Kinetics and Equilibrium again Kinetics and the Rate constant Equilibrium constants Kc and Kp Acids and alkalis and pH Curves and neutralisation Buffers Revised: 6-Jun-14
Need to do: • clarify graphs • Expand rate-limiting step & mechanism
Remember: • 5 factors that affect rate: • Temperature • Catalyst • Surface area (S), Pressure (G), Concentration (L) • They actually affect: • Frequency of collision (all) • Energy of collision (temperature) • Activation energy / path (catalyst)
Need to know definition of Ea • Use Ea in your answers
Measuring rate • How much (similar to speed ) • Unit of time • Real pattern is • ∆concentration of A • ∆time
Pressure and concentration • Both expressed as [species] • Read brackets as “concentration of” • So ∆[NO2] reads “change in concentration of NO2”
How to measure [product] or [reactant] • Collect gas – volume (say temp. constant) • Gas reactions – pressure (say temp. constant) • Colour change –use colorimeter • Change in liquid volume (tiny) (called dilatometry) • pH using meter/data logger • Use radioisotope tracer (rare!)
Methods continued... • Polarisation of light (optical isomers) • Sample + quench at low temperature, then: • Halides – use to displace iodine and do a starch titration OR use colour change • Acid/base – neutralise one and/or titrate • All these need “final” measurement for calibration before you can work out the rate.
Rate equations • The rate of reaction is generally proportional to the concentration of at least one reactant • So a reaction like mA + nB C can be represented by a rate equation of the form: • Rate = k [A]m[B]n
Note: • The multiplier has become a power • k is called the rate constant • k is affected by temperature. • k is affected by the presence (or effectiveness) of a catalyst. • Not by concentration.
Note: the initial gradient is the one we want, after that, concentrations have changed from the ones we set.
Reaction rate equation • General form • Rate = k[A]m[B]n ..... • The m and n terms are described as “the order of reaction with respect to...”
order • Order of 0 means concentration makes no difference to rate • Order of 1 means direct proportion • Order of 2 means square-law proportion • Yes, fractional orders do occur but not at A level.
A good first guess.... • It happens that, for many reactions, a “first guess” for the order of reaction with respect to one of the reactants is its stoichiometric ratio: • E.g. the reaction N2 + 3H2 2NH3 is probably 1st order with respect to N2 and 3rd order w.r.t. H2. • And the backwards reaction is 2nd order w.r.t. NH3
In a concentration-time graph: • A zero-order reaction gives a straight line. • A first-order reaction does not give a straight line, it’s like a half-life curve. • Only first-order reactions have constant half-lives. • A second-order reaction will give a steeper curve.
In a rate-concentration graph: • A zero-order reaction gives a horizontal line. • A first-order reaction gives a straight line. • A second-order reaction will give a curve. • This curve will be a straight line only if you plot rate2 against time.
So the rate equation is... • Rate = k [A]0[B]1 • Or rate = k [B] (when simplified.) • And k is: • From 1st line of table... • 10 = k x 1 • Therefore k = 10
So the rate equation is... • Rate = k [A]1[B]0 • Or rate = k [A] (when simplified.) • And k is: • From 1st line of table... • 10 = k x 1 • Therefore k = 10
So the rate equation is... • Rate = k [A]1[B]1 • Or rate = k [A][B] (when simplified.) • And k is: • From 1st line of table... • 10 = k x 1 • Therefore k=10
So the rate equation is... • Rate = k [A]1[B]2 • Or rate = k [A][B]2 (when simplified.) • And k is: • From 1st line of table... • 10 = k x 1 x 1 • Therefore k=10
So the rate equation is... • Rate = k [A]1[B]1 • Or rate = k [A][B] (when simplified.) • And k is: • From 1st line of table... • 2 x 10-3 = k x 0.02 x 0.3 • Therefore k = 0.333
What affects k? • It is not obvious that k can be changed: • Rate = k[A]m[B]n ..... • If temperature goes up, k goes up • Adding a catalyst increases k.
What about catalysts? • If a catalyst is involved in a reaction it must have an effect on rate, so it will be in the rate equation. • It (probably) won’t be in the balanced (stoichiometric) equation, though you may see it drawn above the arrow.
Units of k • k is independent of moles and volume for order 1 • You need to do the same operations to the units as you have done to the numbers • Or maybe memorise:
Typical (easy) exam question: The following data were obtained in a series of experiments on the rate of the reaction between compounds A and B at a constant temperature. (i) Deduce the order of reaction with respect to A. (ii) Deduce the order of reaction with respect to B. (2 marks)
Mechanism • Reactions usually have more than one step: • A B C • Or written as: • A B followed by B C • Often have a rate-determining step (the slowest)
How to recognise a rate-determining step: • Any step after the rate-determining step will not appear in the rate equation because it can have little or no effect on the overall rate. • E.g. if a reaction proceeds in two steps: • Step 1: A B then step 2: B + C D • If the rate equation is • Rate = k[A] then step 2 is not rate-determining • Rate = k[A][C] then step 2 must be. • .
Remember: • Equilibrium involves: • Forward and backward reactions • Equal rates • Closed system:- nothing leaves or enters, temperature and pressure constant • Constant concentrations, NOT equal concentrations • Note – an equilibrium is not actually much use to us.
Two reactions: • For a hypothetical equilibrium: • aW +bX Ý cY + dZ • The forward and backward reactions have rate equations that must be equal • So rate = k1[W]a[X]b = k2[Y]c[Z]d • We can get an overall equation for the equilibrium constant:
Some nice little points .... • Anything in large excess at the beginning of the reaction, particularly solvents (usually water) and solids don’t count in the calculations and end up included in Kc • Because if their concentrations do not change, they are constant • UNLESS they are part of the reaction, like in hydrolysis • Concentrations must be measured at equilibrium, obviously.
What affects Kc and Kp? • Temperature (and nothing else) • Forward reaction exothermic: increased temp decreases Kp, make more reactant • Forward reaction endothermic: increased temp increases Kp, make more product
Think about units • It is worth “renaming” the unit of mol dm-3 as, say, “c” or “u”, because you can manipulate the unit and then convert the answer back again • E.g. if the equation is • Then you could say the units are c2 / c3 • The c2 cancels with the c3, giving c(2-3), so the units are c-1 • This is (mol dm-3)-1 • This is mol-1 dm3
Let’s visit an old friend..... • For N2 + 3H2Ý 2NH3 • And the units will be: • C2 / C4 or C(2-4) or C-2 • That is mol-2 dm6
If we use Kp.... • We will also be able to use the symbol Kp for gas equilibria, instead of Kc; we use partial pressures instead of concentrations. • For N2 + 3H2Ý 2NH3 • And the units will be: • p2 / p4 or p(2-4) or p-2 • That is, atm-2 or Pascals-2, depending on the question
Partial pressures • To calculate Kp, we need to know the partial pressure of each species, using: • Or alternatively, • Partial pressure = mole fraction x total pressure • E.g. if the total pressure of the air is 1 atm and mole fraction of N2 is 0.78, then pN2(g) is 0.78 atm.
E.g: • (atm – atmospheres – one times standard atmospheric pressure) • In a sample of gas, there is 3 moles of N2, 4 moles of NO and 3 moles of O2. If the total pressure is 12 atmospheres, what are the three partial pressures?
e.g. consider: • 2NO Ý N2O2 • Given: at equilibrium, the partial pressures are: • NO 3.5 atm • (atm – atmospheres – one times standard atmospheric pressure) • N2O2 4 atm • Assuming that the equilibrium is represented by : • What is the value of Kc?
examples • PbSO4(s) + H2O(l)Ý Pb2+(aq) + SO42-(aq) • Remember, water is a solvent so we can ignore it and the solid can also be ignored, as its concentration doesn’t change. • So Kc = [SO42-] [Pb2+] • And the units will be mol2 dm-6
Or • E.g. 2 • In the flame of a burning hydrogen balloon, just after it is ignited, there is 12 moles of H2, 482 moles of H2O and 6 moles of O2. If the total pressure is 22 atmospheres, what are the three partial pressures?
Kp • For the reaction: • N2(g) + 3H2(g)Ý 2NH3(g) • Calculate Kp for the Haber process, given that the number of moles of N2(g) = 1.2, H2(g)=0.4 and NH3(g)=2.3, and the total pressure = 6 atm.
Surprise! • Kp doesn’t change for overall pressure changes. • You’d think it would, but it doesn’t, you just calculate it from the pressures given. • And that’s why Le Chatelier’s principle works, because if Kp can’t change, then the equilibrium must respond to change.
A typical exam question 3 (a) The expression for an equilibrium constant, Kc, for a homogeneous equilibrium reaction is given below. Kc = [A]2[B] [C][D]3 (i) Write an equation for the forward reaction. (ii) Deduce the units of Kc (iii) State what can be deduced from the fact that the value of Kc is larger when the equilibrium is established at a lower temperature. Jan 06 (3 marks)