Unit 4 Flashcards
Initial rate
Rate at start of reaction, where it is fastest.
Calculating order of particular reactant:
Compare 2 experiments where only that reactant is being changed.
If conc. Is doubled and rate stays same:0 order
If conc. Is doubled and rate doubles: 1st order
If conc. Is doubled and rate quadruples: 2nd order.
Working out orders when 2 reactant conc. Are changed at same time.
Where r= k[A][B]^2:
If [A] is x2, rate is x2
If [B] is x3, rate is x3^2 = x9
If this happens at the same time, rate is x2x9=18
Rate determining step
Slowest step in a mechanism.
No. of molecules is same as the order for each substance.
If rate determining step has intermediate, use what made up that intermediate from previous step.
Sn1 and Sn2
Sn1= substitution, nucleophilic, 1 molecule in rate determining step. Sn2= substitution, nucleophilic, 2 molecules in RDS.
Kc
Equilibrium constant
Units depends on equation.
Working out units for Kc
- Put units of conc. Into equation for Kc.
2. Cancel out, including powers.
Calculating Kc
Mol. of reactant at equilibrium= initial mol. - mol. Reacted
Mol. of product at equilibrium= initial mol. + mol. formed.
1. Draw table of mol. and equilibrium mol.
2. Initial mol. Of product is always 0.
3. Work out equilibrium conc. by dividing equilibrium mol. by vol.
4. Put conc. into Kc equation.
When Kc has no units
There are equal no. Of reactants and products, ∴don’t need to divide by vol. to find conc., can just put equilibrium mol. straight into Kc expression.
Calculating units for k(rate constant)
- Rearrange rate equation to make k the subject.
- Insert units and cancel.
Units of k=s^-1
Effect of temp on position of equilibrium and Kc
If position of equilibrium moves to left, value of Kc gets smaller as less products.
Catalysts have no effect on Kc or equilibrium.
Effect of pressure on position of equilibrium and Kc
With increased pressure, position of equilibrium will shift to side with less mol. of gas.
Kc will stay constant, as only changes with temp change.
Catalysts have no effect on Kc or equilibrium.
Bronsted- Lowry Acid
Substance that can donate a proton. (Proton donator).
Brondsted-Lowry Base
Substance that can accept a proton. (Proton acceptor).
Calculating pH
pH= -log [H+]
Calculating pH of strong acid
Strong acids completely dissociate.
Conc. of H+ ions in a monoprotic strong acid will be same as conc. of acid.
Give pH to 2dp.
Calculating [H+] from pH
[H+] = 1x10^-pH (inverse log on calculator)
Ionic product of water, equation
Kw= [H+][OH-]
At 25C, Kw=1x10^-14 mol^2dm^-6
pH of pure water
[H+]=[OH-], ∴Kw= [H+]^2, ∴[H+]=_/Kw (square root of Kw)
pH of strong base
- Find [H+] using Kw.
2. Use pH= -log[H+] to find pH.
Weak acids
Only partially dissociate when dissolved in water, giving equilibrium. HAH^+ + A^- Weak acid dissociation equation: Ka=[H+][A-]/[HA] Larger the Ka, stronger the acid. pKa= -logKa, ∴Ka=10^-pKa.
Calculating pH of weak acid
[H+]= [A-],∴Ka=[H+]^2/[HA]initial [HA]equil = [HA]initial.
Strong acid and strong base neutralisation
- Work out mol. of original acid.
- Work out mol. of base added.
- Work out which one is in excess.
- If excess acid: work out new conc. of excess H+ ions, [H+]= mol. excess H+ /total vol.(vol. of acid + base)
- pH= -log[H+]
- If excess alkali: work out new conc. of excess OH-ions,
[OH-]= mol. of excess OH- /total vol. - [H+]= Kw/ [OH-]
- pH= -log[H+]
Weak acid and strong base neutralisation
- Work out mol. of original acid.
- Work out mol. of base added.
- Work out which is in excess.
- If excess acid: work out new conc. of excess HA, [HA]= initial mol. HA - mol. OH- / total vol.
- Work out conc. of sat formed, [A-]=mol. OH- added/ total vol.
- Rearrange equation for Ka to get [H+].
- pH= -log[H+]
- If excess alkali: same method as with strong acid and strong base.