Download KEY Chem Activity 58 Integrated Rate Laws Brief answers to the

KEY Chem Activity 58 Integrated Rate Laws
Brief answers to the EXERCISES (see below for longer answers):
1. First order because the graph of ln[N2O5] vs. time is a straight line.
Equation is therefore ln[N2O5] = -(6.92 x 10-3) t + ln[N2O5]0.
b. [N2O5] = 0.0933 mol/L
2. Graph ln[BrO3-] vs time and 1/[BrO3-] vs time. Whichever line is straight gives the order of
the reaction. If ln[BrO3-] vs time is linear, it is a first order reaction, if 1/[BrO3-] vs time is
linear it is a second-order reaction.
3. k = 6.58 x 10-4 s-1
T = 2,110 seconds.
4. The rate does not remain constant as the reaction proceeds. We know this because when
[A] is graphed vs. time a straight line is not produced. The slope of [A] vs. time would be a
straight line if the rate of the reaction was constant. If you still doubt this, use the data in
problem number 1 to calculate ΔM/Δt at different stages of the reaction. You’ll get different
numbers.
5. IV is the correct graph.
Model #2:
4. a.0.010 mol/L
b. 0.010 M/.005 M = 2/1
c. 0.005 M/0.0025 M = 2/1
d. 0.0025 M/0.00125 M = 2/1
5) Skip.
Model #3:
6. The t1/2 does not depend on the concentration of chloroethane! No matter how much
chloroethane is present, the half-life is the same. In other words, it always takes the same
amount of time to decrease the concentration of chloroethane by half.
7. .5R0
b. plug numbers into the integrated rate law for a first order reaction:
ln(0.5) = ln1 – kt Solve for t (using knowledge of logarithms) and you get 0.693/k = t!!
8. Skip
Exercises
7. Half life is 1050 seconds (plug k into the t1/2 = 0.693/k equation)
25% is two half-lives (one half life is 50%, the next half-life would be 25%). 2 x 1050 =
2100 seconds. This is consistent with the answer to #3 above.
8. 1/8
9. 86.8% would remain
10. The bowl is 7350 years old.
Long Answer to #1: The graphing was done in class. It is a first-order equation with respect to
N2O5. The rate law is rate = k{N2O5]1.
That means that the equation describing the line is
Ln[N2O5] = -kt + ln[N2O5]0.
There are two ways to determine k. One way is to plug in a set of numbers from the data
table. For instance, you can choose time 50s and plug in the numbers like so:
Ln(0.0707) = -k(50) + ln(0.1000)
Solve for k. k=6.93 x 10-3
OR you can calculate the slope of the line using rise over run OR Δy/Δx.
-2.649-(-2.303) = 6.92 x 10-3
50-0
(k is nearly the same value no matter which method is used!)
That makes the full equation describing this reaction:
Ln[N2O5] = -(6.92 x 10-3) t + ln[N2O5]0.
b. To calculate the concentration of N2O5 at t=10 s, plug 10 s into the equation for time.
Ln[N2O5] = -(6.92 x 10-3) (10) + ln(0.1000)
You have to know your calculator well in order to solve this equation for [N2O5]. It’s easy to
do the first parts:
ln[N2O5] = -2.3722
To take the inverse natural log, push the SHIFT or 2nd Function
key, then the ln button.
[N2O5] = 0.093275 rounded to three sig figs = 0.0933 (although the times have just one sig
fig apiece – really, it should be rounded to only one sig fig!)
Check your work: does this concentration make sense? Look at the data table: 10 seconds
would fall between 0 and 50 seconds, and the concentration present at those times is
0.1000 and 0.0707, respectively. 0.0933 falls between them, so the calculations are correct.
Long Answer to #3:
Assume an initial concentration of 1. Therefore, at 247 s, the concentration would be .85.
Plus those numbers into the equation and solve for k.
ln(0.85) = ln (1) – k(247)
Then to determine what time the concentration of CH3NC is 25% of its original value, you
should use 1 as the initial concentration and .25 as the 25% of the original in the equation.
ln(.25) = -(6.58 x 10-4)(t) + ln(1)