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Final

COMPLETE Principles of General Chemistry Notes -- 93% on final exam

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Chemistry
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CAS CH 171
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4.1 Rates of Reactions -3 -1 Reaction Rate = change in amount of reactants/products per unit time (units: mol dm s ) Following a reaction; • gas volume produced (gas syringe) • mass lost (balance) • colour change (colorimeter) • clock reaction (sudden change at particular time means specific concentration of product has been reached - the shorter the time taken, the faster the rate) • electrical conductivity (number of ions will change as reaction occurs) Concentration-Time Graph Rate at any point can be found by drawing a tangent at that point on the graph and finding the gradient. Orders of Reaction The order of reaction = how the reactants concentration affects the rate INCREASE REACTANT – RATE STAYS THE SAME – ORDER OF 0 INCREASE REACTANT – RATE INCREASES BY 1 FACTOR – ORDER OF 1 INCREASE REACTANT – RATE INCREASES BY 2 FACTORS – ORDER OF 2 You can only find the order of a reaction *experimentally* – there is NO theoretical order system. Shapes of Rate-Concentration Graphs tell you the order. ZERO ORDER FIRST ORDER SECOND ORDER *square brackets indicate concentration. For example [X] = concentration of X. Half-life = time taken for half the reactant to react If the half life is constant = first order If the half life is doubling = second order You can also calculate the half life using reaction rates. For example, if you’re given the rate constant (see below) and the order you can work out half life (you don’t need to know how, just to be aware of it) Rate Equations Rate equation = tell you how the rate is affected by the concentrations of reactants. E.G. Rate = k[A] [B] n Where: m = order ofA n = order of B n+m = overall order k = rate constant (always the same for a reaction at specific temp and pressure, increase temp = increase k = bigger value of k = faster reaction) EXAMPLE + - Propanone + Iodine —> Iodopropanone + H + I (reaction occurs in acid) Info: First order with respect to propanone and H and zero order with respect to iodine Rate equation = k[propanone] [H ] [iodine] 0 Simplify to; + Rate equation = k[propanone][H ] (because anything to the power of 0 is 1) How to calculate rate constant from the orders and rate? Rearrange to make k the subject and calculate. -3 -3 -1 Units of k can be found as you know concentration is moldm and rate is moldm s using a normal “cancelling” method. Using data to deduce the order 1) The experiment: titrate sample solutions against sodium thiosulfate and starch to work out the concentration of the iodine. Repeat experiment, changing only the concentration for ONE REACTANT at a time. experiment 1 2 3 4 5 6 7 [propanone] 0.4 0.8 1.2 0.4 0.4 0.4 0.4 [iodine] 0.002 0.002 0.002 0.004 0.006 0.002 0.002 + [H ] 0.4 0.4 0.4 0.4 0.4 0.8 1.2 Here, first we changed the concentration of propanone for experiments 1, 2 and 3. Then, we changed the concentration of iodine in experiments 4 and 5. Lastly, we changed concentration of H in experiments 6 and 7. 2) From this table we can plot 7 Concentration-Time graphs. Finding the gradient at time zero for each of these plots will give us the INITIAL rate of each. 3) Compare the results e.g. Experiment Change compared to Rate of reaction Change experiment 1 1 --- 0.033 --- 2 [propanone] doubled 0.062 Rate doubled 3 [propanone] trebled 0.092 Rate trebled 4 [iodine] doubled 0.034 No change 5 [iodine] trebled 0.032 No change + 6 [H ] doubled 0.058 Rate doubled 7 [H ] doubled 0.094 Rate trebled *Reaction rates won’t be exactly double or treble due to experimental errors etc. 4) Now we can work out the rate equation: • Rate is proportional to [propanone] so the reaction is of order 1 with respect to propanone. • Rate does not change/is independent of [iodine] so the reaction is of order 0 with respect to iodine. + + • Rate is proportional to [H ] so the reaction is of order 1 with respect to [H ]. Rate determining step = slowest step in a multi-step reaction (if a reactant appears in the rate equation it MUST be a rate determining step including catalysts which may appear in a rate equation) PREDICITIONS The order of a reaction with respect to a reactant shows the number of molecules that the reactant is involved in with regard to the rate-determining step. EXAMPLE: rate = k[X][Y] . 2 Here, one molecule of X and 2 molecules of Y will be involved in the rate determining step. Chlorine free radicals in the ozone consist of 2 steps: Cl•(g) + O 3g) —> ClO•(g) + O (g)2 slow rate determining step ClO•(g) + O•(g) —> Cl•(g) + O (g)2 fast reaction Therefore, Cl• and O m3st be in the rate equation as they are the reactants from the slowest step. RATE = k[Cl•][O ] 3 Predicting Mechanisms: Once you know what the rate determining reactants are, you can think about what reaction mechanism it follows. EXAMPLE: If the rate equation is: rate = k[X][Y] And the two different mechanisms are: 1) X + Y —> Z OR 2) X —> Y + Z From the rate equation, we know that X and Y MUST be in the rate determining step, therefore, it’s mechanism 1 which is the right one. Halogenoalkanes – Nucleophilic Substitution (S ) N Halogenoalkanes can be hydrolysed by OH ions by nucleophilic substitution. This is where a - nucleophile (e.g. :OH) attacks a molecule and is swapped/substituted for one of the attached groups (e.g. Br ). In this case the Carbon (C ) to Halogen (X ) bond is POLAR as halogens are much more electronegative than the carbon so they draw in electrons making the Carbon slightly/delta positive. The bond looks like this: C — X δ- Thus, the carbon can be easily attacked by a nucleophile who likes positive areas. This mechanism occurs: *C-Br bond breaks heterolytically (unevenly) • Primary – react by SN2 where 2 molecules/ions are involved in the rate determining step • Secondary – react by SN1 and SN2 • Tertiary – react by SN1 where 1 molecule/ion is involved in the rate determining step You can see by the rate equation if there are 1 or 2 molecules in the rate determining step, which in turn, tells you if the mechanism is SN1 or SN2. EXAMPLE: Rate = k[X][Y] = 2 molecules in rate determining step = SN2 = primary/secondary halogenoalkane OR Rate = k[X] = 1 molecule in rate determining step = SN1 = tertiary/secondary halogenoalkane Activation Energy We can calculate the activation energy using theArrhenius equation: Where; (you don’t have to learn this, just understand the relationship) k = rate constant E A activation energy (J) T = temperature (K) R = gas constant (8.31 JK mol ) -1 A= another constant Some relationships to note: 1) As E iAcreases, k will get smaller. Therefore large activation energy, means a slow rate – this makes sense! 2) As T increases, k increases. Therefore at high temperatures, rate will be quicker – this makes sense too! If we “ln” both sides of Arrhenius’ equation, we get; ln k = – E ART + ln A (don’t forget, ln Ais just a constant, a number) This looks a bit like: y = mx + c If we plot ln k (y) against 1/T (x), the gradient we produce will be –E /R (A). Then R is just a -1 -1 number that we know (8.31 JK mol ) we can rearrange and find the activation energy. EXAMPLE: Iodine clock reaction S 2 82-(aq) + 2I (aq) —> 2SO 42(aq) I 2aq) Rate of reaction is inversely proportional to the time taken for the solution to change colour i.e. increased rate = decreased time taken k α1/t We can say that 1/t is the same as k (rate constant) and we can substitute 1/t instead of k inArrhenius’ equation and find the gradient again to find a value for E .A Catalysts Catalyst = increases rate of a reaction by providing an alternative reaction pathway with a LOWER activation energy (E )AAcatalyst will be chemically unchanged at the end of a reaction. Adv: Small amount needed to catalyse a lot of reactions, also they are remade, thus reusable. Disadv: High specificity to the reactions they catalyse. There are two types of catalysts: HOMOGENOUS CATALYSTS HETEROGENOUS CATALYSTS These are catalysts in the same state as the These are catalysts in different physical states reactants. to the reactants. They are easily separated from products – E.G. when enzymes catalyse reactions in GOOD your body, all reactants are aqueous, this is Can be poisoned (i.e. a substance clings to a homogenous catalysis. catalyst stronger than the reactant would, preventing reaction speeding up) example: sulphur in the Haber process is a “poison”– BAD Solid catalysts provide a large surface area for the reaction to occur e.g. mesh/powder E.G. vanadium pentoxide in the contact process to make sulphuric acid 4.2 Entropy Entropy = a measure of how much disorder there is in a substance, how many different ways particles can be arranged. Systems are MORE energetically stable when disorder/entropy is HIGH. EXAMPLE:Agas will want to escape its bottle because the room it’s in is much bigger and the particles can be arranged in lots of different ways. SOLID LIQUID GAS No randomness, therefore Some randomness, some Most randomness, highest lowest entropy. entropy. entropy. e.g. S (H O2)(s)7.4 JK mol-1 - e.g. S (H O2)(l)70 JK mol-1 -1 e.g. S (H O2)(g)189 JK - 1 1 -1 mol (see below) (see below) (see below) *Note that zero entropy will only occur in a perfectly ordered crystal Affecting Factors: 1. More quanta (packets of energy) = More ways to arrange particles = More entropy 2. More particles = More arrangements = More entropy. E.G. X -> 2Y 2 moles of Y produced from 1 mole of X therefore entropy has increased 3. Increase in temperature = Increase in energy = More entropy E.G. - from solid to liquid entropy has increased a bit - from liquid to gas entropy has increased a lot 4. Complicated/complex molecules = more entropy DEFINITIONS: ѳ Standard entropy of a substance, S , is the entropy of one mole of a substance under standard conditions of 298K and 1atm. The units are JK mol .-1 -1 We expect exothermic reactions to be the spontaneous ones; however some endothermic reactions are spontaneous too. This is to do with entropy. If entropy is high enough, the reaction will be spontaneous, whether the reaction is exo/endothermic. EXAMPLE: NaHCO (3) + H (aq) —> Na (aq) + CO 2g) + H 2(l) 1 mole 1 mole 1 mole 1 mole 1 mole Solid aqueous ions aqueous ions gas liquid Here, the products have high entropy states (e.g. gas) and there are more moles (e.g. reactants to products = 2:3)And so, overall entropy has increased = SPONTANEOUS (also depends on ΔH – see below) LEARN THESE: ΔS sys = S products - S reactants ΔS total = ΔS sys + ΔS surr Where; ΔS sys= Entropy change of a system, the entropy change between the reactants and the products ΔS surrEntropy change of a surrounding ΔS = Total entropy change, the sum of the entropy changes of the system and the total surroundings EXAMPLE: NH 3g) + HCl(g) —> NH C4(s) Info: ΔH = -315kJmol -1 ѳ -1 -1 ѳ -1 -1 ѳ - S (NH )3(g)92.3 JK mol S (HCl (g) 186.8 JK mol S (NH 4l (s) 94.6 JK 1mol-1 1) Find entropy of the system ΔS sys = SproductsSreactants = 94.6 – (192.3 + 186.8) = - 284.5 JK mol -1 2) Find entropy of surroundings -1 = - (-315000)/298 [Note:ΔH = -315kJmol is in KILOJOULES, therefore x1000] = + 1057 JK mol1 -1 3) Find total entropy ΔS total= ΔS +sys surr = -284.5 + 1057 = + 772.5 JK mol -1 [Note: must include sign (and units) with final answer] When will a reaction be spontaneous? • Total entropy must increase • +ΔS totalkinetically favourable (wants to react; spontaneous) • — ΔS totalkinetically stable (will not react on its own; not spontaneous) * You can predict ionic compound solubility using the same idea; if ΔS is positive √, if total negative X ENDOTHERMIC experiments that are spontaneous: 1) Ba(OH) (s2 and NH Cl(s2 Ba(OH) .82 O(s2 + 2NH C2(s) —> BaCl 2s) + 10H 2(l) + 2NH (3) When you add barium hydroxide to ammonium chloride: • Smell of ammonia gas • Temperature drops below 0˚C 2) Cold pack – NH NO 4s) a3d H O(l) 2 NH N4 (s)3 —H O(2)—> NH (aq) + 4 NO (3q) When you dissolve ammonium nitrate crystals in water: Looking at the states in both these experiments, we have an INCREASE in entropy (from solids to liquids/aqueous). These reactions are spontaneous EVEN THOUGH the ΔS surrs negative (because ifΔH is positive for endothermic reactions the equation of ΔS surreans the overallΔS will be negative – see above equation) the ΔS is GREAT ENOUGH to surr sys overcome it, meaning ΔS totalll be positive still. DEFINITIONS: Thermodynamic stability – where the ΔS total negative, at RTP, the reaction will simply not occur. E.G. limestone –> CaO + CO 2 Kinetic inertness – when the ΔS totalf a reaction is positive, a reaction can happen spontaneously, however the rate of reaction at RTP is so slow because the activation energy needed for it to start is so high. E.G. diamond –> graphite The enthalpy change of hydration,ΔH hyd– the enthalpy change when 1 mole of aqueous ions is formed from gaseous ions. E.G. Na (g) —> Na (aq) + ѳ The standard lattice enthalpy,ΔH lattthe enthalpy change when 1 mole of a solid ionic compound is formed from gaseous ions under standard conditions (298K and 1atm). E.G. Na + (g) + Cl(g)—> NaCl(s) The enthalpy change of solution, ΔH – thsolnthalpy change when 1 mole of solute is dissolved in sufficient solvent, so no further enthalpy change occurs on further dilution. E.G. NaCl(s) —> NaCl(aq) Factors affectingΔH ѳlattD ΔH hydinclude; 1) Ionic charge; = larger charge = more exothermic lattice energy = MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION ѳ -1 ѳ -1 E.G. NaCl hasΔH latt -780kJmol whereas MgCl has ΔH 2 latt -2526kJmol because magnesium has a charge of 2+ which is greater than sodium’s 1+ 2) Ionic radii; = smaller ionic radii = more exothermic lattice enthalpy = higher charge density = MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION E.G. Sodium’s ionic radius is bigger than magnesium’s (because Mg has one more proton which has a stronger positive nuclear attraction to its electrons – see unit 1/2) therefore magnesium will have a more negative lattice enthalpy/hydration enthalpy. Finding the enthalpy of solution where we use a similar principle to Hess’ Law; ΔH =ΔH +ΔH 1 2 3 REMEMBER (for ΔH ): GASEsol IONS DOWN, AQUEOUS IONS UP Guns In Detroit, Apples In Ukraine 4.3 Equilibria RECAP: (for exothermic reaction) LE CHATELIER – Where does equilibrium move and why? oppose the motion! Increase Temperature Toward reactants, therefore less products; Move to the endothermic side. Higher kinetic energy so more chance of successful collision LOW temp = high yield = but slow process... Increase Pressure Toward side with less molecules of gas (only affects gases). Particles are pushed together, which increases chances of successful collision. HIGH pressure = high yield = expensive! Introduce Catalyst NO EFFECT ON EQUILIBRIUM POSITION (will affect rate) At equilibrium the amount of reactants and products is the SAME. Dynamic Equilibrium – a reaction that occurs in both ways at the same time (conditions; in a closed system at constant temperature) Many industrial reactions are reversible; we use this sign for equilibria: E.G. Both these experiments are good economically 1) Contact process – making sulphuric acid 2SO 2g) + O 2g) 2SO (3) USES = fertilisers, dyes, medicines, batteries 2) Haber process – making ammonia N 2g) + 3H 2g) 2NH (3) USES = fertilisers, producing nitrogen-based compounds EXPERIMENT: Hydrogen-Iodine Reaction (REVERSIBLE) There is a relationship between the concentration of initial reactants/products and the equilibrium concentrations which are produced from them E.G. H 2g) + I (2) 2HI(g) Initial concentration: H 2 1.0moldm -3 I2= 1.0moldm -3 -3 -3 Equilibrium concentration: H 2 0.228moldm I2= 0.228moldm From this we can see that the ratio has remained the same, i.e. 1:1 K p K c What is K /pK ? c K /pK iscthe ratio of product concentration to reactant concentration, and is commonly known as the equilibrium constant. For example, in the hydrogen-iodine reaction K will be; c *Note: products are because in a balanced equation, there is a 2 in front – see below E.G. 4X 2Y + 3Z *We can calculate K usipg partial pressures (see below) As long as the equilibrium is HOMOGENOUS (all reactants/products in the same state) then we can use this general rule for finding K ; c If the equilibrium is HETEROGENOUS (where reactants/products are in different states) then you must LEAVE OUT any concentrations that are solid. For K , HOMOGENOUS equilibriums can be calculated using; p If the equilibrium is HETEROGENOUS then you only take into account the gases. *Note: we dont use square brackets for equilibrium partial pressures EXPERIMENT: Fe (aq) +Ag (aq) Fe (aq) +Ag(s) 3 -3 3 -3 1) Add 500cm of 0.1moldm silver nitrate solution to 500cm of 0.1 moldm of iron (II) sulfate solution 2) Leave mixture in stoppered flask at 298K, it will reach equilibrium 3) Take samples and titrate CALCULATION: Reactant/Product Fe (aq) Ag (aq) Fe (aq) Ag(s) Initial concentration 0.05 0.05 0 0 (moldm )-3 Equilibrium 0.0439 0.0439 0.0061 solid concentration (1:1 ratio) (0.05 – (from titre results) 0.0439) Equilibrium constant Units Calculating partial pressures; Minty Fruits Taste Minty EXAMPLE: When 3.0 moles of PCl is5heated in a closed system, the equilibrium mixture has 1.75 moles of Cl. If total pressure of the mixture is 714kPa, what is the partial pressure of P5l ? Step 1) Find moles at equilibrium of all reactants and products; We know 1.75 moles of Cl , 2herefore we must also have 1.75 moles of PCl and 3o (3 - 1.75) will leave us with the moles at equilibrium for PCl which is 1.25 moles.Adding these 5 together we get 1.25+1.75+1.75 = 4.75 total moles at equilibrium. Step 2) Find the mole fraction; 1.25 5 Mole fraction of a gas in mixture =4.75 = 19 = 1.66 Step 3) Find partial pressure; 5 Partial pressure of gas = 714 19 = 187.9kPa Equilibrium and Entropy are related ∆S total R lnK When the total entropy, ∆S total,creases, the equilibrium constant, K, will also increase. If; K = 10 -10= reaction will not occur -5 K = 10 = mostly reactants K = 1 = balanced products and reactants 5 K = 10 = mostly products K = 10 = reaction complete 4.7 Acid/base Equilibria From the timeline we can see the change in definition of acids through history. The main ones to know are: Arrhenius definition – when acids/bases dissolve in water then completely/partially dissociate into charged particles (ions) Brønsted–Lowry definition – an acid is a proton donor and a base is a proton acceptor; acids (proton donors) will never release a H on its own, it is always combined with H O to fo2m + HYDROXONIUM IONS – H O 3 *NOTE:Acid-base equilibria involves the transfer of protons, either donated or accepted. example reason Strong HCl(g) —> H (aq) + Cl(aq) - Strong acids and bases ionise almost Acid + completely in water. NaOH(s) + H O(l2 —> Na (aq) + OH(aq) *HCl has a pH of 0 = completely Base ionised Weak CH C3OH(aq) CH C3O(aq) + Weak acids and bases only slightly Acid + ionise. Equilibrium is set up with H (aq) mostly reactants (to the left) Base NH (3q) + H O(2) NH (4q) + OH - (aq) Conjugate acid base pairs • HA and A are conjugate pairs + • H 2 and H O 3re conjugate pairs WATER is special – it can behave as a base and an acid. You can work out the equilibrium constant in the same manner as we did before e.g. However, the equilibrium is very far left and so the equilibrium constant for this reaction is said to have a constant value; -14 2 -6 At 298K/1atm, the K of wacer is 1.0 x 10 mol dm (We often define this with its own notation – K ) w K =wK x [c O] = 2he ionic product of water = [H ][OH] with UNITS: mol dm 2 -6 pH – “power of hydrogen” - is a measure of the hydrogen ion concentration + pH = - log [H ] CALCULATION: finding the pH of a strong acid -3
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