3.19 RATE EQUATIONS

Change in concentration/time

Surface area
Temperature
Concentration(pressure if gas)
Catalyst

Number that tells you how concentration of a reactant affects the reaction rate

Gives mathematical link between concentration of reactants to ROR
Rate(ROR) = k [A]$^m$[B]$^n$

Carry multiple experiments with reactant in different concentration with other reactants in excess and calculate initial rate(gradient)
Then plot a graph with rate against reactant concentration and find order

0 = no change EG if concentration x2 = rate x 2$^0$
1 = directly proportional EG if concentration x2 = rate x 2$^1$
2 = x$^2$ EG if concentration x2 = rate x 2$^2$

Sum of all order numbers of reactants

Increase ROR without being used so doesnt appear in chemical equations but DOES in rate equation

its concentration will change little so it can be considered constant so unlikely to change ROR so we give zero order to it

rate = k[A]$^0$ but you remove A since 0 so rate = k constant

constant half life

Increasing half life
Calculate order via finding conc and rate(gradient) at two points and compare

Links ROR with concentration of reactants
Increases with temperature exponentially

Re-arrange equation for k
rate units = mol dm$^{-3}$ s$^{-1}$
concentratio units = mol dm$^{-3}$
Cancel anything on top and bottom

Its the slowest step in a reaction which takes multiple steps to do

y is created but then used up so you dont include it (since its hard to calculate its concentration) but instead include the things that created it

1) OH- immediately attacks C this is slow and leads to formation of intermediate where OH- has partially bonded to C while Br has partially broken which is then followed up fast step of it fully breaking bond and forming bond.
Since both OH and haloalkane are part of rate determining step equation is Rate = k[CH3CH2Br][OH-]
2) First Br breaks bond with C forming carbocation then in part 2 OH- attacks carbon and forms bond.
Since only haloalkane is part of rate determining step equation is Rate = k[CH3CH2Br]

1) $S_{N}$ 2
2) $S_{N}$ 1
Primary haloalkanes tend to undergo $S_{N}$ 2 while tertiary $S_{N}$ 1

K = Ae$^{-Ea/RT}$
K = rate constant
A = pre-exponential factor constant(arrhenius constant) ( mol$^{-1}$dm$^{3}$ s$^{-1}$ )
Ea = Activation energy (J mol$^{-1}$)
R = constant gas (8.314 J mol)
T = temperature (kelvin)

take natural log of both removing the e part
Turns K = Ae$^{-Ea/RT}$
INTO InK = InA - Ea/RT

Rearrange for ea by first adding Ea/RT to both sides making it go left side then subtract InK then multiply RT

-Ea/R

R = 8.31