Solutions and Tests for Exploring Creation With Chemistry
SOLUTIONS TO T H E E X T R A P R A C T I C E P R O B L E M S F O R M O D U L E #14
1. Reaction rate has units of M/s. When solving for the rate constant, you divide by the concentrations
of the reactants raised to their orders. To get these units for the rate constant, then, the rate must be
divided by M^. That means the overall order is 2.
2. I f the order is three, then you must raise the concentration to the third power. I f you double the
concentration, that would be like raising 2 to the third power, which is 8. Thus, the rate increases by a
factor of 8.
3. The rate equation for this reaction will look like:
R = k[N0f[Cl2f
To figure out k, x, and y, we have to look at the data from the experiment. The value for x can be
determined by comparing two trials in which the concentration of NO changes, but the concenfration
of CI2 stays the same. This would correspond to trials 1 and 2. In these two trials, the concentration of
NO doubled, and the rate went up by a factor of 4. This means that x = 2, because the only way you
can get a 4-fold increase in rate from a doubling of the concentration is by squaring the concenfration.
The value for y can be determined by looking at trials 2 and 3, where the concenfration of NO stayed
the same but the concentration of O2 doubled. When that happened, the rate increased by a factor of 2.
Since rate doubled when concentration doubled, that means y = 1. Thus, the rate equation becomes:
R = k[N0]'[Cl2]
Now that we have x and y, we only need to find out the value for k. We can do this by using any one
of the trials in the experiment and plugging the data into the equation. The only unknown will be k,
and we can therefore solve for it:
R = k[N0]'[Cl2]
M
0.113— = k-(0.050 M ) ' -(0.050 M )
s
k=
0.1131
,1
=9.0x10^
(0.0025 M')-(0.050 M )
'
M'-s
Thus, the final rate equation is:
R = (9.0xlO^^)-[NOf[ClJ
4. The rate equation will take on the form:
R = k[C3H6Br2]''[r]y
Extra Practice Problem Solutions
To determine x and y, we look at trials where the concentration of one reactant stayed the same and the
concentration of the other reactant changed. In trials 1 and 2, the concentration of T remained the
same but the concentration of C3H6Br2 doubled. When that happened, the rate doubled. This means
that X = 1. In trials 1 and 3, the C3H6Br2 concentration remained constant, but the V concentration
doubled. When that happened, the rate doubled. This means y = 1. The rate equation, then, looks
like:
R = k[C3H6Br2][r]
To solve for k, we can use the data from any frial and plug it into our rate equation. We can then solve
fork:
R = k[C3H,BrJ[r]
M
0.234 — = k • (0.100 M ) • (0.100 M )
M
0.234 —
,
k =
^
= 234
—
(0.100 M)-(0.100 M )
• Ms
The overall rate equation, then is
5. Since chemical reaction rate doubles for every 10 °C increment, then to increase the rate of the
reaction by a factor of 8,1 just need to raise the temperature by 3 ten degree increments. That way, I
will multiply the old rate by 2x2x2, which equals 8. Thus, to increase the reaction rate by a factor if 8,
I just raise the temperature by 30 degrees. Therefore, the new temperature should be 25 °C + 30°C =
55 °C.
6. A catalyst is used up in an early step and remade in a later step so that its concenfration does not
change. This is happening with Clg. It is a homogeneous catalyst, as it is in the same phase as the
reactants.
7. It lowers the size of the hill