Kinetic
Energy Diagrams
Pg 250
#2
Pg 267
#2, 3, 4, 5, 6, 7
Answers
Pg 250
#2 a) 1 b) 3 c) 2 d) 4
Pg 267
#2 Curve 3. This curve demonstrates the highest mean kinetic energy of the
particles. Given that kinetic energy increases proportionally to temperature,
this curve represents the highest reaction temperature.
#3 a) There would be no effect on the overall reaction rate, since the
rate determining step is step 2. The graph would show the first peak lower
down, since the activation energy of this step would decrease.
b) The overall reaction rate would increase. The graph would show the
second peak lower down, since the activation energy of this step would
decrease.
c) There would be an effect if the inhibitor increased the activation
energy higher than step 2, which would lead to a decrease in the overall
reaction rate. The graph would show the first peak higher up, since the
activation energy of this step would increase.
d) There would be a decrease in the rate (since it is happening slower)
because this change would affect the rate determining step. The graph would
show the second peak higher up, since the activation energy of this step would
increase.
#4 No, the addition of a catalyst in a reaction decreases the activation
energy, but does not play a role in the final concentration of the
products.
#5 a) Reaction 2. The substances in the three reactions are all at he same
phase but reaction 2 has fewer bonds to break than the other reactions.
Therefore, it will be the fastest.
b) Reaction 3. It has the most bonds to break.
#6 The dotted line would move to the right because there would be fewer
particles that would have the necessart kinetic energy for the reaction to
occur. An inhibitor increases the activation energy of a reaction.
#7 a) 1: addition of an inhibitor; 2:without the addition of a catalyst; 3:
addition of a catalyst.
b) Under condition 3, since by reading the graph from right to left, the
reverse activation energy is lowest in the case of curve 3.
Calculating Reaction Rate
Pg 219
#3, 5, 6 a and c,
Answers
#3 The rate of water vapour production is 0.102 mol/L∙s
#5 a) Rate HCl consumption: 9.052 x 10-4 mol/s
b) The rate of hydrogen production is 0.0109 L/s
#6 a) Rate = -(-6 mol)/45 s = 0.133 mol/s
b) Rate of formation of D = 0.399 mol/s
t =15.3 s
Pg 228
#4
Answer
#4 a) It is a reactant because the concentration decreases.
b) 0.3 mol/L∙s
c) r (10 s) = 0.5 mol/L∙s
r (20 s)= 0.25 mol/L∙s
r (50 s) = 0.0667 mol/L∙s
Pg 258
# 2, 3, 4, 5, 7, 9, 10, 11, 15
Answers
#2. a) r = k [H2SO4][Ca(OH)2]
b) r = k [I-]5 [H+]6
[IO3- ]3
c) r = k
[HCl]2
d) r = k [N2O4][H2O]
#3 The waiting time will be shorter, because the concentration of hydrogen
peroxide will be greater in this product. The greater the concentration of a
reactant, the higher the reaction rate will be.
#4 Reaction Rate: 8.90 x 10- 7
mol/L∙s
#5 r = k [NO]2[Br2], k = 0.8 L2/mol2∙s
#7 r2= 32r1, the reaction rate of r2 is 32 times greater than r1
#9 a) 3C(s) + 2Fe2O3(aq) → 4Fe(s) + 3CO2(g)
b) r = k [Fe2O3]2 L/mol∙s
#10 a) r = k [O2]8
b) k = 4.08 x 10- 6L7/mol7∙s
c) There will be no change because
sulfur is in solid form.
d) 20.96 mol/L∙s
#11 a) 7.98 x 10 5 L5/mol5∙s
b) 3.4 x 10- 5mol/L∙s
#15 a) S2O82-(aq) + 3I-(aq) → 2SO42-(aq) + I3-(aq)
b) r = k [S2O82][ I- ]3
c) k = 0.0217L3/mol3∙s
Energy Diagrams
Pg 250
#2
Pg 267
#2, 3, 4, 5, 6, 7
Answers
Pg 250
#2 a) 1 b) 3 c) 2 d) 4
Pg 267
#2 Curve 3. This curve demonstrates the highest mean kinetic energy of the
particles. Given that kinetic energy increases proportionally to temperature,
this curve represents the highest reaction temperature.
#3 a) There would be no effect on the overall reaction rate, since the
rate determining step is step 2. The graph would show the first peak lower
down, since the activation energy of this step would decrease.
b) The overall reaction rate would increase. The graph would show the
second peak lower down, since the activation energy of this step would
decrease.
c) There would be an effect if the inhibitor increased the activation
energy higher than step 2, which would lead to a decrease in the overall
reaction rate. The graph would show the first peak higher up, since the
activation energy of this step would increase.
d) There would be a decrease in the rate (since it is happening slower)
because this change would affect the rate determining step. The graph would
show the second peak higher up, since the activation energy of this step would
increase.
#4 No, the addition of a catalyst in a reaction decreases the activation
energy, but does not play a role in the final concentration of the
products.
#5 a) Reaction 2. The substances in the three reactions are all at he same
phase but reaction 2 has fewer bonds to break than the other reactions.
Therefore, it will be the fastest.
b) Reaction 3. It has the most bonds to break.
#6 The dotted line would move to the right because there would be fewer
particles that would have the necessart kinetic energy for the reaction to
occur. An inhibitor increases the activation energy of a reaction.
#7 a) 1: addition of an inhibitor; 2:without the addition of a catalyst; 3:
addition of a catalyst.
b) Under condition 3, since by reading the graph from right to left, the
reverse activation energy is lowest in the case of curve 3.
Calculating Reaction Rate
Pg 219
#3, 5, 6 a and c,
Answers
#3 The rate of water vapour production is 0.102 mol/L∙s
#5 a) Rate HCl consumption: 9.052 x 10-4 mol/s
b) The rate of hydrogen production is 0.0109 L/s
#6 a) Rate = -(-6 mol)/45 s = 0.133 mol/s
b) Rate of formation of D = 0.399 mol/s
t =15.3 s
Pg 228
#4
Answer
#4 a) It is a reactant because the concentration decreases.
b) 0.3 mol/L∙s
c) r (10 s) = 0.5 mol/L∙s
r (20 s)= 0.25 mol/L∙s
r (50 s) = 0.0667 mol/L∙s
Pg 258
# 2, 3, 4, 5, 7, 9, 10, 11, 15
Answers
#2. a) r = k [H2SO4][Ca(OH)2]
b) r = k [I-]5 [H+]6
[IO3- ]3
c) r = k
[HCl]2
d) r = k [N2O4][H2O]
#3 The waiting time will be shorter, because the concentration of hydrogen
peroxide will be greater in this product. The greater the concentration of a
reactant, the higher the reaction rate will be.
#4 Reaction Rate: 8.90 x 10- 7
mol/L∙s
#5 r = k [NO]2[Br2], k = 0.8 L2/mol2∙s
#7 r2= 32r1, the reaction rate of r2 is 32 times greater than r1
#9 a) 3C(s) + 2Fe2O3(aq) → 4Fe(s) + 3CO2(g)
b) r = k [Fe2O3]2 L/mol∙s
#10 a) r = k [O2]8
b) k = 4.08 x 10- 6L7/mol7∙s
c) There will be no change because
sulfur is in solid form.
d) 20.96 mol/L∙s
#11 a) 7.98 x 10 5 L5/mol5∙s
b) 3.4 x 10- 5mol/L∙s
#15 a) S2O82-(aq) + 3I-(aq) → 2SO42-(aq) + I3-(aq)
b) r = k [S2O82][ I- ]3
c) k = 0.0217L3/mol3∙s
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