© Prep 101 Chem 203 Term Test 2 Exam Solutions
Prep 101 – Chemistry 203 Term Test 2
Kinetics
Problem 1.
Solution:
v1 k [ I  ]1m [ S2O82 ]1n

v2 k [ I  ]2m [ S2O82 ]2n
1.25  105 k (0.080)m (0.040)n

 2  2m  m  1
6.25  106 k (0.040)m (0.040)n
v1 1.25  105 k (0.080)m (0.040)n


 2  2n  n  1
v3 6.25  106 k (0.080)m (0.020)n
rate  k [ I  ]1[ S2O82 ]1
k
rate
[ I ][ S2O82 ]

1.25  105
k1 
 3.91  103 M 1s 1
(0.080)  (0.040)
Problem 2.
Solution:
Rate = k[H2 O]x [CH3 Cl]y
Rate1 k (0.0100) x 3.6  104


Rate2 k (0.0200) x 1.44  104
0.5x = .25, therefore x = 2 and the reaction is second-order in H2O
Rate1 k (0.0100) x 3.6  104


Rate2 k (0.0200) x 1.44  104
5.54(2/3)y=3.69
(2/3)y = 2/3, therefore y = 1 and the reaction is first-order in CH3Cl
Rate = k[H2 O]2 [CH3 Cl]1
Problem 3.
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
 [ A]0 
ln[ A]t1/2  ln[ A]0  kt1/2 , ln 
 kt1/2
 [ A]t 
1/2 

[ A]0
for t1/2 [A]t1/2 
2
 [ A]0 
 [ A]0 
ln 
 ln 
  ln 2  kt1/2
 [ A]t 
 [ A]0 2 
1/2 

ln  2 
 t1/2 
k
Problem 4.
Solution: If 20% decomposes, then 80% of the sample remains.
ln[ A]  ln[ A]0  kt rearrange
0.8[ A]o 
ln 
 k(50s)
 [ A]o 
k  4.463103 sec1
ln 2
t1/ 2 
 155 sec
k
Problem 5.
Solution: (a)
v1 k [CH 3 NNCH 3 ]1n

v2 k [CH 3 NNCH 3 ]2n
2.8  106 (5.13  102 ) n

 0.25  0.25n  n  1
5
1 n
1.1  10
(2.05  10 )
Therefore the reaction is the first order.
2.8  106 M / s
Rate = k [CH3 NNCH3(g) ], therefore k =
= 5.46 x 10-5 s-1
5.13  102 M
(b)
ln[A] = ln[A]o – kt
ln[0.15.13102] = ln[5.13102] – (5.4610-5)t
-5.27 = -2.97 - (5.4610-5)t
(5.4610-5)t = 2.30
t = 4.21104 s = 11.7 hours
(c)
[A]10% = 0.15.13102 = 5.1210-3M
Rate = k[A] = (5.4610-5 s-1)( 5.1210-3 M) = 2.8010-7 M/s
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 6.
Solution:
v2 k [CH 3COCH 3 ]2m [ Br2 ]2n [ H  ]l2

v1 k [CH 3COCH 3 ]1m [ Br2 ]1n [ H  ]1l
8.00  105 k (0.02)m (0.01)n (0.01)l

 2  2m  m  1
5
m
n
l
4.00  10
k (0.01) (0.01) (0.01)
v3 8.00  105 k (0.02)m (0.02)n (0.01)l


 1  2n
5
m
n
l
v2 8.00  10
k (0.02) (0.01) (0.01)
ln 0.5  ln 2n  n ln 2  n  0
v4 8.00  105 k (0.01)m (0.02)n (0.02)l


 2  2l  l  1
v1 4.00  105 k (0.01)m (0.01)n (0.01)l
rate  k [CH 3COCH 3 ]1[ Br ]0 [ H  ]1  k [CH 3COCH 3 ][ H  ]
k
rate
[CH 3COCH 3 ]2 [ H  ]
4.00  105
k1 
 4.00  103 M 1s 1
(0.01)  (0.01)
Problem 7.
Solution: Because this reaction is second order
1
1

 kt if 1/[A] was plotted vs. t then the y[ A] [ A]o
intercept would be 1/[A]0 and the slope would be k
1/[A]
slope = +k
1/[A]0
t
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 8.
Solution: I is the catalyst as it is used up in the first reaction and is produced in the second. It does not show
up in the overall reaction.
Problem 9.
Problem 10.
Solution: Since Hrxn > 0, then Hproducts > Hreactants
Thus, Ea rev = 66 – 41 = 25 kJ/mol
(draw an energy diagram to see this better)
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 11.
Solution:
Let T1 = 96 C = 369 K, and T2 = 25 C = 298 K.
𝑘
Using the Arrhenius Equation 𝑙𝑛 (𝑘1 ) =
2
𝐸𝑎
𝑅
1
1
2
1
(𝑇 − 𝑇 )
 3.55 x 1010 s 1 
Ea
1 
 1
ln 



10 1 
1
1 
8.314 JK mol  369 K 298K 
 1.2 x 10 s 
and solve for Ea, Ea = 12.8 kJ/mol
Problem 12.
k 
E  1
1
Solution: from Arrhenius equation ln  2    a   
R  T2 T1 
 k1 
since vk, then
k 
v  E  1 1 
ln  2  ln  2   a   
R T2 T1 
k1 
v1 
E  1
1 
ln(3)   a 

 Ea  77433.3 J / mol  77.4 kJ / mol
8.314 325 313
Problem 13.
2
Solution: Since we have an elementary process, then rate of the reaction is rate  k[ A] , This reaction is
second order therefore
1
1

 kt
[ A] [ A]o
1
1

 k(35.2 min)
0.1M 0.2 M
5 M-1 = k(35.2 min)
k = 1.42 x 10-1 M-1min-1

Problem 14.
The steps are elementary, and the first step is the rate determining step. So the overall rate law only
depends on the first step:
rate = k[O3][NO]
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 15.
Solution:
Slow step rate = k2[E][B]
For the fast equilibrium, rate = k1[A][B] = k-1[E]
Rearrange to:
[E] 
k1
kk
[A][B] , therefore, rate  1 2 [ A][B]2 This is inconsistent with the rate law.
k1
k 1
(ii): slow step rate = k2[E][A] and fast equilibrium rate = k1[A][B] = k-1[E]
[E] 
kk
k1
[A][B] , therefore, rate  1 2 [ A]2[B] This is consistent with the rate law
k1
k1
(iii) The first step is the rate determining step. The rate law would be
rate = k1[A]2 This is inconsistent with the proposed rate law.

Problem 16.
Solution: N2O2 is an intermediate, therefore cannot be present in the rate law.
Rate is found from the slower of the two reactions:
Rate = k2[N2O2][O2], where k-1[N2O2] = k1[NO]2
Therefore, rate 
k1k2
[NO]2[O2 ]  k [NO]2[O2 ]
k1

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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Equilibrium
Problem 17.
Solution:
1.
CO2 3eq
K eq 
C3H 8 1eq O2 5eq

2. Keq  CH 3COO   Na 
eq
eq
Problem 18.
Solution:
(a) HBr(g)
½ H2(g) + ½ Br2(g)
From H2(g) + Br2(g)
2 HBr(g), K1 = 3.5 x 104, you must reverse the reaction:
2 HBr(g)
H2(g) + Br2(g), K-1 = 1/K1 = 2.9 x 10-5, and then half the reaction:
HBr(g)
½ H2(g) + ½ Br2(g), K = K 1 = 5.3 x 10-3
(b) H2(g) + ½ Cl2(g) + ½ Br2(g)
HCl(g) + HBr(g)
This reaction is the sum of half of each of the reactions above.
H2(g) + Br2(g)
2 HBr(g), with K1 = 3.5 x 104
½ H2(g) + ½ Br2(g)
HBr(g), with K3 = K1 = 187
H2(g) + Cl2(g)
2 HCl(g), with K2 = 9.5 x 103
½ H2(g) + ½ Cl2(g)
HCl(g), with K 4 = K 2 = 97
H2(g) + ½ Cl2(g) + ½ Br2(g)
HCl(g) + HBr(g), K = K3 x K4 = 1.8 x 104
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 19.
Solution:
The expression for K 
 P4O10 eq
 P4 eq O2 5eq
3
 P4O10 init  1 M
 1  3 , the reaction proceeds to the right.
5
5
1 M  1 M 
 P4 init O2 init
 P4O10 init 
5.0 M
In (b) Q 
 3.7  3 , the reaction proceeds to the left.
5
 P4 init O2 init 8.0 M   0.70 M 5
In (a) Q 
Problem 20.
Solution:
To determine the final concentrations, the first thing needed are the initial reactant concentrations and an
expression for the reaction coefficient Q.
[A2(g) ] =
and Q
1.00 mol
2.00 mol
= 4.00 M, [AB(g) ] = [B2(g) ] =
= 8.00 M,
0.250 L
0.250 L
2
AB init


 A2 init  B2 init

(8.00)2
 2.0
(4.00) (8.00)
The direction of the reaction needs to be determined. To do this, we compare Q vs K.
Since Q = 2.0 > K = 0.5, the reaction proceeds to the left.
To determine what final concentrations will be from initial concentrations, a handy tool – the Initial,
Change, Equilibrium (ICE) table can be used.
All compounds involved in the reaction are included in an ICE table as follows, with the species that will be
consumed on the left side, and the species that will be produced on the right side. Since the reaction
proceeds to the left, A2 and B2 will be formed and AB will be consumed.
A2(g)
B2(g)
Concentration (M) 2 AB(g)
Initial
8.00 M
4.00 M
8.00 M
Change
-2x
+x
+x
Equilibrium
8.00 – 2x
4.00 + x
8.00 + x
As the reaction proceeds, x moles of A2(g) and B2(g) are formed as 2x moles of AB(g) are consumed. Make
sure to consider stoichiometric coefficients appropriately.
The equilibrium values are simply the sum of the initial + change concentrations.
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the
reaction in the way that it was initially described):
K
 AB 2eq

 A2 eq  B2 eq

(8.00  2 x )2
 0.5
(4.00  x ) (8.00  x )
so,
(8.00  2 x )2  0.5(4.00  x )(8.00  x )
64  32 x  4 x 2  0.5(32  12 x  x 2 )
64  32 x  4 x 2  16  6 x  0.5x 2
3.5x 2  38 x  48  0
To solve for x, the quadratic formula must be used,
x
b  b2  4ac 38  ( 38)2  4(3.5)(48)

 1.46 or 9.39,
2a
2(3.5)
Since 2 x 9.39 = 18.78 > 8.00 (which would leave the equilibrium concentration of AB at equilibrium
negative), then the x = 1.46
So, final concentrations are:
[A2] = 4.00 + 1.46 = 5.46 M
[B2] = 8.00 + 1.46 = 9.46 M
[AB] = 8.00 – 2(1.46) = 5.08 M
To check, substitute these concentrations into the Equilibrium constant expression,
 AB 2eq
(5.08)2
K 

 0.5 , matches up.
 A2 eq  B2 eq (5.46) (9.46)
Problem 21.
[ NO ]4
Answer: K c 
[ H  ]4 [ NO3 ]4
Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over
Reactants, and coefficients in the equation are written as superscripts.
Problem 22.
Solution:
K c = [CO2(g) ]
Remember: Solids and liquids are not included in the expression.
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 23.
Solution: Equation 2 is equal the double and reverse of equation1 therefore
1
K p2 = K −2
) 2 = 40.6
p1 = (
0.157
Problem 24.
Answer:
1
−1/2
Equilibrium Constant = K1
=
√K1
Solution: The second equation if reversed (1/K) and halved (K1/2). Combining these two gives 1/K1/2.
Problem 25.
Answer:
So the reaction equations is:
2 CO(g) + O2(g) ⇌ 2 CO2(g) K c = 3.3 × 1091
The relationship between K c and K p is:
K p = K c (RT)∆n gas In this case there are 3 moles of gas in the reactants and 2 moles of gas in the products,
so ∆n = -1
So solving for K p :
−1
L atm
K p = K c (RT)∆n gas = (3.3 × 1091 ) [(0.0821
) (298 K)] = 1.35 × 1090
mol K
Problem 26.
Solution:
N2(g) +
i:
c:
e:

1.00
+x
1.00+ x
Q=1>K so then rxn goes to the left
C2 H2(g) ⇌
1.00
+x
1.00 + x
2 HCN(g)
1.00
-2x
1.00-2x
(1.00  2 x )2
(1.00  2 x )
 Kc 
 Kc
2
(1.00  x )
(1.00  x )
x
1  Kc
2  Kc
 0.488
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 27.
Solution:
COCl2(g)
initial:
change:
equil:
x2
Kc =
0.04  x
x2 + Kcx – 0.04Kc = 0
x
0.04
-x
0.04-x
⇌ CO(g)
+Cl2(g)
0
+x
x
0
+x
x
 Kc  Kc 2  (4)(0.04)( Kc)
 5.29  103
2
Problem 28.
Solution: Kc 
I
C
E
[CO2 ][ H 2 ]
 5.10
[CO][ H 2O]
CO +
0.1M
-x
0.1-x
H2 O ⇌
0.1
-x
0.1-x
CO2
0.1
+x
0.1+x
+H2
0.1
+x
0.1+x
(0.1  x )2 0.01  0.2 x  x 2
Kc 

 5.10
(0.1  x )2 0.01  0.2 x  x 2
0.01  0.2 x  x 2  0.051  1.02 x  x 2
0  0.041  1.22 x
1.22 x  0.041
x  0.034 M
Therefore at equilibrium [H2] = 0.1 + x = 0.1 + 0.034M = 0.134M
Problem 29.
Solution:
Only C is true
An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO
at equilibrium.
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© Prep 101 Chem 203 Term Test 2 Exam Solutions
Problem 30.
Solution:
a) shift to the left
b) no effect
c) shift to the right
d) shift to the right
Problem 31.
Solution:
The reaction will shift left forming Ni(CO)4(g) to reach equilibrium
Problem 32.
Answer: C
Solution: Q = [NO]2[Cl2]/[NOCl]2 = (1.2)2(0.56)/(1.3)2 = 0.51 = K, therefore we are already at equilibrium.
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