Due: November 6, 2007
1.
In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to
nitric acid by the high temperature reaction:
4NH3(g)
a.
+
4NO(g)
+
6H2O(g)
How is the rate of consumption (disappearance) of O2 related to the rate of production
(appearance) of NO and H2O?
Rate = -∆[O2] = ∆[NO]
5 ∆t
4∆t
b)
→
5O2(g)
∆[H2O]
6∆t
=
It was found that at a particular instant, NO was produced at a rate of 0.305 mol
L •s
(ie., ∆[NO] = 0.305 mol )
∆t
L •s
i) What is the rate of formation of H2O(g)?
∆[NO] = 0.305 mol
4 ∆t
4 L •s
Rate =
∆[H2O]
6∆t
-∆[O2]
5∆t
=
= 0.07625 mol
L•s
(divide both sides by four – to include stoichiometry)
∆[NO]
4∆t
=
∆[H2O]
6∆t
therefore,
= 0.07625 mol
L•s
∆[H2O] = 0.458 mol
∆t
L•s
ii) What is the rate of consumption of O2(g)?
∆[O2]
5∆t
= 0.07625 mol
L•s
therefore,
∆[O2] =
∆t
- 0.381 mol
L•s
2.
The oxidation of Iodide ion by peroxydisulfate ion is described by the equation:
3IӨ(aq)
+
a) If -∆[S2O82-]
∆t
S2O82-(aq)
→
I3 Ө(aq)
+
2SO42-(aq)
= 1.5 x10-3 mol for a particular time interval, what is the value of
L •s
- ∆[IӨ] for the same interval?
∆t
-∆[S2O82-]
∆t
= 1.5 x10-3 mol
L •s
= -∆[IӨ ]
3∆t
∆[IӨ] = - 4.5 x 10-3 mol
∆t
L •s
b) What is ∆[SO42-] for the same interval?
∆t
-∆[S2O82-]
∆t
= 1.5 x10-3 mol
L •s
=
∆[ SO42-]
2∆t
∆[ SO42-] = 3.0 x 10-3 mol
∆t
L •s
3. The reaction of gaseous chloroform and chlorine is described by the equation:
CHCl3(g) +
Cl2(g)
→
CCl4(g) +
HCl(g)
The rate law is rate = k[CHCl3][Cl2]½. What is the order of the reaction with respect each of
the reactants? What is the overall order of the reaction?
Reaction order with respect to CHCl3(g) = 1
Reaction order with respect to Cl2(g) = ½
Overall reaction order = 1½
4. The oxidation of BrӨ by BrO3Ө in acidic solution is described by the equation:
5 BrӨ(aq)
BrO3Ө(aq) + 6H⊕ (aq) → 3Br2(aq)
+
+ 3H2O (l)
The reaction is first order in BrӨ, first order in BrO3Ө, and second order in H⊕
a)
Write the rate law.
rate = k[BrӨ][BrO3Ө][ H⊕]2
b)
What is the overall reaction order?
1 + 1+ 2 = 4
How does the reaction rate change if the concentration of H⊕ is tripled?
c)
The rate goes up by 9 times.
What is the change in the rate if the concentrations of both BrӨ and BrO3Ө are halved
d)
The rate will decrease by a factor of ¼
5. At 600oC, acetone (CH3COCH3) decomposes to ketene (CH2=C=O) and various
hydrocarbons. Initial rate data are given in the table below:
Experiment
1
2
a)
Initial [CH3COCH3] (M)
6.0 x10-3
9.0 x10-3
Initial rate of decomposition (M•s-1)
5.2 x 10-5
7.8 x 10-5
Determine the rate law.
Rate = k[CH3COCH3]m
Rate 2
Rate 1
7.8 x 10-5 =
5.2 x10-5
1.5
=
k[CH3COCH3]m experiment 2
k[CH3COCH3]m experiment 1
(9.0 x 10-3)m
(6.0 x 10-3)m
= (1.5)m
Log 1.5 = m log 1.5
m=1
Rate = k[CH3COCH3]1
b)
What is the value of the rate constant?
Using experiment 1, fill in the variables:
Rate = k[CH3COCH3]1
5.2 x10-5 M•s = k {6.0 x10-3 M)
k = 5.2 x10-5 M•s
6.0 x 10-3M
= 8.7 x10-3 s-1
What is rate of decomposition when the acetone concentration is 1.8 x 10-3 M
c)
Rate = k[CH3COCH3]1
Rate = k[CH3COCH3]1
= (8.7 x10-3 s-1)(1.8 x10-3M) = 1.57 x10-5 ~ 1.6 x 10-5 M•s-1
6.
For the reaction:
NH4⊕(aq) + NO2Ө(aq) →
N2(g)
+
2H2O(l)
Initial rate data at 25oC are listed in the table below:
Experiment
1
2
3
a)
Initial [NH4⊕]
0.24 mol/L
0.12 mol/L
0.12 mol/L
Initial [NO2Ө]
0.10 mol/L
0.10 mol/L
0.15 mol/L
Determine the rate law.
rate = k[NH4⊕]m[NO2Ө]n
rate2 = k[NH4⊕]m[NO2Ө]n (from experiment 2)
rate 1 k[NH4⊕]m[NO2Ө]n (from experiment 1)
3.6 x10-6 M•s-1 = k(0.12 mol/L)m(0.10 mol/L)n
7.2 x10-6 M•s-1
k(0.24 mol/L)m(0.10 mol/L)n
0.5 = 0.5m
m =1
Initial rate of consumption of NH4⊕]
7.2 x10-6 M•s-1
3.6 x10-6 M•s-1
5.4 x10-6 M•s-1
Rate3 = k[NH4⊕]m[NO2Ө]n (from experiment 3)
rate 2 k[NH4⊕]m[NO2Ө]n (from experiment 2)
5.4 x10-6 M•s-1 = k(0.12 mol/L)m(0.15 mol/L)n
3.6 x10-6 M•s-1
k(0.12 mol/L)m(0.10 mol/L)n
1.5 = 1.5n
n =1
rate = k[NH4⊕]1[NO2Ө]1
b)
What is the value of the rate constant?
rate = k[NH4⊕]1[NO2Ө]1
from experiment 3,
5.4 x10-6 M•s-1 = k(0.12 mol/L)1(0.15 mol/L)1
k = 3.0 x10-4 M-1s-1
c)
What is the reaction rate when the concentrations of [NH4⊕] =0.39 M and [NO2Ө] =
0.052 M
rate = k[NH4⊕]1[NO2Ө]1
rate = 3.0 x10-4 M-1•s-1 (0.39 M)(0.052M) = 6.1 x10-6M•s-1