10.A – Thermo: Endo/Exo and ΔH
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. For each of the following laws of thermodynamics, what does each actually mean in terms of the behavior of energy and particles?
1st Law of Thermodynamics?
Energy is neither created nor destroyed in a chemical reaction
AKA “Law of Conservation of Energy”
2nd
Law of Thermodynamics?
Disorder/chaos of a system always increases in terms of particles/energy distribution
AKA Probability states that disorder increased as order is less probable
3rd Law of Thermodynamics?
Disorder/chaos of a system becomes constant as temperature approches absolute zero
AKA 0 K = order
Instructions: Classify each statement as talking about an [EXO]thermic or [ENDO]thermic reaction:
2.
3.
4.
5.
[EXO]
[EXO]
[EXO]
[ENDO]
surroundings get hot
dynamite explodes
energy is a product
H is positive
6.
7.
8.
9.
[EXO]
[EXO]
[ENDO]
[ENDO]
reactants have more energy
H is negative
absorbing sunlight to make sugar
surroundings get “cold”
10. [ENDO]
11. [ENDO]
12. [ENDO]
13. [ENDO]
products have more energy
energy is a reactant
conversion of sugar to fat
wood logs on a fire
Instructions: ON A SEPARATE SHEET OF PAPER, Preform the following thermo calcualtions that follow using Dimensional Analysis. Remember NW = NC
14. How much heat is released when 9.22 grams of glucose C6H12O6 in your body
reacts with according to the following equation?
C6H12O6 + 6 O2  6 CO2 + 6 H2O + 2803 kJ
17. Calculate the heat released when 74.6 grams of SO2 reacts according to the
following equation.
2 SO2 + O2  2 SO3 + 99.1 kJ
9.22 g C6H12O6
74.6 g SO2
1 mol C6H12O6
–2803 kJ
180.18 g C6H12O6
1 mol C6H12O6
= – 143 kJ
1 mol SO2
–99.1 kJ
64.07 g SO2
2 mol SO2
= – 57.7 kJ
15. How much heat is absorbed during photosynthesis when 9.22 grams of glucose
C6H12O6 is produced?
2803 kJ + 6 CO2 + 6 H2O  C6H12O6 + 6O2
18. Calculate the heat release when 266 grams of white phosphorus P4 burns in air
according to the following equation.
P4 + 5 O2  P4O10 + 3013 kJ
9.22 g C6H12O6
266 g P4
1 mol C6H12O6
+2803 kJ
= + 143 kJ
180.18 g C6H12O6
1 mol C6H12O6
1 mol NO2
–138 kJ
46.01 g NO2
3 mol NO2
–3013 kJ
123.88 g P4
1 mol P4
= – 6470 kJ
19. Calculate the amount of heat released when 1.26 x 104 grams of ammonia NH3
are produced according to the following equation.
N2 + 3 H2  2 NH3 + 92.6 kJ
16. How much heat is released when 147 grams of NO2(g) is dissolved in excess
water?
3NO2 + H2O  2HNO3 + NO + 138 kJ
147 g NO2
1 mol P4
1.26 x 104 g NH3
= – 147 kJ
1 mol NH3
–92.6kJ
= – 3.42 x 104 kJ
17.04 g NH3
2 mol NH3
10.B – Thermo: Entropy and Hess’s Law
1.
Identify which one of the following pairs of samples has the higher entropy?
2.
a.
Br2(l) or Br2(g)
Br2 (g)
gas has >S than liquid
b.
c.
MgO(s) or NaCl(aq)
NaCl (ag) solid has <S than uniform solid d.
C2H6(g) or C3H8(g)
C2H6 (g)
lower MM >S than higher MM
KOH(s) or KOH(aq)
KOH (ag)
dissociated has >S than solid
Predict the entropy change (ΔS) for the following processes:
a.
O2(g)  2O(g)
+∆S
1 part. to 2
b.
2O3(g)  3O2(g)
+∆S
2 part. to 3
c.
CH4(s) + 2O2(g)  CO2(g) + 2H2O(g)
+∆S
(s) to (g)
d.
NaCl(s)  Na+(aq) + Cl-(aq)
+∆S
(S) to (aq) dissociation
e.
C2H5OH(l)  C2H5OH(g)
f.
Ag+(aq)
 AgCl(s)
−∆S
(aq) dissociation to (s)
+∆S
(l) to (g)
+
Cl-(aq)
Instructions: Solve the following Hess’s law problems. Remember NW = NC
3. Calculate the ∆H for the following reaction:
Sn + 2Cl2  SnCl4
You are given these two equations:
Sn + Cl2  SnCl2
∆H = -325 kJ
SnCl2 + Cl2  SnCl4
∆H = -186 kJ
Sn + Cl2  SnCl2
SnCl2 + Cl2  SnCl4
Sn + 2Cl2  SnCl4
∆H = −325 kJ
∆H = −186 kJ
∆H = – 511 kJ
Leave same
Leave same
4. Calculate the ∆H for the following reaction:
NO + O  NO2
You are given these three equations:
O2  2O
∆H = +495 kJ
2O3  3O2
∆H = -427 kJ
NO + O3  NO2 + O2
∆H = -199 kJ
2O  O2
3O2 2O3
2NO + 2O3  2NO2 + 2O2
2 NO + 2 O  2NO2
Have to simplify
∆H =
∆H =
∆H =
∆H =
∆H =
− 495 kJ
+ 427 kJ
− 398 kJ
− 466 kJ
– 233 kJ
As “O” is a react = FLIP
FLIP to cancel “O2 & O3”
x2 To Cancel
5. Calculate the ∆H for the following reaction:
2H2O2  2H2O + O2
You are given these two equations:
2H2 + O2  2H2O
∆H = -572 kJ
H2 + O2  H2O2
∆H = -188 kJ
2H2 + O2  2H2O
2 H2O2  2 H2 + 2 O2
2H2O2  2H2O + O2
7. Calculate the ∆H for the following reaction:
CS2 + 2H2O  CO2 + 2H2S
You are given these two equations:
CS2 + 3O2  CO2 + 2SO2 ∆H = -1075 kJ
H2S + 1.5O2  H2O + SO2 ∆H = -563 kJ
FLIP to have “H2O” as react
/“H2S” as prod and cancel
“O2” by x2
3H2O + 3SO2  2H2S + 3O2
∆H = +1126 kJ
CS2 + 3O2  CO2 + 2SO2
∆H = −1075 kJ
CS2 + 2H2O  CO2 + 2H2S
∆H = + 51 kJ
FLIP to have “H2O2” as react
and x2 to cancel “H2 & O2”
∆H = −572 kJ
∆H = + 376 kJ
∆H = – 196 kJ
6. Calculate the ∆H for the following reaction:
2CO + 2NO  2CO2 + N2
You are given these two equations:
2 CO + O2  2CO2
∆H = -566.0 kJ
N2 + O2  2NO
∆H = 180.6 kJ
FLIP to have “N2” as a prod.
/”NO” as a react. As well
as cancel “O2”
2 CO + O2  2CO2
∆H = −566.0 kJ
2NO  N2 + O2
∆H = −180.6 kJ
2CO + 2NO  2CO2 + N2 ∆H = – 746.6 kJ
8. Calculate the ∆H for the following reaction:
4Al + 3MnO2 2Al2O3 + 3Mn
You are given these two equations:
4Al + 3O2  2Al2O3
∆H = -3352 kJ
Mn + O2  MnO2
∆H = -521 kJ
FLIP to have “MnO2” as
react/“Mn & O2” as prod
and cancel “O2” by x3
4Al + 3O2  2Al2O3
3MnO2  3Mn + 3O2
4Al + 3MnO2 2Al2O3 + 3Mn
∆H = −3352 kJ
∆H = + 1563 kJ
∆H = – 1789 kJ
10.C – Kinetics: Collision Theory
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. What conditions must be satisfied in order for a reaction to occur?
(1) Particles must collide
If they do not…the other two conditions do not matter
(2) When they collide they must posses sufficent KE to start the reaction
(3) Must collide w/proper orientation
Vocabulary Term = Activation Energy = Ea
All 3 together, vocabulary term = Effective Collision
2. What makes a collision “effective”?
(1) Particles MUST COLLIDE. When they collide MUT be w/
(a) sufficent energy
If they do not…the other conditions do not matter
ALL MUST BE TRUE IN ORDER TO BE EFFECTIVE
(b) high frequency of collisions
(c) correct orientation for rxn
3. What makes a collision “ineffective”?
(1) Particles MUST COLLIDE. When they collide CAN be w/
(a) insufficent energy
If they do not…the other conditions do not matter
ANY CAN BE TRUE IN ORDER TO BE INEFFECTIVE
(b) low frequency of collisions
(c) incorrect orientation for rxn
4. Define reaction rate:
Is proportional to the # of effective collisions and is defined as how fast (or slow) reactants are converted to product
5. What is the difference between a reaction rate and the reaction mechanism?
Rate = Speed of rxn along pathway
Mechanism = pathway/setps taken to go from reactants to products
6. What factors affect the rate of reaction?
Nature of Reactants
Temperature
Concentration
Particle size / Surface Area
NOT CATALYSTS OR INHIBITORS, they effect the mechanism
7. What effect, and how, do the following factors have upon reaction rate?
Nature of Reactants?
Refers to complexity and # of bonds that must be broken and reformed in same manner in the course of the rxn. The more bonds that mus be broken or
formed, the more Ea required to do so
Temperature?
Higher the temp, higher rate of rxn. Higher the temp, the more KE, the more movement = higher probability of collisions and collision possessing Ea
Concentration?
More particles in smaller area incr the likelyhood of collisions. More collisions = more changes to collide w/proper orientation = higher rate of rxn
Particle Size?
Smaller the particle, the more surface area exposed/available for rxn. More particles possible of reacting, the more that can collide w/proper orientation
= higher rate of rxn
Cayalyst?
DOES NOT DIRECTLY CHANGE THE RXN RATE, CHANGES RXN MECHANISM. This lowers the Ea, making more energy available for more rxn to occur.
PRECIEVED as an incr in rxn rate, rather than shortening the mechanism.
Inbitor?
DOES NOT DIRECTLY CHANGE THE RXN RATE, CHANGES RXN MECHANISM. This raises the Ea, making less energy available for less rxn to occur.
PRECIEVED as an decr in rxn rate, rather than extending/lengthing the mechanism.
10.D – Kinetics: Reaction Pathway (Potential Energy) Diagrams
Instructions: Use the reaction pathway diagram below to answer questions 1 – 9 for the forward reaction.
1. Which of the letters in the diagram represents the PE of the prods?
_e_
2. Which letter indicates the potential energy of the activated complex? _c_
3. Which letter indicates the potential energy of the reactants?
_b_
4. Which letter indicates the activation energy of the A + B→C?
_a_
5. Which letter indicates the heat of reaction?
_f_
6. Is the reaction exothermic or endothermic?
_endothermic_
7. Which letter indicates the activation energy of the reverse reaction? _d_
8. Which letter indicates the heat of reaction of the reverse reaction?
9. Is the reverse reaction exothermic or endothermic?
Instructions: Use the reaction pathway diagram on the right to answer questions 10 –21. Remember N3
10. The heat content of the reactants of the forward reaction is about _80 kJ_
11. The heat content of the products of the forward reaction is about _160 kJ_
12. The heat content of the activated complex of the forward reaction about _250 kJ ±10 kJ
13. The activation energy of the forward reaction is about _170 kJ ±10 kJ
14. The heat of reaction (ΔH) of the forward reaction is about _250 kJ ∆ = need sign
15. The forward reaction is (endothermic/exothermic).
16. The heat content of the reactants of the reverse reaction is about _160 kJ
17. The heat content of the products of the reverse reaction is about _80 kJ
18. The heat content of the activated complex of the reverse reaction is about _250 kJ ±10 kJ
19. The activation energy of the reverse reaction is about _90 kJ ±10 kJ
20. The heat of reaction (ΔH) of the reverse reaction is about _– 80 kJ ±10 kJ
21. The reverse reaction is (endothermic/exothermic).
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
22. Chemical reactions occur when reactants collide. For what reasons may a collision fail to produce a chemical reaction?
(1) incorrect orientation
(2) not enough energy available for activation
23. If every collision between reactants lead to a reaction, what determines the rate at which the reaction occurs?
Nature of reactants
The complexity and # of bonds that must be broken / released
24. What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
Ea = Energy requeared to breakt eh bonds of reactant to enable new bonds to be formed in products.
Activated Complex = Represents the total PE that must be present for the rxn to occur.
Difference = The PE of the reactants and the activated complex is the Ea
25. What happens when a catalyst is used in a reaction?
Alters the mechanism of the rxn, lowering the Ea as the mechanism is “shorter”
26. Name 4 things that will speed up or slow down a chemical reaction.
Nature of Reactants
Temperature
Concentration
NOT CATALYSTS OR INHIBITORS, they effect the mechanism
More rxns w/same amount of energy
Particle size / Surface Area
_f_
_exothermic_
27. Sketch an energy diagram for a reaction. (Label the axes.)
Potential energy of reactants = 350 kJ
Activation energy = 100 kJ
Potential energy of products = 250 kJ
28. Is the reaction in #27 exothermic or endothermic? Explain.
Exothermic, the PE of the products is less than the PE of the reactants as the PE is being REALEASED = exothermic
29. How could you lower the activation energy for the reaction in #27
Add catalyst
10.E – Equilibrium: Le Chatelier’s Principle
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. What is the difference between a physical and chemical equilibrium?
Phys = a closed @ constant temp/presurre in which rate of forward and reverse physical changes are equal
Chem = a closed system w/constant temp/pressure/concentration in which rate of forward and reverse chemical change are equal
Instructions: Complete the following chart by writing “left”, “right”, or “none” for equilibrium shift given the. Write “increases”, “decreases”, or “remains the same” for the
concentration of reactants and products, and for the value of K (will need to come back and do on day 6)
N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) + 22.0 kcal
Stress
NaOH (s) ↔ Na+ (aq) + OH- (aq) + 10.6 kcal
Equalibirum
Shift
[ N2 ]
[ H2 ]
[ NH3 ]
K
right
right
----
Decr.
----
Incr.
Incr
same
same
add N2
add H2
Stress
Equalibirum
Shift
Amount
of
NaOH(s)
---incr
[ Na+ ]
[OH- ]
K
Add NaOH
right
incr
incr
same
Add
NaCl
left
---decr
same
(adds Na+)
add NH3
left
Incr
Incr
---same
Add
KOH
left
incr
decr
---same
(adds OH-)
remove N2
left
---Incr
Decr
same
Add
H+
right
decr
incr
---same
(removes OH-)
remove H2
left
Incr
---Decr
same
incr. temp.
left
incr
decr
decr
decr
remove NH3
right
Decr
---same
decr. temp.
right
decr
incr
incr
incr
incr. temp.
left
Incr
Incr
Decr
Decr
incr. Press.
none
same
same
same
same
decr. temp.
right
Decr
Decr
Incr
Incr
decr. Press.
none
same
same
same
same
incr. Press.
right
Decr
Decr
Incr
Incr
Add Ca(OH)2
left
incr
decr
---decr
decr. Press.
left
Incr
Incr
Decr
Decr
Add Na3PO4
left
incr
---decr
decr
Instructions: Consider the following equilibrium and answer the questions when these changes are made chart by writing “left”, “right”, or “none” for equilibrium shift given
the. Write “increases”, “decreases”, or “remains the same” for the concentration of reactants and products.
5. Predict which way these equilibrium systems will shift when the total pressure is increased.
(A) N2 (g) + O2 (g) ↔ 2 NO (g) no shift
(B) 2 SO2 (g) + O2 (g) ↔ 2 SO3 (g) right/product
C) 4 NH3 (g) + 5 O2 (g) ↔ 4 NO (g) + 6 H2O (g) left/react
6. Heat + CH4 (g) + 2 H2S (g) ↔ CS2 (g) + 4 H2 (g) and CH4 gas is added:
(A) direction of equilibrium shift?
(C) [CS2]?
right/product
incr _
(B) [H2S]?
decr _
(D) [H2]?
incr _
(B) [CH4]?
decr _
(D) [H2]?
incr _
7. Heat + CH4 (g) + 2 H2S (g) ↔ CS2 (g) + 4 H2 (g) and CS2 gas is removed:
(A) direction of equilibrium shift?
(C) [H2S]?
right/product
incr _
8. Heat + CH4 (g) + 2 H2S (g) ↔ CS2 (g) + 4 H2 (g) and H2 gas is added:
(A) direction of equilibrium shift?
left/react
(B) [CH4]?
decr _
(C) [CS2]?
decr _
(D) [H2S]?
incr _
(B) [H2S]?
decr _
(D) [H2]?
decr _
9. Heat + CH4 (g) + 2 H2S (g) ↔ CS2 (g) + 4 H2 (g) and temperature is increased:
(A) direction of equilibrium shift?
right/product
(C) [CS2]?
incr _
(E) [CH4]?
incr _
10.F – Equilibrium: Equilibrium Constant (K)
Instructions: ON A SEPARATE SHEET OF PAPER, write the Keq expressions for the reactions below.
1.
____N2 (g) + 3 H2 (g) ↔
Keq =
2.
2 KClO3 (s) ↔
Keq =
2 NH3 (g)
4.
2 KCl (s) +
[O2]3
Keq =
3 O2 (g)
3. ____H2O(l) ↔ ____ H+ (aq) + ____OH-(aq)
Keq =
2 CO(g) + ____O2 (g) ↔ 2 CO2 (g)
[NH3]2 _
[N2] [NH3]
[
H+
][
OH-
[CO2]2 _
[CO]2 [O2]
5. ____ Li2CO3 (s) ↔ 2 Li+ (aq) +
Keq =
[Li+]2 [CO32-]
____CO32- (aq)
6.
4 CS2 (g)
____CH4 (g) +
]
2 H2S (g)
Keq =
↔
+
____ H2 (g)
[H2]4
[CS2] _
[H2S]2 [CH4]
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
7. In considering the following equilibriums, which equilibrium favors products to the greatest extent? Reactants to the greatest extent? Notice that only 2 responses are req
a) 2NO2 (g) ⇄ N2O4 (g)
b)
Cu2+(aq)
Keq = 2.2
+ 2Ag(s) ⇄ Cu(s) +
c) Pb2+ (aq) + 2 Cl- (aq) ⇄
d) SO2(g) + O2 (g) ⇄
2Ag+ (aq)
PbCl2(s)
SO3 (g)
Keq = 1 x
NA _
10-15
react to greatest extent _
Keq = 6.3 x 104
prod to greates extent _
Keq = 110
NA _
8. What is the only way to change the value of the Keq?
Changing the heat content of the rxn (i.e. temp) _
OR Pressure for gas based systems
9. In the reaction: A + B ⇄ C + D + 100kJ, what happens to the value of Keq if we increase the temperature?
Will decr _
10. If the value of Keq decreases when we decrease the temperature, is the reaction exothermic or endothermic?
Endothermic _
11. In the reaction; W + X + 100kJ ⇄ Y + Z, what happens to the value of Keq if we increase the [X]? Explain your answer.
Keq constant as only to change the equilibirum is by changing the temperature _
12.
If the value of Keq increases when we decrease the temperature, is the reaction exothermic or endothermic?
13.
Predict whether reactants of products are favored in the following equilibrium systems
Exothermic _
(a)
CH3COOH(aq) ⇋ H+(aq) + CH3COO-(aq)
Keq = 1.8 x 10-5
Reacts _
(b)
H2O2(aq) ⇋ H+(aq) + HO2(aq)
Keq = 2.6 x 10-12
Reacts _
(c)
CuSO4(aq) + Zn(s) ⇋ Cu(s) + ZnSO4(aq)
1037
Prods _
Keq =
Instructions: ON A SEPARATE SHEET OF PAPER, use the information below to calculate the equilibrium constant (Keq) for the following reactions and tell whether
equilibrium lies favors the reactants or the products.. NW = NC. Hint: Common mistakes will be made here.
14. H2 (g) + Cl2 (g) ↔ HCl (g)
At equilibrium [H2] = 0.42 M, [Cl2] = 0.075 M, and [HCl] = 0.95 M
Keq =
[0.95 M]2 _
[0.42 M] [0.75 M]
=
29, prod favored
15. N2 (g) + H2 (g) ↔ NH3 (g)
At equilibrium [N2] = 0.34 M, [H2] = 0.13 M, and [NH3] = 0.19 M.
Keq =
[0.19 M]2 _
= 48, prod favored
[0.13 M]3 [0.34 M]
16. NO (g) + O2 (g) ↔ NO2 (g)
At equilibrium [NO] = 2.4×10-3 M, [O2] = 1.4×10-4 M, and [NO2] = 0.95 M.
Keq =
[0.95 M]2 _
= 1.1x109, prod favored
-3
-4
[2.4x10 M] [1.4x10 M]
17. C (s) + CO2 (g) ↔ CO (g)
At equilibrium [CO2] = 8.3 × 10-6 M, [CO] = 5.4 × 10-5 M
Keq =
[5.4x10-5 M ] _
= 3.5x10-4, react favored
-6
[8.3x10 M]
18. 2NO (g) + Br2 (g) ↔ 2 NOBr (g)
At equilibrium [NO] = 0.5 M, [Br2] = 0.25 M, and [NOBr] = 3.5 M.
Keq =
[3.5 M]2 _
= 200, prod favored
[0.5 M]3 [0.25 M]
19. 2 Fe (s) + 3H2O (g) ↔ Fe2O3 (s) + 3 H2 (g)
At equilibrium [H2O] = 1.0 M, and [H2] = 4.5 M.
Keq =
[4.5 M]3 _
= 91, prod favored
[1.0 M]3
20. CaCO3 (s) ↔ CaO (s) + CO2 (g)
At equilibrium [CO2] = 4.0 × 10-3 M.
Keq =
[4.0x10-3 M]
=
4.0x10-3, react favored
21. PCl5 (g) ↔ PCl3 (g) + Cl2 (g)
Equilibrium concentrations are: [PCl5] = 0.25 M, [PCl3] = 9.7 × 10-4 M, and [Cl2]
= 3.2 × 10-3 M.
Keq = [3.2x10-3 M] [9.7x10-4 M]
= 1.2x10-5, react favored
[0.25 M]