CHAPTER 13 REGULAR HOMEWORK
KINETICS II
Question 1
Select one or more:
Explanation
Question 2
Determine the rate constant from the concentration-time dependence (The units for the initial rate of reaction are Ms-1):
[A] | [B] | Init. Rate |
0.020 | 0.15 | 0.055 |
0.035 | 0.15 | 0.096 |
0.050 | 0.35 | 0.75 |
Explanation
Question 3
Explanation
- Convert torr to atm (this is done because our k value is in atm)
- Treat the atm values as you would concentration and use the second order integrated rate law to solve for t (seconds needed)
Question 4
Explanation
Question 5
The decomposition of formic acid (see below) is measured at several temperatures. HCOOH(g) → CO2(g) +H2(g) The temperature dependence of the first-order rate constant is: T(K) . . . . k(s-1)
800 ....0.00027 825 ....0.00049 850 ....0.00086 875 ....0.00143 900 ....0.00234 925 ....0.00372
Calculate the activation energy, in kJ/mol. Use all data points and do a linear regression using calculator or Excel. Do not pick 2 data points. This is less accurate and assumes all data points are equally good.
Explanation
Question 6
Explanation
Question 7
Explanation
Question 8
Explanation
Question 9
Explanation
Question 10

Explanation
Question 11
Explanation
N2O4 is the intermediate because it is created and consumed within the reaction-is not a reactant nor a product. This question is slightly different than the previous one in that the slow step depends on the products of the fast step to occur–an important distinction.
- N2O4 is made from two molecules of NO2 (in the reversible fast reaction) and so its concentration will depend on the concentration of NO2.
- That means that you can replace [N2O4] by [2NO2]
- Our initial rate equation for the slow reaction was:
- Our new rate equation (with replacement) is: