12.2&12.3 rate law

Question 1

Why use rate law?

Answer 1

Sometimes, ∆[A] gets complicated if there is reverse reaction occurring. Using rate law helps us calculate the instantaneous rate without the ∆[A].

Question 2

What is the equation? What does each symbol mean? Add the unit of measure

Answer 2

Rate= - ∆[A] / ∆t = k [A}m [B]n
[A]= concentration of the REACTANT (mol/L)
k= rate constant (variable unit)
m and n= reaction order
Rate= mol/L*sec or M/s

Question 3

How can we determine the rate law of a reaction?

Answer 3

It has to be determined experimentally. cannot be predicted

Question 4

What must we determine when trying to find the rate law?

Answer 4

k and n(and m)

Question 5

What does it mean to have an order of 0, 1 ,2, 3?

Answer 5

0 order:no matter how much we change [ ] the rate will stay the same (anything^0=1)
1st order: means that if we multiply the [ ] by 2, the rate will also double by 2. (2^1=2)
2nd order: if [ ] x2, rate x 4 (2^2=4)
3rd order: [ ]x2 = ratex8 (2^3=8)

Question 6

2 NO (g) + O2 (g) → 2 NO2(g)
initial [NO]: 0.015, 0.030, 0.015
initial [O2]: 0.015, 0.015, 0.030
Initial rate of the formation of NO2: 0.048, 0.192, 0.096

Calculate the rate law, rate constant and rate constant unit.

Answer 6

Rate = k [NO]2 [O2]
k= 1.4*104 1/M^2*s

Question 7

What is the rate of decomposition of N2O5 at 55°C when its decomposition is 0.030M? (k= 1.7*10-3 s-1 )

Answer 7

Because the unit of “k” is s-1 so it has to be 1st order.

Rate= - Δ [ N2O5]/Δt = k [N2O5] = (1.710-3 s-1) (0.030 M) = 5.110-5 M/s