1
Sample Exam 1 solutions
1.
(a)
To find the reaction orders, look at the dependence of rate on
concentrations.
From rows 1 and 3, cutting the [OCl
–
] in half cuts the rate in half,
so the reaction is
1st order in OCl
–
.
Similarly, from rows 2 and 3, cutting the [I
–
]
in half cuts the rate in half, so it's also
first order in I
–
.
From rows 3, 4 and 5 it is
clear that cutting hydroxide concentration in half doubles the rate, so it is
inverse
first order in OH
–
.
(b) The overall reaction order will then be 1+1–1 =
1
(c) The rate law is
rate = k [OCl
–
][I
–
]/[OH
–
]
To find the rate constant k, plug in some data.
From row 1, for example,
rate = 4.8x10
–4
Ms
–1
= k [0.0040M] [0.0020M]/[1.00M]
k = 60s
–1
2.
Note for 2003, 2004, 2005, etc.
If you found this one difficult, don't worry
too much.
...we have not looked at similar 'pressure' problems this year.
First define P = total pressure, P
o
= initial pressure of C
2
H
4
O, P
a
= pressure of
C
2
H
4
O, P
b
= P
c
= pressure of CH
4
= pressure of CO, and x = number of moles of
C
2
H
4
O consumed at a given time in the reaction.
You need to convert the total
pressure to the pressure of A (P
a
), and then do the usual plots of this pressure
(which is proportional to concentration) vs time to find the reaction order.
First, find the initial pressure of A.
At "infinite" time, when you can assume all the
C
2
H
4
O has reacted, P = P
b
+ P
c
= 2P
b
= 249.88 torr.
Therefore, P
b
= 124.94
torr.
From the stoichiometry of the reaction, that means the initial pressure P
o
=
124.94 torr.