Review Powerpoint

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Question One

 What does it mean if the Kc is greater than one?
 The reaction is product favored
Question Two

 Write the expression for the equilibrium constant for
the following reaction: carbon monoxide reacts with
oxygen to form carbon dioxide
 2 CO (g) + O2 (g)   2 CO2 (g)
Keq =
[CO2]2
[CO]2 [O2]
Question Three

 Write the expression for the equilibrium constant for
the production of HI gas from hydrogen gas and
iodine gas.
H2 (g) + I2 (g)   2 HI (g)
K =
[HI]2
[H2] [I2]
Question Four

 When the following reaction is at equilibrium at a
certain temperature, sulfur dioxide + oxygen <->
sulfur trioxide, [SO2] = 0.20 M, [O2] = 0.20 M, and
[SO3] = 0.40 M. Calculate Keq.
2 SO2 (g) + O2 (g)   2 SO3 (g)

[0.20]
K =
[SO3]2 =
[SO2]2[O2]
[0.20]
[0.40]
(0.40)2
= 20.
(0.20)2 (0.20)
Question Five

 2 NF2 (g)  N2F4 (g) 0.500M N2F4 is mixed with no NF2
initially present. The equilibrium concentration of N2F4 is
0.200M. Calculate the Keq value.

initial
change
equilibrium
Keq =
2 NF2 (g)   N2F4 (g)
[0]
[0.500]
+0.600
[0.600]
-0.300
[0.200]
[0.200] = 0.556
[0.600]2
Question Six

 Consider the following reaction:
2 HCl (g)  H2 (g) +
Cl2 (g) The initial concentration of HCl is 2.0M and there is no H2
or Cl2 present. After equilibrium conditions have been established,
the concentration of Cl2 is 0.10M. What are the equilibrium
concentrations for HCl and H2? Calculate Kc
2 HCl (g)  H2 (g)
Cl2 (g)
[2.0]
0
0
-0.20
+0.10
+0.10
[1.80]
[0.10]
[0.10]
Keq =
[0.10] [0.10]
[1.80]2
=
0.0031
Question Seven

 Calculate the enthalpy of reaction for the combustion
of C3H6.
 2C3H6 + 9O2  6CO2 + 6H2O
 -3809 kJ
Question Eight

 What three things are necessary for a successful
reaction to occur?
 Reactant molecules must collide
 The collision must occur with sufficient energy to get
over the activation energy hump
 The collision must occur with the proper orientation
Question Nine

 What are four ways to speed up a reaction?





Increase concentration of reactants
Increase temperature
Increase pressure of a gaseous system
Add a catalyst
Increase Surface Area
Question Ten

 The reaction, A + B   C + D has a Keq of 6.57 x 10-3. Determine the final
equilibrium concentrations of all substances if 0.200M C and 0.200M D are mixed.
Initial
Change
Eq
A
+
0
+x
x
[0.185M]
B
 C +
D
0
[0.200]
[0.200]
+x
-x
-x
x
0.2 - x
0.2 - x
[0.185M] [0.015M] [0.015M]
6.57 x 10-3 = [C] [D] =
[A] [B]
(0.200 -x)2
x2
x = 0.185
*simplify by taking the square root immediately. So:
0.0811 = 0.2 – x
x
or
0.0811x = 0.20 – x
1.0811x = 0.20
x = .185
Question Eleven

 How does increasing the temperature increase
reaction rate?
 Molecules have more energy and therefore there will
be more collisions and the increased energy means a
greater fraction of the molecules will have enough
energy to get over the activation energy hump
Question Twelve

 What is entropy?
 A measure of the disorder in a system
Question Twelve.Five

 How many grams is required to make a 0.250 M
solution of plumbic dichromate if I want to make a
450. mL solution?
 Pb(Cr2O7)2
 0.25 x 0.45 = 0.1125 moles x 639 = 71.9 g
Question Thirteen

 For the reaction, N2 (g) + O2 (g)   2 NO (g), the
equilibrium constant is found to be 1.0 at room
temperature. If 1.00M N2 is mixed with 1.00M O2, find
the eventual equilibrium concentration of all substances.
N2
+ O2

2 NO

initial
1.0
1.0
0
change
-x
-x
+2x
eq.
1–x
1–x
2x
1 – 0.33
[0.67]
 1.0 =
(2x)2
(1-x)(1-x)
1 – 0.33
[0.67]
= x = 0.33
2(0.33)
[0.67]
Question Fourteen

 How does adding a catalyst affect the reaction rate?
 It speeds up a reaction by lowering the activation
energy so that more molecules have sufficient energy
to get over the activation energy.
Question Fifteen

 Kc is 0.50 for synthesis of nitrogen monoxide gas from its
elements. Find the eventual equilibrium concentration of
all substances if 0.25 moles of NO are placed in a 4.0L
container.
N2
+ O2

2 NO
initial
0
0
[0.0625]
change
+x
+x
-2x
equilibrium
x
x
0.0625 – 2x
[0.023] [0.023]
[0.017]

0.50 = (0.0625-2x)2
x2
Question Sixteen

 Which has more entropy, solid water or water
vapor? Why?
 Water vapor has more entropy because there is more
disorder in the gaseous state than in the solid state due
to the increased molecular motion of gases
Question Seventeen

 What is the definition of rate of reaction?
 Change in concentration of reactant or product per
unit time
Question Eighteen

 In the reaction of hydrogen gas and chlorine gas to form
hydrogen monochloride gas, the Keq = 1.00 x 102. The initial
concentrations of both the hydrogen gas and chlorine gas are
1.0M. What are the equilibrium concentrations of all three
species?
H2
+
[1.0]
-x
1–x
[0.17]
Initial
Change
Equilibrium
1.00 x 102 =
10 =
Cl2

[1.0]
-x
1–x
[0.17]
(2x)2
( 1 – x)(1 – x)
2x
(1 – x)
x = 0.83
2 HCl
0
+ 2x
2x
[1.7]
take the square root so:
Question Nineteen

 In the reaction CO (g) + H2O (g)  CO2 (g) + H2 (g) (the
Keq is 5.0), if the initial concentrations for all species is [1.00]
find the equilibrium concentrations of each substance.
CO
+ H2O
[1.0]
[1.0]
-x
-x
1–x
1–x
[0.62]
[0.62]
Initial
Change
Equilibrium
  CO2
[1.0]
+x
1+x
[1.4]
+
H2
[1.0]
+x
1+x
[1.4]
*because Keq is greater than 1, the reaction proceeds toward products.
5.0 = (1 + x)(1 + x)
(1 – x)(1 – x)
2.24 = 1 + x
1–x
x = 0.38
*take the square root so:
Question Twenty

 For an exothermic reaction, which is higher – the
enthalpy of the bonds in the reactants or the
enthalpy of the bonds in the products?
 The enthalpy of the bonds in the products
N2 (g)
Question Twenty One
+ 3H2 (g)

2NH3 (g)
DH = -230 kJ

1. Write the equilibrium constant expression
Keq = [NH3]2
[N2][H2]3
2. Calculate the value of Keq if at equilibrium nitrogen is
2.0M, hydrogen is 1.0M, and ammonia is 0.50M.
Keq = [0.50]2
= 0.13
[2.0][1.0]3
3. Will there be more reactant or product?
Favors the reactant
Question Twenty Two
N2 (g) + 3H2 (g)  2NH3 (g) DH = -230 kJ

Which direction will the reaction proceed if:
a. the pressure is decreased.
reactant
b. the reaction is cooled.
product
c. a catalyst is added.
neither
d. NH3 is added.
reactant
e. H2 is added.
product
f. some nitrogen is removed.
reactant
Question Twenty Three

 Initially 4.0M H2 and 2.0M N2 are placed in a container at a
certain temperature. After the equilibrium has been established
1.0M NH3 has been produced. What are the equilibrium
concentrations of H2 and N2? What is Kc?
3 H2 + N2  
2 NH3
initial
[4.0]
[2.0]
0
change
- 1.5
- 0.50
+1.0
equilibrium
2.5 M
1.5 M
1.0M
Keq = [1.0]2
[2.5]3[1.5]
= 0.043
Question Twenty Four

For the following reaction: N2O4 (g) + 58.9 kJ   2 NO2 (g) A 1.0L flask at 55oC is found to
contain 3.6 moles of N2O4 and 1.75 moles of NO2 at equilibrium. What is the value of Keq?
Keq =
[1.75]2
= 0.85
[3.6]
 If the reaction is heated, in what direction would the equilibrium shift?

The reaction would shift toward the products.

 Would the Keq change? If so, would it increase or decrease?

Yes, the Keq would increase (products are in the numerator).

 If NO2 is removed from the flask, in what direction would the equilibrium shift?

The reaction would shift toward the products.

 Would the Keq change? If so, would it increase or decrease?

No, the Keq would not change.

 What effect would increasing the pressure have on this equilibrium?

The reaction would shift toward the reactants.
Question Twenty Five

 Write the equilibrium expression for the following
reactions:
 CaCl2 (s) D
Ca+2(aq) + 2Cl- (aq)
 Kc = [Ca+2] [Cl-1] 2
 ZnO(s) + CO(g) ) D
 Kc = [CO2] / [CO]
Zn(s) + CO2 (g)
Question Twenty Seven

 The colourless gas dinitrogen tetroxide decomposes to the
brown coloured air pollutant nitrogen dioxide and exists in
equilibrium. A 0.125 mol sample of dinitrogen tetroxide is
introduced into a 1.00 L container and allowed to decompose at
a given temperature. When equilibrium is reached, the
concentration of the dinitrogen tetroxide is 0.0750 mol/L.
What is the value of Keq for this reaction?
 N2O4 D 2NO2
0.125
0
- 0.05
+2(0.05)
0.075
0.1
Keq = .12/0.075 = 0.133
Question Twenty Seven

 A 0.921 mol sample of dinitrogen tetroxide is placed
in a 1.00 L vessel and heated to 100°C. At
equilibrium it is found that 20.7 % of the dinitrogen
tetroxide has decomposed to nitrogen dioxide.
Calculate the Keq for this reaction.
 N2O4 D 2NO2
0.921
- 0.191
0.73
0
+2(.191)
0.382
.207(.921) = .191
Kc = .3822/.73 = 0.200
Question Twenty Nine

 The following reaction has Keq value of 85.0 at 460°C:
SO2(g) + NO2(g) ⇌ NO(g) + SO3(g) If a mixture of sulfur
dioxide and nitrogen dioxide is prepared, each with an
initial concentration of 0.100 mol/L, calculate the
equilibrium concentrations of nitrogen dioxide and
nitrogen monoxide at this temperature.

SO2(g) + NO2(g) ⇌ NO(g) + SO3(g)
0.10
-x
0.10 –x
0.10
-x
0.10 –x
0
+x
x
[SO2] = [NO2] = 0.0098 M
[NO] = [SO3] = 0.0902M
85 = x2/(.1-x)2
0
9.22 = x/(.1-x)
+x .922 – 9.22x = x
x
.922 = 10.22x
x = 0.0902 M
Question Thirty

 Hydrogen and iodine gases react to form hydrogen iodide gas. If
6.00 mol of H2 and 3.00 mol of I2 are placed in a 3.00 L vessel and
allowed to come to equilibrium at 250°C calculate the equilibrium
concentrations of all species. The Keq for the reaction is 4.00 at 250
°C.
 H2 + I2 D 2HI
2
1
0
-x
-x
+2x
2-x 1-x
2x
[H2] = 1.33 M
[I2] = 0.33M
[HI] = 1.34M
Keq = (2x)2/(2-x)(1-x)
4.00 = 4x2 / (2 -3x + x2)
8 – 12 x + 4x2 = 4x2
8 – 12x = 0
8 = 12x
x = 0.67 M
Question Thirty One

 The equilibrium constant for the reaction below is 0.11.
Calculate all equilibrium concentrations if 0.33 mol of
iodine chloride gas is placed in a 1.00 L vessel and
allowed to come to equilibrium.
2 ICl(g) ⇌ I2(g) + Cl2(g)
0.11 = x2 / (0.33 – 2x)2
0.33
0
0
0.33 = x / (.33 – 2x)
-2x
+x
+x
0.1089 – 0.66x = x
0.33 -2x
x
x
0.1089 = 1.66x
x = 0.0656
[ICl] = 0.20 M
[I2] = [Cl2] = 0.066M
Question Thirty Two

 The dissociation of nitrogen monoxide gas to nitrogen
and oxygen gases has a Kc of 2.63 x 10-2 at 27°C. If 2.00
mol of nitrogen monoxide is placed in a 2.00 L vessel and
allowed to reach equilibrium, what is the concentration of
oxygen and nitrogen?
 2NO D N2 + O2
1.00 0
0
-2x +x +x
1-2x
x
x
[N2] = [O2] = 0.122M
2.63 x 10-2 = x2 / (1-2x)2
0.162 = x / (1-2x)
0.162 – 0.324 x = x
0.162 = 1.324x
x = 0.122 M
Question Thirty Three

For the reaction below, which change would cause the equilibrium to shift to the right?
 CH4(g) + 2H2S(g) ↔ CS2(g) + 4H2(g) DH = -43.0 kJ
(a) Decrease the concentration of dihydrogen sulfide.
Shift to the Left
(b) Increase the pressure on the system.
Shift to the Left
(c) Decrease the temperature of the system.
Shift to the Right
(d) Increase the concentration of carbon disulfide.
Shift to the Left
(e) Decrease the concentration of hydrogen.
Shift to the Right
Question Thirty Four

 Explain WHY a reaction will shift towards products
if a reactant is added
 More reactant molecules present will result in more
effective collisions causing the forward reaction to
occur where reactants will collide and form products
Question Thirty Five

 How many grams of calcium phosphate are needed
to make 250. mL of a 3.5 M solution?
 0.35 x 0.25 = 0.0875 mol Ca3(PO4)2 x 310 = 27g
Question Thirty Six

 What is the volume of an ammonium acetate
solution that has a mass of ammonium acetate of 45.0
grams and a concentration of 1.2 M?
 45 g NH4C2H3O2 / 77 = 0.584 mol
 1.2 = 0.584 / v
 V = 0.49 L
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