AQA Chemistry Unit 4 with Anon_98Watch
I'll tag the people who previously said they were interested once I've finished. Really hoping it won't take me too long bc I only have 2..and a bit..ish days + I'd like to spend one of those days memorising the info + doing the past papers.
Edit: Sorting out my notes will take me way too long + I don't have time. As suspected, I am actually missing quite a lot of info on a few of the topics. I basically have all of the physical stuff but lacking in about 50% of the organic section... which isn't great.
Anyway, I decided to check my school website instead + luckily there is lots of material which I can use so I think (or at least, hope) that's pretty much saved me. I've also decided that I'm going to bed rn instead so I can wake up v early to work on this.
Okay, so for a reaction:
A + D -----------------> products.
And you need to know that the basic rate equation is:
r = k[A]x[D]y[C]z
r = rate equation, I.E. The change in concentration of a reactant (say A), per unit time.
Units: moldm-3 s-1
k= rate constant for the reaction; this is temperature dependant; it's units depend on the value of x, y and z.
[A] and [D] are the concentrations of the two reactants in the chemical equation.
[C] is the concentration of any catalyst present. [C] MUST be a catalyst bc it does not appear as a reactant in the chemical equation.
x, y and z are orders of reaction with respect to each of the reactants and the catalyst. Values can be typically 0, 1 or 2. These values can only be found from experimental data.
NOTE: Overall order of reaction = x + y + z.
Alright, so there are kinda 3 steps you need to remember to help you work out the rate equation from any data they may provide you with. IF THERE IS:
NO EFFECT in the data given. I.E. You double [A] and the rate is constant? Conclusion = 0 order.
AN EFFECT in the data given. I.E. You double [A] and the rate also doubles? Conclusion = 1st order.
AN EFFECT in the data given. I.E You double [A] and the rate quadruples? Conclusion = 2nd order.
Changes in temperature on the rate constant, k.
A reduction in temperature reduces the rate of reaction bc fewer particles have the activation energy or greater and so fewer particles have successful collisions.
The rate of reaction is directly proportional to the rate constant, k.
The graph of rate constant, k, against temperature looks as follows:
Species that feature in the rate equation + the number of reactant species multiplied by their orders of reaction must be involved in the rate determining step or any steps before it.
Kc is the equilibrium constant calculated from equilibrium concentrations for a system at constant temperature.
Homogeneous - All reactants + products in the same phase. A phase is a physically distinct part of a system with a boundary.
Dynamic - rate of forward reaction equals the rate of the backward reaction.
Equilibrium - constant concentrations of all reactants and products in a system.
For a reaction:
aP + bQ ---------> cR + dS
The equilibrium constant = Kc= [R]c [S]d / [P]a[Q]b
[R] = concentration of R in moldm-3
NOTE: A particular value of Kc refers to a particular temperature.
NOTE 2: If Kc has no units, then it's bc the number of moles on both sides of the equation are the same therefore they cancel out.
NOTE 3: When given data to work out the value of equilibrium constant, Kc then you must first draw up the table thingy. (Initial mol, amount reacted, equilibrium amount).
Then you must use the equation to construct an expression for Kc then substitute the equilibrium amount values you've just worked out.
The value of the equilibrium constant is not affected by changes either in concentration or the addition of a catalyst. The equilibrium constant is ONLY changed by changes in temperature.
To be able to predict the effect of changes in temperature on the value of the equilibrium constant, you must remember the following:
An increase in temperature moves the equilibrium position to the endothermic side of the equation to oppose the change.
A decrease in temperature moves the equilibrium position to the exothermic side of the equation to oppose the change.
Alright so, the Hydrogen ion, H+ is a proton and bc of this it has a v high charge density + does NOT exist in solution. It reacts with water to from the hydroxonium ion, H3O+. The proton is said to be labile which means it moves from one water molecule to another as the water molecules collide:
H3O+ (aq) + H2O(l) <------> H2O(l) + H3O+ (aq)
Acid – Proton donor.
Base - Proton acceptor.
When an acid/base reaction occurs, the acid becomes a base therefore creating a conjugate acid-base pair. For example:
CH3COOH(aq) (Acid 1) + H2O(l) (Base 2) <------------------> CH3COO- (aq) (Base 1) + H3O+ (aq) (Acid 2)
The ethanoic acid donates a proton to the water + the water becomes a base bc it accepts the proton. Therefore CH3COOH and CH3COO- are a conjugate acid-base pair, as are H2O and H3O+ .
In general, an acid HA (monoprotic acid) has a conjugate base A-.
A substance which can act as both an acid and a base is said to be amphiprotic.
pH = - log10[H+(aq)] (Where “[H+(aq)]” = Concentration of H+ in moldm-3 )
H+ = 10- PH
Alright so, PH is measured using a PH meter in the following way:
The pH meter is first calibrated with at least two buffer solutions of accurately known pH. This means putting the electrode into the buffer solution and adjusting the reading until it gives the pH of the buffer solution.Once calibrated the pH meter electrode is placed in a solution of unknown pH and a reading taken – usually to 2 decimal places.
Strong acid: Fully dissociated (e.g. HCl, H2SO4, HNO3)
Weak acid: Only slightly dissociated (e.g. CH3COOH)
Strong base: Fully dissociated or ionised (e.g. NaOH, KOH, BaOH)
Weak base: Only slightly dissociated or ionised (e.g. NH3, CO32-)
Water is said to be weakly dissociated, and even when pure has the slightest electrical conductivity. This is bc some of the water molecules ionise. The equilibrium reaction for the dissociation of pure water is:
H2O(l) + H2O(l) <--------------------> H3O+ (aq) + OH-(aq)
(The equilibrium position is way over to the left hand side so that’s why the dissociation is very slight.)
The equation can be simplified to:
H2O(l) <---------------------> H+ (aq) + OH-(aq)
Kw = [H+] [OH-] -- the ionic product of water.
The value of Kw (at 298K) is 1.0 x 10-14mol2dm-6
IF asked to work out calculate the pH of a strong base from its concentration then remember that [OH-] represents the base and the steps are as follows:
1) So, substitute the concentration given into the Kw formula.
2) Substitute the value of Kw which is given.
3) Rearrange the equation to get [H+] as the subject.
4) Use the PH equation to substitute the value of[H+] and find PH to 2.d.p.
Kw, like all equilibrium constants is temperature dependant. As the temperature rises so does the value of Kw.
What is the sign of DH for the dissociation of water and why?
So if we know that Kw increases with increasing temperature, then the equilibrium position must move over to the right hand side. According to Le Chatelier’s principle this must be because the forward reaction is endothermic.I.E. An increase in temperature is opposed by moving the position of equilibrium in the endothermic direction.
Therefore, the sign of DH is positive.
basically represent the reversible reaction. - I don't know how else to do it. :'3