kinetics answers

Hera are the answers to the Kinetics problems, the superscripts and subscripts don't translate well, neither did the diagram. If you need clarification then I can email you the answers.

3. Tripling the [OH] causes the rate to triple so the order with respect to OH is 1
4. Doubling the [Cl2] causes the rate to double, doubling the [Cl2] and doubling the [NO] causes the rate to increase 8 fold so the rate law would be:
Rate = k[NO]2[Cl2]
5. Insert the given data for say expt. 1:
0.269M min-1 = k [0.5M] [1.8M]2
solve for k gives 0.269 M min-1/ 0.5M x 1.8M2
k = 0.166M-2 min-1, however we are asked for units of L2 mol-2 min-1 M = mol l-1
so, k = 0.166 L2 mol-2 min-1
Intergrated rate laws
1. Half life does not depend on the initial concentration for a first order reaction
1/8 of original value = 3 half lives so (500s x 3) = 1500s

2. Initial concentration [A]o = 1.00M, at t= 480s the concentration of B was 0.385M, so the concentration of A at time t [A]t is 1-0.385 = 0.615M
For a first order reaction:
ln[A]t = -kt + ln[A]o ln(1) =0

ln[A]t = -kt

ln 0.615 = -k x 480s

so k = 0.001s-1

3. For a first order reaction t 1/2 = 0.693/k
If k = 1.01 x 10-3 s-1 then t1/2 = 0.693/1.01 x 10-3
t 1/2 = 686s
4. After 10 half lifes the remaining sample = 1/210 = 1/1024 = 0.000977


Reaction mechanisms and rate laws, reaction pathways
1.

1. HBr(g) + O2(g) → HOOBr(g)
2. HOOBr(g) + HBr(g) → 2HOBr(g)
3. 2HOBr(g) + 2HBr(g) → 2H2O(g) + 2Br2(g)
a. Overall: 4 HBr(g) + O2(g) → 2H2O(g) + 2Br2(g)
b. Rate = k[HBr][O2] the rate law = the slowest, rate determining step so the rate law is the rate law for step 1. Step 1 is therefore the slow (rate determining) step
c. Intermediates are produced in 1 step and consumed in another so: HOOBr(g) and HOBr(g)
d. No, the intermediate produced in the step is consumed very rapidly in the next step

2.
a. Intermediates are produced in 1 step and consumed in another so: NO3
b. rate law is the rate law for the slow step so: rate = k[NO2]2 {uses the stoichiometric coefficients of the elementary steps (not the overall reaction)}
c.
Energy
Reaction Pathway





Something like this overall exothermic, first step big Ea endothermic, 2nd step small Ea exothermic
3. The catalyst decreases the Activation Energy. The equilibrium is not affected, it is just reached quicker.
4. Done in class:

Overall reaction:

Cl2(g) + CHCl3(g) → CCl4(g) + HCl(g)
The reactive intermediates are: Cl(g) and CCl3(g)

Rate Law is equal to the rate law for the slowest step so rate = k[Cl][CHCl3]

But Cl is a reactive intermediate so we must solve for it with something in the overall equation:

Step 1 is at equilibrium, so rate = k1[Cl2] = rate =k-1[Cl]2

So [Cl]2 = k1/k-1[Cl2]

So k1/k-1[Cl2]1/2

So overall rate law is rate = k1k2/k-1[CHCl3][Cl2]1/2 = k [CHCl3][Cl2]1/2