This page has all of the required homework for the material covered in the third exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12^{th} edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
rate = k[NO]^{2}[Cl_{2}]
What are the orders of NO and Cl_{2} and what is the overall order of the reaction?NO is second order or the order is equal to 2 for nitric oxide.
Cl_{2} is first order or the order is 1 for chlorine.
The overall order for the reaction is equal to two plus one or three. Overall order = 3.
C_{12}H_{22}O_{11} + H_{2}O → C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6}
2NO_{2} + F_{2} → 2NO_{2}F
Experiment | [NO_{2}] (M) | [F_{2}] (M) | Initial Rate (M/sec) |
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1 | 0.15 | 0.15 | 0.855 |
2 | 0.30 | 0.15 | 1.71 |
3 | 0.30 | 0.30 | 3.42 |
In both cases doubling the concentration doubles the rate and so they are both first order. The general rate expression is rate = k[NO_{2}][F_{2}] and the rate constant can be determined from one of the experiments. Using the first experiment gives k = (0.855 M/sec)/(0.15 M)^{2} = 38 M^{-1}sec^{-1}.
The rate law would be rate = (38 M^{-1}sec^{-1})[NO_{2}][F_{2}]
Here is an illustration of how this can be done mathematically. Divide the rate expression for the second experiment by the rate expression for the first experiment, cancel those factors that are constant and then solve for the exponent (the exponent is the order for that reactant).
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⇒ ln(2) = (x)ln(2) ⇒ x = 1
Experiment | [A] (M) | Initial Rate (M/sec) |
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1 | 0.2 | 3.04 x 10^{-5} |
2 | 0.4 | 1.20 x 10^{-4} |
When the concentration is doubled, the rate is four times as fast. The reaction is second order in A. Mathematically:
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⇒ ln(4) = (x)ln(2) ⇒ x = 2 ⇒ rate = k[A]^{2}
From experiment #1: 3.04 x 10^{-5} = k(0.2) ⇒ k = 1.52 x 10^{-4}
⇒ rate = (1.52 x 10^{-4})[A]^{2}
3ArSO_{2}H → ArSO_{2}Ar + ArSO_{3}H + H_{2}O
Experiment | [ArSO_{2}H] (M) | Initial Rate (M/sec) |
---|---|---|
1 | 0.25 | 4.75 x 10^{-5} |
2 | 0.3 | 6.84 x 10^{-5} |
2 | 0.35 | 9.31 x 10^{-5} |
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⇒ ln(1.44) = (x)ln(1.2) ⇒ 0.346 = x(0.182) ⇒ x = 2
⇒ rate = k[ArSO_{2}H]^{2}
From experiment #1: 4.75 x 10^{-5} = k(0.25)^{2} ⇒ k = 7.6 x 10^{-4}
⇒ rate = (7.6 x 10^{-4})[ArSO_{2}H]^{2}
3A + 2B → A_{3}B_{2}
Experiment | [A] (M) | [B] (M) | Initial Rate (M/sec) |
---|---|---|---|
1 | 1.200 | 1.200 | 6.912 |
2 | 1.500 | 1.200 | 10.800 |
3 | 1.200 | 3.100 | 17.856 |
The general rate expression would be rate = k[A]^{x}[B]^{z}
To get x, divide the rate expression for the second experiment by the rate expression for the first experiment, cancel those factors that are constant and then solve for the exponent. The same is done to find z using experiments #1 and #3. The data from one of the experiments is then used to find k and the overall rate law can then be written.
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⇒ rate = k[A]^{2}[B]; ⇒ k = (6.912)/[(1.2)^{2}(1.2)] = 4 (using expt. #1)
⇒ rate = (4)[A]^{2}[B]
Get the rate constant: k = -[ln(0.2/1)]/3.2 sec = 0.503 sec^{-1}
⇒ t = -[ln(0.01/1)]/(0.503 sec^{-1}) = 9.2 sec2NO(g) + Cl_{2}(g) → 2NOCl(g)
2NO(g) + Cl_{2} → 2NOCl(g)
(a) ΔH = (2 mole)(52.6 kJ/mole) - (2 mole)(90.37 kJ/mole) - 0 = -75.5 kJ
ΔH is negative which suggests that the reaction goes from high energy to low energy and that according to energy the reaction should be spontaneous.
ΔS = (2)(264 J/K) - (2)(210.62 J/K) - (1)(222.96 J/K) = -116 J/K
ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.
ΔG = (2 mole)(66.3 kJ/mole) - (2 mole)(86.71 kJ/mole) - 0 = -41 kJ
ΔG is negative which suggests that the reaction is spontaneous. In this case the energy has more influence than the entropy.
(b) ΔG = ΔH - TΔS = (-75.5 kJ) - (298 K)(-0.1162 kJ/K) = -41 kJ
(c) At equilibrium ΔG = 0 = ΔH - TΔS
⇒ T = ΔH/ΔS = (-75.5 kJ)/(-0.116 kJ/K) = 651 K
Below 651 K the reaction is spontaneous. Above 651 K the reaction is not spontaneous.
(a) ΔS = (1)(282.7 J/K) - (1)(219.4 J/K) - (1)(188.83 J/K) = -125.53 J/K
(b) ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.
(c) There are two moles of gas particles on the reactant side and only one mole of gas particles on the product side. It is going from disorder to order so that a negative change in entropy is correct.
CH_{3}OH(l) + CO(g) → CH_{3}COOH(l)
(a) ΔH = (1 mole)(-487 kJ/mole) - (1 mole)(-238.6 kJ/mole) - (1 mole)(-110.5 kJ/mole) = -138 kJ
ΔH is negative which suggests that the reaction goes from high energy to low energy and that according to energy the reaction should be spontaneous.
ΔS = (1)(159.8 J/K) - (1)(126.8 J/K) - (1)(197.9 J/K) = -165 J/K
ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.
ΔG = (1 mole)(392.4 kJ/mole) - (1 mole)(-166.23 kJ/mole) - (1 mole)(-137.2 kJ/mole) = -89 kJ
ΔG is negative which suggests that the reaction should be spontaneous. In this case the energy has more influence than the entropy.
You could also calculate ΔG as follows:
ΔG = ΔH - TΔS = (-138 kJ) - (298 K)(-0.165 kJ/K) = -89 kJ
(b) At equilibrium ΔG = 0 = ΔH - TΔS
⇒ T = ΔH/ΔS = (-138 kJ)/(-0.165 kJ/K) = 836 K
Below 836 K the reaction is spontaneous. Above 836 K the reaction is not spontaneous.