Lucky 13 and Equilibrium

1

• Lucky 13 and Equilibrium
• To this point, when dealing with chemical
  reactions, we have considered that the reaction
  go to __________________. By completion, we
  meant that one reactant (the _____________
  reactant) was consumed by the reaction. However,
  for many reactions, none of the reactant
  substrates are consumed by the reaction.
• An example 2 NO2 (g) ? N2O4 (g) fig. 13.1 pg.
  610
• This reactions will reach a point where the
  forward rate (rate of NO2 consumption) is
  ________ to the backward rate (rate of NO2
  production from N2O4). At this point the
  reaction is said to have reached a point of
  dynamic ____________________. An application of
  pressure will shift the equilibrium to the
  _________ for there are less product than
  reactant particles (thus less pressure) pg. 640.
  An increase in temperature will shift the
  equilibrium ________ due to this being an
  exothermic process (meaning heat is a product and
  the system will shift to reduce the stress of
  more product). pg. 641
• Sec 13.1 The Equilibrium Condition
• 1. At equilibrium the forward and backward rates
  of a reaction are ____________ (meaning that the
  amounts of reactant and product molecules do not
  change).
• There is a constant interconversion of reactant
  and product molecules so the equilibrium is said
  to be _____________.
• Sec 13.2 The Equilibrium Constant
• 1. The ratio of products raised to their
  stoichiometric powers and reactants raised to
  their stoichiometric powers is a constant.
• jA kB ? lC mD
• K
  Cl Dm means molarity
• AjBk

completion
limiting
equal
equilibrium
right
left
equal
dynamic
2
initial

• See table 13.1 pg. 616 to see that K is constant
  at a given temperature regardless of the
  ___________ set of concentrations of reactants or
  products (at equilibrium, the concentrations are
  called the equilibrium positions). There are
  an _____ number of equilibrium positions but only
  ___ value of equilibrium constant for a given
  reaction at a given temperature (as seen earlier
  temperature can shift the equilibrium).
• 2. The larger the equilibrium constant, the more
  the reaction will tend toward completion (but the
  size of K says nothing of the ______ of the
  reaction which is determined by the size of the
  activation energy barrier).
• Haber Process N2 (g) 3 H2 (g) ? 2NH3 (g) See
  Summary Table pg. 615
• Sample Exercise 13.1-13.3

8
1
rate
13.2b.
K N2H23 NH32
13.2c. K NH3
N21/2H23/2
13.2a. K NH32
N2H23
13.1 K NO24H2O6
NH34O27
K 1 3.795.104
K 0.0312
0.850.00313
K (3795)1/2
K 190
K 3.8.104
K 2.6.10-5
Sec 13.3 Equilibrium Expressions Involving
Pressure 1. Concentration is related to pressure
through PV nRT or n / V P / RT or C
P / RT (could that be an ice cream P/RT?) so P
CRT. 2. An equilibrium expression can be written
as pressures. Kp PNH32 Kc NH32 Kp NH3
2 (RT)2 Kp Kc (RT)-2 PN2 PH23
N2H23
N2(RT)H23(RT)3 (see pg. 619) Kp
KC (RT)?n ?n ?n products - ?n reactants
3
Sample Exercise 13.4 pg. 618-620
Sample Exercise 13.5 pg. 618-620
Kp PNOCl2 PNO2 PH2
Kp KC( RT)?n
Kc Kp /( RT)?n
Kp (1.2 atm) 2 (0.050 atm)2(0.30 atm)
KC 1920 atm-1 / (0.08206 Latm/molK(298K))2-3
Kp 1.9.103 atm-1
KC 1920 / (0.08206(298))-1
KC 4.7.104

• Sec 13.4 Heterogeneous Equilibria
• 1. So far we have examined equilibria where all
  species are gases (one phase). A homogeneous
  equilibrium consists of ________ phase (such as
  gases). Many equilibria contain more than one
  phase (any combination of solids, liquids or
  gases) and are called _______________________
  equilibria.
• 2. Experimental results have shown the
  equilibrium constants do not depend upon
  _________ of solids or liquids (as their
  concentrations of pure substances cannot change).
  
• 3. Since the concentrations do not change, pure
  solids and liquids are _______________ from the
  equilibrium constant expressions.
• Sample Exercise 13.6 pg. 621-622

one
heterogeneous
amount
excluded
Homework pg. (dont need to write 9, 10, 11
15) write 19a,b 20 c,d 21 25 29 31 42 (use
the chart technique
4

• Sec 13.5 Applications of the Equilibrium
  Constant
• Knowledge of the equilibrium constant indicated
  the tendency for a reaction to occur. The larger
  the equilibrium constant the ______ likely a
  reaction is to occur (as discussed yesterday K
  gives no indication of rate of reaction).
• Reaction Quotient
• 1. If the reactants of a reaction are mixed with
  no products, the system will shift to the
  _________ to produce products until the
  equilibrium condition is attained. If only
  products are mixed then the system will shift
  ________ to produce reactants until an
  equilibrium is attained. How do we see which way
  some set of initial concentrations of reactants
  and products will shift? Answer Reaction
  quotient.
• 2. The reaction quotient (Q) is calculated in the
  _________ same fashion as the equilibrium
  constant (K), the only difference is the Q is
  calculated at the time the reactants and products
  are originally mixed (before the shift).
• 3. If Q gt K, then there is too large of a ratio
  of product to reactant and the system will shift
  _________.
• If Q lt K, then there is too large of a ratio
  of reactant to product and the system will shift
  _________.
• If Q K then the system is in a state of
  ___________________ (no shift).
• Dont worry to show the units on K, Kp or Q.
• See example 13.7 pg. 625

more
right
left
exact
left
right
equilibrium
5

• Exercise 13.7
• Q NH32
• N2 H23
• Q 1.0.10-32
• 1.0.10-52.0.10-33
• Q 1.3 .107
• Since Q gt K system will shift left to attain
  equilibrium
• b. Q 2.00.10-42
• 1.50.10-53.54.10-13
• Q 6.0 .10-2
• Since Q K system is at equilibrium (no shift)
• c. Q 1.0.10-42
• 5.0 1.0.10-23
• Q 2.0 .10-3
• Since Q lt K system will shift right to attain
  equilibrium

6

• Equilibrium Problems
• Sample Exercises 13.8 13.11 pg. 626-635
• When equilibrium constants are small the amount
  removed or added to some initial concentration
  can be neglected (use the 5 rule).

7

• Sample Exercise 13.8
• N2O4 (g) ? 2 NO2(g) Kp 0.133
• 2.71 atm ?
• KP PNO22
• P N2 O4
• P NO2 (Kp P N2O4)1/2
• P NO2 (0.133(2.71 atm))1/2
• P NO2 0.600 atm
• Sample Exercise 13.9
• PCl5 (g) ? PCl3 (g) Cl2
  (g) K ?
• Initial 0.00870 mol 0.298 mol 0 mol (no
  product so shift right)
• - 0.00200 mol
  0.00200 mol
• final 0.006700 mol 0.300 mol 0.00200
  mol
• Final concentrations
• 0.006700 mol / 1.00 L 0.300 mol /
  1.00 L 0.00200 mol / 1.00 L
• K 0.30000.002000
• 0.006700
• K 8.96.10-2

8

• Exercise 13.10
• CO (g) H2O (g) ? CO2(g) H2 (g) K
  5.10
• Initial 1.000 mol 1.000 mol
  1.000 mol 1.000 mol
• mixed in a 1.000 L flask so all initial
  molarities are 1.000 M
• Q CO2H2
• CO H2O
• Q 1.00 Since Q lt K system will shift right
• CO (g) H2O (g) ?
  CO2(g) H2 (g)
• Final 1.000 M -x 1.000 M -x
  1.000 M x 1.000 M x
• K 1.00 x2
• 1.00 x2
• (K)1/2 (1.000 x) / (1.000 x)
  
• (5.10)1/2 - (5.10)1/2 x 1.000 x
• 1.2583 3.3583 x
• x 1.2583 / 3.3583 0.38618 M

CO H2O (1.000 M 0.38618 M) 0.614 M
CO2 H2 (1.000 M 0.38618 M) 1.386 M
Check K CO2 H2 CO
H2O K 1.38618 2
0.613822
K 5.10
9

• Exercise 13.11
• H2 (g) F2 (g) ? 2 HF (g) K 115
• 3.000 mol 3.000 mol
  3.000 mol
• For initial concentrations you must divide by the
  volume (1.500 L)
• initial 2.0000 M 2.0000 M
  2.0000 M
• Q HF2
• H2 F2
• Q 2.00002
• 2.00002.0000
• Q 1.000
• Since Q lt K system will shift right
• H2 (g) F2 (g) ? 2 HF (g)
• final 2.0000 x 2.0000 x
  2.0000 2x

115 (2.0000 2x)2 / ((2.0000 x)(2.0000
x)) 115 (2.0000 2x)2 / (2.0000 x)2 (115)1/2
(2.0000 2x) / (2.000 x) 21.448 10.724 x
2.0000 2x 12.724 x 19.448 x 19.448 /
12.724 x 1.5285
Final 0.472 M 0.472 M 5.51 M
10

• Exercise 13.12
• H2 (g) I2 (g) ?
  2 HI (g) Kp 100.
• initial 0.01000 atm 0.005000 atm
  0.5000 atm
• final 0.01000 atm x 0.005000 atm x
  0.5000 atm - 2x
• 0.04548 atm 0.04048
  atm 0.4290 atm
• Q (0.5000 atm)2
• (0.01000 atm) (0.005000 atm)
• Q 5000.
• Since Q gt K system will shift left
• 100. (0.5000 -2x )2
• (0.01000 x)(0.005000 x)
• 100. 0.25000 2.0000 x 4x2
• 0.000050000 0.015000 x x2
• 0.0050000 1.5000 x 100.x2 0.25000 2.0000
  x 4x2
• 96 x2 3.5000 x 0.2450 0
• x2 0.0364583 x 0.0025521 0 use Quadratic
  x (-b /- (b2-4ac)1/2)/2a
• x 0.0354775 (lots of /- rule so dont worry
  about it here)

11
Homework Day 1 pg. 645-647 s 13 35 37a,b
39 41 43 45 Additional (Hint when reactions
equations are added, equilibrium constants are
multiplied) Determine the equilibrium constant
for the following Na2O (s) ½ O2 (g) ? Na2O2
(s) K ? Na2O (s) ? 2Na(l) ½ O2 (g) K1
2.10-25 NaO (g) ? Na (l) ½ O2 (g) K2
2.10-5 Na2O2 (s) ? 2Na (l) O2 (g) K3
5.10-29 NaO2 (s) ? Na (l) O2 (g) K4
3.10-14 Day 2 pg. 645-650 s 20a,b 30 36
44 47 50 51a (neglect 2x taken from NOCl) 76a
Additional (Hint when reactions equations are
added, equilibrium constants are
multiplied) Determine the equilibrium constant
for the following a. NaO (g) Na2O (s) ? Na2O2
(s) Na (l) K ? (see reactions of Day 1
additional) b. 2 NaO (g) ? Na2O2 (s) K ?
12

• Shifting Principles
• Sec 13.7 Le Chateliers Principle
• 1. When dealing with a particular reaction that
  exists in an equilibrium condition, chemists try
  to find ways to _________ the equilibrium so as
  to obtain as much product as possible. In order
  to know how to do this it is important to
  understand ____________ which can shift the
  equilibrium.
• These factors include concentration, pressure and
  temperature.
• 2. Le Chateliers principle states that if a
  stress is imposed on a system in equilibrium, the
  position of the equilibrium will _________ to
  reduce the stress placed upon the equilibrium.
  For 2 of these factors (concentration and
  pressure) the position of the equilibrium may be
  shifted but the value of __ will still be
  constant.
• 3. Changing temperature can shift the
  equilibrium, but __ is also dependent upon the
  temperature.
• The Effect of a Change in Concentration
• 1. The effect of concentration has already been
  explored through the use of ___.
• 2. An increase in concentration of reactant will
  shift the equilibrium _______ (if there is more
  than one reactant, the other reactant will
  decrease in concentration, the products will
  increase in concentration).
• 3. An increase in concentration of product will
  shift the equilibrium ________ (recall addition
  of pure solid or liquid will not alter the
  equilibrium).
• See pg. 637 for an example.
• Sample Exercise 13.13 pg. 638.

shift
factors
shift
K
K
Q
right
left
13

• The Effect of a Change in Pressure
• 3 ways of changing the pressure on a reaction
  system
• 1. _____ (or remove) gaseous reactant or product
  (same effect as we explored above when adding
  reactant or product).
• 2. Addition of an ________ gas (one that is not
  involved in the reaction). No effect on
  equilibrium.
• 3. Change the volume of the container. When
  decreasing volume, equilibrium will shift to the
  side of the reaction with the __________ number
  of gaseous particles (since the volume of the
  container is reduced, the system tries to reduce
  its own volume). Increasing volume has the
  opposite effect.
• Sample Exercise 13.14 pg. 640.
• The Effect of a Change in Temperature
• 1. Altering the temperature will alter ___. By
  a knowledge of the enthalpy of a reaction (?H),
  we can predict the direction the equilibrium will
  shift.
• a. When ?H is __ (an exothermic reaction), heat
  is a ___________ of the reaction. Therefore,
  adding heat will shift the equilibrium to the
  _______ (removal of heat has the opposite effect)
• b. When ?H is ___ (an endothermic reaction), heat
  is a __________. The application of heat will
  shift the equilibrium to the __________ (removal
  of heat will have the opposite effect).
• Sample Exercise 13.15 pg. 642
• Table 13.4 pg. 642

add
inert
least
K
-
product
left
reactant

right
14
Homework pg. 645-649 s 16 (dont have to
write) 57-61 63 64 Additional at 25oC, K
0.090 for the reaction H2O (g) Cl2O (g) ?
2HOCl Determine the concentrations of all species
at equilibrium for each of the following
cases a. 1.0 g H2O and 2.0 g Cl2O are mixed in a
1.0 L flask. b. 1.0 mole pure HOCl is placed in a
2.0 L flask.