Name ____Mr. Perfect____________________________________ Date ______Sp 16_______________
1. Consider the following reaction shown below: (15 pts)
2ClO2 + 2OH- → ClO3- + ClO2- + H2O
[ClO2]
0.05
0.10
0.10
[OH-]
0.10
0.10
0.05
Initial Rate (M/s)
5.75 x 10-2
2.30 x 10-1
1.15 x 10-1
a. Write the rate law for the reaction. For full credit, state the order for each reactant.
Dependence on [ClO2]
Dependence on [OH-]
2.30 𝑥 10−1 𝑘(0.10)𝑥 (0.10)𝑦
2.30 𝑥 10−1
𝑘(0.1)𝑥 (0.1)𝑦
=
=
5.75 𝑥 10−2 𝑘(0.05)𝑥 (0.10)𝑦
1.15 𝑥 10−1 𝑘(0.10)𝑥 (0.05)𝑦
𝑥
4=2
𝑥=2
2 = 2𝑦 𝑦 = 1
Rate = k[ClO2]2[OH-]
second-order in [ClO2] and first-order in [OH-]
b. Calculate the rate constant for the reaction. Include the correct units.
𝑟𝑎𝑡𝑒
2.30 𝑥 10−1 𝑀/𝑠
𝑘=
=
= 𝟐𝟑𝟎/𝑴𝟐 ∙ 𝒔
[𝐶𝑙𝑂2 ]2 [𝑂𝐻 − ] (0.10 𝑀)2 (0.10)
c. What would be the initial rate for the reaction when [ClO2]o = 0.175 M and [OH-]o = 0.0844 M?
Initial rate = (230/M2s)(0.175 M)2(0.0844 M) = 0.59 M/s
2. The following reaction experimentally exhibits the rate law: rate = k[NO]2[O2]
2NO + O2 → 2NO2
Which of the following mechanisms is consistent with the experimental rate law? Show the validation
steps to receive full credit. (10 pts)
Mechanism A
Mechanism B
2NO ⇄ N2O2
N2O2 → NO2 + O (Slow)
O + NO → NO2
(reaction does not equal the one above)
NO + O2 ⇄ NO3
NO3 + NO → 2NO2 (Slow)
Slow step:
rate = k2[N2O2]
Solve for the intermediate N2O2
from the equilibrium step:
Slow step:
rate = k2[NO3][NO]
Solve for the intermediate NO3
from the equilibrium step:
Rate = k[NO]2
Not consistent (not valid)
rate = k[NO]2[O2]
Consistent (Valid Mechanism)
Chemistry 102 Exam 1
Name ____Mr. Perfect____________________________________ Date ______Sp 16_______________
3. Rate Expressions. (10 pts)
a. Write the possible rate expressions for the following reaction:
O3 + O → 2O2
−
∆[𝑂3 ]
∆[𝑂] 1 ∆[𝑂2 ]
=−
=
∆𝑡
∆𝑡
2 ∆𝑡
b. Write a balance equation for the following rate expressions:
−
1 ∆[𝑃𝐻3 ] ∆[𝑃4 ] 1 ∆[𝐻2 ]
=
=
4 ∆𝑡
∆𝑡
6 ∆𝑡
4PH3 → P4 + 6H2
4. A first-order reaction is 38.5 % complete in 480 s. (15 pts)
a. Calculate the rate constant k. 100 % - 38.5 % = 61.5 % remains
[A]t = 0.615[A]o
0.615[𝐴]0
𝑙𝑛
= −𝑘(480 𝑠)
[𝐴]0
k = 1.0 x 10-3 s-1
b. Calculate the half-life for this reaction.
𝑡1/2 =
ln 2
0.693
=
= 𝟔𝟗𝟑 𝒔
𝑘
1.0 𝑥 10−3 𝑠 −1
c. How long will it take for the reaction to go to 25 % completion?
𝑙𝑛
0.75[𝐴]0
= −(1.0 𝑥 10−3 𝑠 −1 )𝑡
[𝐴]0
t = 288 s
Chemistry 102 Exam 1
Name ____Mr. Perfect____________________________________ Date ______Sp 16_______________
5. The first order reaction, rate = k[A], has a rate constant of 4.6 x 10-2 s-1 at 0 C. When the reaction
temperature is increased to 20 C, the rate constant increases to 8.1 x 10-2 s-1. Use the 2 point form of
the Arrhenius Equation to calculate the activation energy in kJ/mol. {R = 8.314 J/mol K) (10 pts)
𝑙𝑛
𝑘2 𝐸𝑎 𝑇2 − 𝑇1
=
(
)
𝑘2
𝑅 𝑇1 ∙ 𝑇2
𝑙𝑛 (
8.1 𝑥 10−2
𝐸𝑎
293 𝐾 − 273 𝐾
(
)
)=
−2
𝐽
4.6 𝑥 10
273 𝐾 ∙ 293 𝐾
8.314
𝑚𝑜𝑙 ∙ 𝐾
Solve for the Activation Energy (Ea):
Ea = 18800 J/mol = 18.8 kJ/mol
6. At a given temperature, the equilibrium constant, K, is equal to 278 for the following reaction:
2SO2(g) + O2(g) ⇄ 2SO3(g)
Calculate the values of K for the following reactions: (10 pts)
a. SO2(g) + ½ O2(g) ⇄ SO3(g)
1
1
𝐾 ′ = (𝐾)2 = (278)2 = 𝟏𝟔. 𝟕
b. 2SO3(g) ⇄ 2SO2(g) + O2(g)
𝐾 ′′ =
1
1
=
= 𝟑. 𝟔𝟎 𝒙 𝟏𝟎−𝟑
𝐾 278
7. A sample of solid ammonium chloride was heated in a closed container to produce a total pressure of
4.4 atm. Calculate Kp for the following decomposition reaction. (10 pts)
I
C
E
NH4Cl(s) ⇄ NH3(g) + HCl(g)
--0
0
--+x
+x
--x
x
PT = PNH3 + PHCl = 4.4 atm ; therefore x = 2.2 atm
Kp = PNH3PHCl = x2 = (2.2)2 = 4.8
Chemistry 102 Exam 1
Name ____Mr. Perfect____________________________________ Date ______Sp 16_______________
8. The equilibrium constant, Kp, for the following reaction is 0.090 at 25 C. Under the following
conditions, clearly state in which direction will the reaction shift or if it is at equilibrium: PH2O = 200 torr,
PCl2O = 49.8 torr, PHOCl = 21 torr. Show work to receive full credit. (10 pts)
H2O(g) + Cl2O(g) ⇄ 2HOCl(g)
Convert torr to atm and then solve for the Reaction Quotient:
𝑄𝑃 =
2
𝑃𝐻𝑂𝐶𝑙
(0.0276)2
=
= 0.044
𝑃𝐻2𝑂 𝑃𝐶𝑙2𝑂 (0.263)(0.0655)
Qp (0.044) < Kp (0.090)
The reaction will shift towards the products until equilibrium is established.
9. The equilibrium constant for the following reaction is 40 at 448 C. What is the equilibrium
concentration of the product if the starting concentrations are [SbCl3] = 0.180 M and [Cl2] = 0.062 M?
Hint: try the 5 % rule. (10 pts)
I
C
E
𝐾=
SbCl3(g) + Cl2(g) ⇄ SbCl5(g)
0.180
0.062
0
-x
-x
+x
0.180-x 0.062-x
x
[𝑆𝑏𝐶𝑙5 ]
𝑥
= 40 =
[𝑆𝑏𝐶𝑙3 ][𝐶𝑙2 ]
(0.180 − 𝑥)(0.062 − 𝑥)
𝑥
− 0.242𝑥 + 0.011
Quadratic Equation: 40x2 - 10.68x + 0.44 = 0
40 =
40x2 - 9.68x + 0.44 = x
𝑥=
or
𝑥2
−(−10.68) ± √(−10.68)2 − 4(40)(0.44)
= 0.216 𝑀 𝑜𝑟 𝟎. 𝟎𝟓𝟏 𝑴
2(40)
At equilibrium, [SbCl5] = 5.1 x 10-2 M
10. Extra Credit. Given the following rate law, determine the units for the rate constant k. (5 pts)
rate = k[Cl2] ½[CHCl3]
units for k are M-½·s-1
Chemistry 102 Exam 1