gas law.
 Van der Waal’s equation
   a = 6.49 atm L2 /mol2
   b = 0.0562 L/mol

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