DETERMINATION OF ORDER USING THE INTEGRATED RATE LAWS
HOW MUCH REMAINS OVER TIME?
Salvador Dali,
“Apparition of Face and Fruit Dish on a Beach,” 1938
LINEARIZING DATA
For the reaction A → B,
here are integrated rate laws we developed in the last lecture when the order of the reaction is:
LINEARIZING DATA
If we rearrange the equations a little, we get this:
WHY DID WE WRITE THE INTEGRATED RATE LAWS THIS WAY?
WHY DID WE WRITE THE INTEGRATED RATE LAWS THIS WAY?
WHY DID WE WRITE THE INTEGRATED RATE LAWS THIS WAY?
Consequences:
If you graph
the one that gives a linear plot indicates the order of the reaction.
This is called “linearizing data”
WHY DID WE WRITE THE INTEGRATED RATE LAWS THIS WAY?
PRACTICE
The concentration vs time data for the decomposition of NO2 is shown above. Using graphical methods, a) determine if the reaction is 0th, 1st, or 2nd order with respect to NO2, b) determine the value of the rate constant, k, and c) write the rate law for the reaction.
Time (s) | [NO2] (M) |
0 | 1.00 × 10−2 |
60 | 6.83 × 10−3 |
120 | 5.18 × 10−3 |
180 | 4.18 × 10−3 |
240 | 3.50 × 10−3 |
300 | 3.01 × 10−3 |
360 | 2.64 × 10−3 |
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
To do this, lets first determine the ln[NO2] and 1/[NO2] for each data point.
Time (s) | [NO2] (M) |
0 | 1.00 × 10−2 |
60 | 6.83 × 10−3 |
120 | 5.18 × 10−3 |
180 | 4.18 × 10−3 |
240 | 3.50 × 10−3 |
300 | 3.01 × 10−3 |
360 | 2.64 × 10−3 |
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
To do this, lets first determine the ln[NO2] and 1/[NO2] for each data point.
Time (s) | [NO2] (M) | ln[NO2] | 1/[NO2] (M−1) |
0 | 1.00 × 10−2 | | |
60 | 6.83 × 10−3 | | |
120 | 5.18 × 10−3 | | |
180 | 4.18 × 10−3 | | |
240 | 3.50 × 10−3 | | |
300 | 3.01 × 10−3 | | |
360 | 2.64 × 10−3 | | |
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
To do this, lets first determine the ln[NO2] and 1/[NO2] for each data point.
Time (s) | [NO2] (M) | ln[NO2] | 1/[NO2] (M−1) |
0 | 1.00 × 10−2 | −4.605 | 100 |
60 | 6.83 × 10−3 | −4.986 | 146 |
120 | 5.18 × 10−3 | −5.263 | 193 |
180 | 4.18 × 10−3 | −5.477 | 239 |
240 | 3.50 × 10−3 | −5.655 | 286 |
300 | 3.01 × 10−3 | −5.806 | 332 |
360 | 2.64 × 10−3 | −5.937 | 379 |
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
Now we should graph each variation of the concentration vs. time to determine which gives a linear relationship.
Time (s) | [NO2] (M) | ln[NO2] | 1/[NO2] (M−1) |
0 | 1.00 × 10−2 | −4.605 | 100 |
60 | 6.83 × 10−3 | −4.986 | 146 |
120 | 5.18 × 10−3 | −5.263 | 193 |
180 | 4.18 × 10−3 | −5.477 | 239 |
240 | 3.50 × 10−3 | −5.655 | 286 |
300 | 3.01 × 10−3 | −5.806 | 332 |
360 | 2.64 × 10−3 | −5.937 | 379 |
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
Now we should graph each variation of the concentration vs. time to determine which gives a linear relationship.
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
2 NO2(g) 🡪 2 NO(g) + O2(g)
PRACTICE
This means the rate law for the reaction is:
rate = 0.79M-1s-1[NO2]2
2 NO2(g) 🡪 2 NO(g) + O2(g)
GUIDED PRACTICE TIME ON YOUR OWN COMPUTER!
Time (s) | [N2O5] (mol/L) |
0 | 0.0365 |
600 | 0.0274 |
1200 | 0.0206 |
1800 | 0.0157 |
2400 | 0.0117 |
3000 | 0.00860 |
At a temperature of 225oC, the data to the right was collected for the reaction, 2 N2O5(aq) 🡪 4 NO2(aq) + O2(g)
a) Graph the data and appropriate variations of the data to determine the order of the reaction with respect to N2O5 (g).
b) Determine the value of the rate law constant. Don’t forget the units!
c) Write the general rate law equation for this reaction.
GUIDED PRACTICE TIME ON YOUR OWN COMPUTER!
Time (s) | [N2O5] (mol/L) |
0 | 0.0365 |
600 | 0.0274 |
1200 | 0.0206 |
1800 | 0.0157 |
2400 | 0.0117 |
3000 | 0.00860 |
At a temperature of 225oC, the data to the right was collected for the reaction, 2 N2O5(aq) 🡪 4 NO2(aq) + O2(g)
a) Graph the data and appropriate variations of the data to determine the order of the reaction with respect to N2O5 (g).
ln[N2O5] vs t is linear
so,
the reaction is 1st order
GUIDED PRACTICE TIME ON YOUR OWN COMPUTER!
Time (s) | [N2O5] (mol/L) |
0 | 0.0365 |
600 | 0.0274 |
1200 | 0.0206 |
1800 | 0.0157 |
2400 | 0.0117 |
3000 | 0.00860 |
At a temperature of 225oC, the data to the right was collected for the reaction, 2 N2O5(aq) 🡪 4 NO2(aq) + O2(g)
b) Determine the value of the rate law constant. Don’t forget the units!
-k = m
k = -m
so,
k = 4.79x10-4 s-1
GUIDED PRACTICE TIME ON YOUR OWN COMPUTER!
Time (s) | [N2O5] (mol/L) |
0 | 0.0365 |
600 | 0.0274 |
1200 | 0.0206 |
1800 | 0.0157 |
2400 | 0.0117 |
3000 | 0.00860 |
At a temperature of 225oC, the data to the right was collected for the reaction, 2 N2O5(aq) 🡪 4 NO2(aq) + O2(g)
c) Write the general rate law equation for this reaction.
rate = k[N2O5]x
we know k and x
so,
rate = 4.79x10-4 s-1[N2O5]
GUIDED PRACTICE TIME ON YOUR OWN COMPUTER!
Time (s) | [N2O5] (mol/L) |
0 | 0.0365 |
600 | 0.0274 |
1200 | 0.0206 |
1800 | 0.0157 |
2400 | 0.0117 |
3000 | 0.00860 |
At a temperature of 225oC, the data to the right was collected for the reaction, 2 N2O5(aq) 🡪 4 NO2(aq) + O2(g)
d) Write the integrated rate law equation for this reaction. Don’t forget the units on constants!
ln[N2O5]t = -kt + ln[N2O5]0
so,
ln[N2O5]t = -(4.79x10-4 s-1) t + ln(0.0365)
ln[N2O5]t = -(4.79x10-4 s-1) t - 3.31
GUIDED PRACTICE TIME ON YOUR OWN COMPUTER!
Time (s) | [N2O5] (mol/L) |
0 | 0.0365 |
600 | 0.0274 |
1200 | 0.0206 |
1800 | 0.0157 |
2400 | 0.0117 |
3000 | 0.00860 |
At a temperature of 225oC, the data to the right was collected for the reaction, 2 N2O5(aq) 🡪 4 NO2(aq) + O2(g)
e) Calculate the [N2O5] at 2.70x104 s after the start of the reaction.
ln[N2O5]t = -(4.79x10-4 s-1) t - 3.31
so,
ln[N2O5]t = -(4.79x10-4 s-1)(2.70x104 s) + 0.0365 M
ln[N2O5]t = -12.90
[N2O5]t = e-12.90
[N2O5]t = 2.51x10-6 M