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A- Level Chemistry A OCR Summary Notes, Post 2015 Spec
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Summary notes of the official OCR A Level Chemistry A OCR Student Book by Rob Ritchie and Dave Gent. Made by an A* student.
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CHEMISTRY A2Rates of Reactions
Orders, Rate Equations and Rate Constants
Rate = change in concentration/ change in time. [A] is shorthand for ‘concentration of A’.
Rate of Reaction: The change in concentration of a reactant or product per unit time.
Units: Usually it is mol dm-3 s-1. -1 in superscript means ‘per’ or ‘divide’.
Orders of Reaction: The order with respect to a reactant is the power to which the
concentration of the reactant is raised in the rate equation. Common orders are…
Zero Order: When concentration of reactant has no effect on rate. Rate ∝ [A]0 . Any number
raised to power zero is 1. Know zero order if reactant is in excess as conc doesn’t change much.
First Order: Rate ∝ [A]1 . If concentration A x3, reaction rate increased by x3 1 = x3.
Second Order: Rate ∝ [A]2. If concentration of A x3, reaction rate increased by x3 2 = x9.
Rate Equation: Rate = k [A]m [B]n
Rate Constant k: The constant that links the rate of reaction with the concentrations of the
reactants raised to the powers of their orders in the rate equation.
M and n superscript are order of reaction. Multiply A and B.
If question asks to ‘write the rate equation’- write equation above but omit zero order reactants
completely and the 1 for first order. Include state symbols.
Write 7.0 x 10n for standard form and not just 7.
Overall Order: Overall order gives overall effect on concentrations of all reactants of rate of
reactions. Overall order = sum of orders with respect to each reactant = m + n from equation
above.
Units of Rate Constant: Make k the subject. Remember rate has different units to concentration.
Cancel top and bottom. Move any units in bottom to top by inversing sign in powers. Put positive
indices before negative in units.
Multiply when power outside bracket OR add powers e.g. (moldm -3 )(moldm-3) = mol2dm-6.
K = rate/ [A]m [B]n
- CO x2, rate of reaction also x2 so reaction is 1 st order with respect to CO.
- Make sure to say which experiments. In table ignore a reactant if zero order, even if values
change when comparing.
- Write rate equation once found out all orders, with chemical symbols. Even if not asked to.
- Remember to write the units of k rate constant. K is just a number- a constant.
- Only reactants involved in rate equation and power is not the balancing number but the
order.
- Zero order wrt to reactant when rate does not change with concentration.
- If something zero order, still prove it is zero order.
Experimental Results: When comparing effect of concentrations, measure rate for all at initial
rate when t = 0.
Change standard form into normal numbers to make it less confusing.
Or just leave out standard form bit in calculator but adjust any standard form differences if
needed.
, Look out for standard form differences.
Concentration- Time Graphs
Colorimeter: Monitoring colour change/ concentration. Amount of light absorbed by solution.
Experiment Method: Prepare standard solutions of known concentrations of solution. Choose
filter with complementary colour to coloured chemical. Zero colorimeter with water. Measure
absorbance readings of standard solutions. Plot calibration curve of absorbance against
concentration of solution.
Carry out reaction. Take absorbance readings of solution at time intervals. Use calibration curve
to determine concentration of solution at each time interval. Plot second graph of concentration
of solution against time to determine order of reaction
Concentration- Time Graphs: Gradient of concentration- time graphs is rate of reaction. Can use
graph to find order if all other reactant concentrations remain constant.
Zero vvOrder: A zero order reaction produces straight line with negative gradient. Reaction rate
does not change during reaction. Value of gradient = rate constant k.
First Order: First order reaction produces downwards curve. Reaction rate
slows down with time. If successive half- lives are constant, reaction is first
order, whereas second would vary. This pattern called exponential decay.
Half-life does not vary with initial concentration.
Second Order: Also downward curve but steeper at start and tailing off
more slowly. Will not be asked to find second order in exam.
All are negative gradient here, whereas ones below are through origin and
positive gradient.
Half- Life: t1/2 is the time taken for concentration of reactant to decrease to half its original value.
Calculating Half- Life: Draw across graph where concentration is half way. Once hit curve, draw
down to find time and work out time difference. Draw successive half- life at 0.5, 0.25 etc. Units
of half- life are seconds.
Or if know just the value of rate or concentration, divide it by two find its value after a half-life.
To find half-life, t1/2, from just knowing rate and concentration use t1/2 = ln 2/ k.
Rate Constant from Graph: Draw tangent on concentration- time graph. Calculate gradient to
get rate of reaction. Sub in rate and concentration where tangent drawn at in the rate equation
to find rate constant- if tangent drawn at 0s, sub in rate and conc at 0s, NOT the conc from
gradient drawn. Gradient = change in concentration/ change in time.
Can also this equation. k = ln2/ t1/2. This is more accurate.
If asks to do find rate of reaction after 40s, draw tangent at 40s.
Learn these graphs and lines.
Rate- Concentration Graphs
Rate- Concentration Graphs:
Zero Order: Straight horizontal line with zero gradient. Rate = k[A]0 so rate =
k = Y intercept. Reaction rate does not change with concentration. Can confirm first order of
reaction of graph by calculating two half-lives and constant. Therefore no change to half- life if
initial concentration increased.
First Order: Straight line through origin. Rate = k[A]1 so rate = k[A].. Rate constant determined
from gradient of line.
, Second Order: Increasing gradient curve starting at origin. Rate = [A]2. Have to plot second graph
of rate against concentration squared. Will get straight line through origin. Rate constant
determined from gradient of this line.
Finding Initial Rate of Formation: Balancing number of substance forming x rate.
Initial Rate: Initial rate can be found by drawing tangent at t = 0.
Clock Reaction: Clock reaction can obtain initial rate. The time is measured for colour change or
precipitate to appear.
Average Rate: Assumed that average rate of reaction over this time will be same as initial rate.
The shorter the time measured, less the rate changes over that time and closer to initial rate.
Initial rate is proportional to 1/ t. Clock reaction repeated with different concentration and 1/ t
calculated for each. Vary concentration of one reactant whilst keeping concentration of other
reactants constant. Plot rate against concentration to determine order.
Use 1/t when finding initial rate.
Iodine Clocks: Starch added to make iodine formed turn blue- black colour. Aqueous iodine is
orange- brown. Aqueous sodium thiosulfate removes the iodine first formed to delay. Solution
colourless at start.
Methods to Measure Rate of Reaction: Mass change with scales e.g. If gas escapes. Gas Syringe/
Measuring cylinder to measure the volume of gas produced. Colorimeter to measure the change
in colour intensity or amount of precipitate formed.
Rate- Determining Step
Reaction Mechanism: Reactions take place in series of steps called reaction mechanism.
- Collision unlikely between more than 2 species in one step.
- Also stoichiometry in rate equation may not match overall equation.
Rate- Determining Step: The steps in reactions at different rates. Slowest step is rate-
determining step.
Creating a Reaction Mechanism: Overall equation (CH3)3CBr + OH- -> (CH3)3COH + Br-. Know
overall equation by cancelling common terms.
What doesn’t appear in rate equation is zero order, has no effect on reaction rate so must be in
fast step, OH-.
Create the two steps. Step 1- (CH 3)3CBr -> Br- + (CH3)3C+ Slow
+ -
Step 2- (CH3)3C + OH -> (CH3)3CBr Fast
Rate equation gives reactants and stoichiometry/ orders of slow step.
Step 1 forms two species- one of them is other product in overall reaction. Other is an
intermediate which also appears in step 2.
If equation quite doesn’t work out, check don’t have to balance equation.
Avoid putting in charges, unless asks for it.
If 2NO in overall equation, one NO in step 1 and another NO in step 2.
H+ is not a catalyst because it is used up and not regenerated. H+ appears in overall equation.
If have to put charges make sure charges balance. For example both sides must be zero, so can
have + charge and - charge together on side as cancel to 0. OR both sides can have -1 charge.
What can be deduced from a rate equation and an overall equation: Power of concentration of
reactant shows --- moles in rate determining step. If this doesn’t match overall it’s a multistep
reaction. If something doesn’t appear in rate equation it’s zero order.
, Sometimes have to be flexible if equations do not match rate equation and include a few fast
step species. Question will direct me to it though.
- Step 1 and 2 give rate equation. Even though step 1 is fast step.
Rate Constants and Temperature
Temperature Increase: As temperature increases, rate increases and the value of rate constant k
increases. This is due to two factors…
Kinetic Energy: As temperature increases, particles move faster and collide more frequently.
Activation Energy: Increasing temperature shifts Boltzmann distribution to right, increasing
proportion of particles that exceed activation energy E a. Change in rate mainly determined by E a.
Arrhenius Equation: This the exponential form of equation.
k = Ae-Ea/ RT R = gas constant = 8.314 J mol -1 K-1. T in Kelvin.
-Ea/ RT
e , Exponential Factor: The exponential factor represents proportion of molecules that
exceed Ea for reaction to take place.
A, Pre- Exponential Factor: Pre- exponential factor or frequency factor is the frequency of
collisions with correct orientation. A is a constant giving the rate if no activation energy.
Ea
Logarithmic Form of Arrhenius Equation: ln k = - + lnA.
RT
1
Gives negative gradient straight line graph. Y = mx + c. =x
T
Ea
- = gradient = dy/ dx. lnA = y intercept. Put ln k on y axis and 1/ T on x axis.
R
Sub T into 1/T and k into lnk.
Pick two points to work out gradient and use this to work out activation energy by multiplying by
R, 8.314. May need to divide by something else such as 10 -3 depending on units.
If need extend graph to find y intercept. lnA = 25 use inverse e 25 = A.
If axis does not start at 0, cannot always extend line of best fit to find y intercept or other points
outside range of data. Instead once know gradient of line, substitute in values to find unknown
ln A.
Equilibrium
Kc
Kc: Remember Kc = products/ reactants. Remember the larger the value, further away from
position equilibrium and towards products- AS.
Know how to work out Kc if given equilibrium concentrations. Now looking at how Kc calculated
from experimental results.
Kc Expression: Include symbol formula, state symbol with square brackets around it. Have
balancing number as power.
Any solids and liquids omitted in Kc expression. Only include gas and
aq.
Units of Kc: Sometimes no units. Add powers when brackets expanded.
, Homogeneous Equilibria: Homogeneous equilibrium contains equilibrium species that all have
the same state.
Heterogeneous Equilibria: Heterogeneous equilibrium contains equilibrium species that have
different states.
Calculating Kc from Experimental Results:
Question will ask ‘find amounts of X in mol in the equilibrium mixture’ or ‘mixture was allowed
to reach equilibrium’ etc.
1.6 mol of NO and 1.4 mol of O 2 mixed in container with volume 4 dm 3. At equilibrium, 1.2 mol
NO2 formed (moles when equilibrium reached). Bold given in question. Work backwards.
Equation 2NO + O 2 -> 2NO2
Reacting Amounts 2 1 2 what is in equation
Initial/ mol 1.6 1.4 0 the normal moles, amounts used, given in
Q
Change/ mol -1.2 -0.6 +1.2 use moles reacting amounts
Equilibrium/ mol 0.4 0.8 1.2 product given, work out reactants.
These equilibrium amounts divided by total volume x1000 = equilibrium concentrations mol dm -
3
.
Then write Kc expression and sub in these equilibrium concentrations as normal. Remember
balancing number as power.
Watch out if volume given in dm3. Do not need 1000 into calculations.
Check which equilibrium amount given, might not be products like usual.
Remember to minus for reactants but add for products to find equilibrium moles.
If given equilibrium moles straight away, don’t need to do table above, just find equilibrium
concentrations.
May not need volume sometimes as volume cancels in Kc expression.
Sometimes says uses 4 dm3 container- use this value as volume in calculations.
Can do above with Kp and others.
Should never get a large number of these calculations but a decimal.
May require calculations to find initial moles and equilibrium moles. If says ethanoic acid at
equilibrium required 72.5 cm3 of a 1.50 moldm–3 solution of NaOH for complete reaction. Work
out moles of NaOH, which equals moles of ethanoic acid at equilibrium = 0.1. However since
flask contains 0.2 moles of ethanoic acid, 0.2 – 0.1 = 0.1 which is the moles of ethanoic acid
reacted- like table above. Use this and molar ratio to find moles of others.
If H2 initial amount increases, the H2 equilibrium amount increases. So the equilibrium shifts left
and HI increases but I2 decreases.
Ag+ reacts with I– to form AgI. Yellow precipitate forms. Equilibrium 2 shifts to the left.
Equilibrium 1 shifts to left. I2 comes out of solution, I2 precipitates. – Write full equations and
talk about both equilibrium shifts.
Kp
Kp: Use Kp with gases. Same as Kc but equilibrium constant in terms of partial pressure instead
of concentration. Products/ reactants. Balancing numbers is the power.
, If asked to write Kp expression, no square brackets and remember p… Only include gases, omit
other states.
Units for partial pressure are kPa, Pa (pascals) or atm (atmospheres). Have to be same units for
all in calculation.
Moles of Gas: Under the same temperature and pressure, same volume of different gases, are
same moles.
Mole fraction x(A) = number of moles A/ total number of moles in gas mixture
Partial Pressure: In a gas mixture, the partial pressure of a gas is the contribution the gas makes
towards total pressure.
Partial pressure p(A) = moles fraction A x total pressure
Remember sum of partial pressure equals total pressure.
An Example: An equilibrium mixture at 400C contains 18 mol of N 2, 54 mol of H2 and 48 mol of
NH3. Total equilibrium pressure is 200 atm. Find Kp.
Mole fractions- x(N2) = 18/ 120.
Partial Pressure- p(N2) = 18/ 120 x 200 = 80 -> use this value in Kp.
1 mol of carbon dioxide, 3 mole of hydrogen. At equilibrium, 0.86 mol of methanol produced at
pressure of kPa.
- Draw table like do with Kc and use molar ratio to find equilibrium moles of reactants.
- Find total equilibrium moles. Find mole fraction for each reactant and product.
- Multiply by total pressure- 500 kPa to find partial pressure for each reactant and product. Do
mole fraction x total pressure in one calculation – quicker.
- Sub this value into kPa.
Controlling Position of Equilibrium
Already know le Chatelier’s principle…
Concentration: If concentration increase, position of equilibrium shifts in direction that reduces
concentration right/ left, so towards product/ reactants.
Pressure: If pressure increased, position of equilibrium shifts towards side with fewer gaseous
molecules right/ left, to reduce pressure, so towards product/ reactants.
Rate of the forward and reverse reactions both increase if concentration/ pressure increases or
catalyst added. Rate increases if temperature increases.
At equilibrium rate of forward and reverse reactions are the same.
Smaller container/ decrease volume = pressure increases.
Equilibrium K (KP or KC): Greater than 1 is towards products.
- Equilibrium counteracts changes in conc and pressure by changing position of equilibrium
and keeps K constant.
- But K does change if temperature changes. K = Kc or Kp.
Forward Exothermic Reactions: Increasing temp, makes equilibrium shift left towards
endothermic direction. Product decreases and reactants increase.
Make sure to name products and reactants as well.
Forward Reaction Endothermic:
- If temperature increased, system no longer in equilibrium. The ratio now less than Kp or Kc.
, - Equilibrium position shifts right in the endothermic direction, towards products to reduce
temperature.
- Named Products increases and reactants decrease. Top increases more than the bottom in
Kc expression.
- So Kc/ Kp equilibrium constant increases in value.
Concentration and Pressure in Terms of Kc: Not AS.
- Kc does not change with pressure/ concentration. -> Write in all Kc questions.
- Increased pressure/ conc, increases named reactant. Denominator (bottom) of Kc/ Kp
expression increases more than top.
- System no longer in equilibrium.
- So equilibrium shifts right so numerator (top) increases and bottom decreases to restore
equilibrium Kc. Named product increases/ reactant decreases.
Pressure of container doubled. Explain in terms of Kc.
- Kc does not change with pressure.
- [NO2]2 increases more than [N2O4] OR concentration term on bottom of Kc increases more
that concentration term on top of Kc.
- Concentration of N2O4 increases AND concentration of NO 2 decreases to restore equilibrium
Kc.
Catalysts: Catalyst affects rate of reaction. Does not affect equilibrium constant or position of
equilibrium. Catalysts speed up both forward and reverse reactions in equilibrium equally.
Equilibrium is reached quicker with catalysts.
Yield/ position of equilibrium is difficult to determine when two things changes as do not know
the relative effect of two opposing factors- pressure/ conc/ temp changes.
When both reactant and product changed in pressure/ conc. If powers of top and bottom are
the same, Kc does not change, no equilibrium shifts and equilibrium conc stays the same.
Acids, bases and pH
Bronsted- Lowry Acids and Bases
Acid: An acid dissociates and releases H+ ions in aqueous solution. A Bronsted- Lowry acid is a
proton donor.
A strong acid completely dissociates AND a weak acid partially dissociates.
Can measure current through an acid to determine if strong and weak acid. Strong conducts
more electricity as more ionised.
Alkali: An alkali dissociates and releases OH- ions in aqueous solution. An alkali is a soluble base.
A Bronsted- Lowry base is a proton acceptor.
Neutralisation: This is neutralisation. H+ (aq) + OH- (aq) -> H 2O (l)
Conjugate Acid- Base Pairs: A conjugate acid- base pair contains two species that can be
interconverted by gain or loss of proton. For example in this dissociation HCl (aq) -> H+ (aq) + Cl-
(aq). Combining dissociation and neutralisation ionic equation above get and acid- base
equilibria…
HCl is acid as donates H+ and OH- is base as
accepts H+ in forward reaction.
, In reverse reaction H2O an acid as donates H+ and Cl- base as accepts H+.
Equilibrium Sign: Single arrow in dissociations indicates forward reaction complete if strong
acid. Equilibrium sign for weaker acids.
Hydronium Ion: Dissociation only takes place when water present as requires proton transfer
from acid to base. Water can act as base and acid. H 2O below accepts a proton to form H 3O+.
Hydronium ion present in all aqueous acids.
H+ in equation is actually the simplified version H 3O+ so can interchange. H3O+ + OH- -> 2H2O.
When constructing acid- base equilibrium, instead of forming OH- or H+, form H 3O+.
Two acids can react together. The stronger acid will act as a proton donor (acid) the weaker acid
will act as a proton acceptor (base).
Reactants were given. To find products take a H and then add a H. No water involved here in
reactant so none in product. Must have negative charge for base formed and positive charge
for acid formed. Unless one of them already has a charge, may form neutral or double charge,
always +/- 1.
Monobasic: Monobasic, dibasic and tribasic aka monoprotic refer to number of hydrogen ions in
acid that can be replaced per molecule in an acid- base reaction.
In organic compounds, do not replace H on carbon chain e.g. CH 3COOH is monobasic. H3BO3 is
tribasic.
Neutralisation Equations: Decide whether acid is mono, di or tribasic. Then write equation using
as many Na in NaOH needed e.g. H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Then balance as normal. Now
can this equation to work out the volume of NaOH requires to neutralise 25cm 3 of acid.
Ionic Equation: Split all ionic compounds (solid/ aqueous) into ion. Leave solids, liquids and
gases. Cancel ions that do not change- spectator ions. Rewrite.
Redox Reaction: Acids- H2SO4, HCl, H3PO4 and HNO2. These form sulphate, chloride, phosphate
and nitrate salts.
Acid (aq) + metal (s) -> salt + hydrogen. Solid dissolves and fizzing.
Ionic equation- Mg + 2H+ -> Mg2+ + H2.
Neutralisation Reactions between Acids and Bases…
Acid + metal carbonate -> salt + water + carbon dioxide. Effervescence and if soluble carbonate,
the solid dissolves.
2H+ + CO32- -> H2O + CO2.
Acid + base (metal oxide) -> salt + water. Solid dissolves.
2H+ + O2- -> H2O.
Acid + alkali -> salt + water. No observations. Alkali can be NaOH or KOH.
Acid + ammonia -> Ammonium salt.
H2SO4 + 2NH3 -> (NH4)2SO4. No observations.
Be careful using sodium carbonate as formula is Na 2CO3 so needs two moles of acid and salt.
Always check formula of alkali by working out.
, If any kind of acid reacts with base (question may say they are acid/ base), the ionic equation will
always be H+ + OH- -> H2O. And in normal equation, ONLY two products are salt and water if
acid + alkali.
pH Scale and Strong Acids
pH Scale: Uses negative logarithm of powers of 10. More manageable scale.
H Ion Concentration: Low value of [H+ (aq)] = high value of pH. A change in one pH number is
10x the [H+ (aq)].
Measuring pH: The electrode in pH meter measures electrical potential of H+ ions.
pH = - log[H+ (aq)]
[H+ (aq)] = 10-pH
Strong acid nothing to do with pH but with extent of dissociation. pH relates to concentration of
H+- the amount of solute, in moles, dissolved in 1dm 3 of solution.
Strong Acid: [H+ (aq)] = [HA (aq)].
Changes to Dilution: Doubling volume = concentration halved as solution more diluted.
gdm-3: Mass/ Mr = moles. So will get moldm-3 which is concentration, so do as normal. Use Mr
of acid or base.
2 SF for pH values is 2.62 and not 2.6, as negative logarithm. Start counting SF for pH values after
decimal point like DP.
Acid Dissociation Constant Ka
Acid Dissociation Constant Ka: Tells us extent of dissociation. No Ka value for strong acid as
completely dissociate. Units are moldm-3.
[ H +(aq)][ A−(aq)]
Ka = Use this one if Q asks for Ka expression.
[ HA (aq)]
Equilibrium Constant: Ka changes with temperature so Ka standardised at 25C. Larger Ka means
equilibrium further to right, more H+ dissociates so pH decreases.
pKa: Convert Ka to pKa as Ka has negative indices like [H+ (aq)] so pKa easier to compare relative
numbers. The larger the Ka value but smaller the pKa value.
pKa = - logKa
Ka = 10-pKa
pH of Weak Acids
, Ka can be calculated from equilibrium amounts.
Approximations: Equation above can be simplified to equation below by assuming two things…
Dissociation of Water: There is a small concentration of H+ from dissociation of water which is
neglected. Therefore when HA dissociates, H+ and A- ions formed in equal quantities.
Exception- Approximation above does not work for very weak weak acids or dilute weak acids. If
pH more than 6, then [H+(aq)] from dissociation of water more significant compared to
dissociation of weak acid. More dilute solution means more water.
Decrease in Concentration: The equilibrium concentration of HA is smaller than undissociated
concentration of HA as [HA (aq)]eqm = [HA (aq)]start - [H+ (aq)]eqm. Can neglect any decrease
in concentration of HA from dissociation as very small in weak acids.
Exception- Approximation above does not work for stronger weak acids and for dilute solutions.
[HA (aq)]start - [H+ (aq)]eqm difference is more significant as [H+ (aq)] more significant.
Therefore can simplify to…
[ H +(aq)]2
Ka = [HA (aq)]
Calculate pH using Ka:
[ H + ( aq ) ] =√(Ka x [ HA ( aq ) ] ) Make sure to include brackets around the expression.
All aqueous weak acids contains hydroxide ions as well since water dissociates. H 2O ⇌ H+ + OH-.
Therefore if know [H+], can find concentration of hydroxide ions in an acid.
Kw
[OH-] =
H+ ¿ ¿
pH of Bases
Determine if acid or alkali.
Check if weak or strong before doing method.
Also always check if 2OH or 2H as would be dibasic and would need to multiply conc x 2.
Check in questions if given pKa or Ka.
Kw: Ionic product of water- the ions in water H+ and OH- multiplied together. Kw increases with
temperature.
[H+ (aq)] [OH- (aq)] = 1.00 x 10-14 mol2dm-6 = (Kw)
at 298K.
Water: Water produces same number of H+ and OH-, [H+ (aq)] = [OH- (aq)] so pH = 7.
- So Kw = [H+]2.
Acids and Alkalis: When [H+ (aq)] > [OH- (aq)] solution is acidic. When [H+ (aq)] < [OH- (aq)]
solution is alkaline.
Dissociation of water is endothermic because Kw increases with temperature OR dissociation
involves bond breaking.
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