Mathematics in Sixth Form Chemistry: Building Confidence

Transcription

1 Mathematics in Sixth Form Chemistry: Building Confidence DR MARYJANE TREMAYNE SCHOOL OF CHEMISTRY, UNIVERSITY OF BIRMINGHAM, EDGBASTON, BIRMINGHAM, UK

2 Outline Units length, volume, temperature, energy Ideal gas equation Gibbs free energy, enthalpy & entropy Percentage yield & stoichiometry Plotting graphs Arrhenius expression Other mathematical applications

3 Why are units important? NASA lost $125 million Mars Climate Orbiter Thrusters fired incorrectly Air Canada flight ran out of fuel halfway through flight Conversion between kg & litres LM data in lbs, NASA in Newtons Tokyo Disneyland s Space Mountain derailed roller coaster as axle broke Wrong size axles ordered Inches to centimetres

4 Units (length) How many cm are there in a dm? a) 0.1 b) 1 c) d) 100 e) 1000

5 Units (length) How many cm are there in a dm? 1 cm 10 cm 10 cm = 1dm

6 Units (length). so which of the following statements are true? a) 1cm = 0.01 dm b) 1cm = 0.1 dm = 10-1 dm c) 1cm = 1 dm d) 1cm = 10 dm e) 1cm = 100 dm

7 Units (length) How many cm are there in a dm? 1 cm 10 cm 10 cm = 1dm 1 cm = 1 dm = 0.1 dm 10

8 Units (volume) How many cm 3 are there in a dm 3? a) 10 b) 100 c) 300 d) e) 3000

9 Units (volume) How many cm 3 are there in a dm 3? 1 cm 10 cm 1cm x 1cm x 1cm = 1 cm 3 10 cm x 10 cm x 10 cm = 1000 cm 3 = 1 dm 3

10 Units (volume). so which of the following statements are true? a) 1cm 3 = dm 3 = 10-3 dm 3 b) 1cm 3 = 0.1 dm 3 c) 1cm 3 = 1 dm 3 d) 1cm 3 = 10 dm 3 e) 1cm 3 = 1000 dm 3

11 Units (volume) How many cm 3 are there in a dm 3? 1 cm 10 cm 1 cm 3 = 0.1 dm x 0.1 dm x 0.1 dm = dm 3 = 10-3 dm 3

12 Units (ideal gas equation) Describes the dependence of pressure, volume and temperature for an ideal gas p V = n R T K N m -2 m 3 mol J K -1 mol -1 S I UNITS (J Nm)

13 Units (ideal gas equation) Describes the dependence of pressure, volume and temperature for an ideal gas p V = n R T At constant temperature (isotherm): p = nrt V V = nrt p

14 Units (volume) How many dm 3 are there in a m 3? a) 10 b) 100 c) 300 d) e) 3000

15 Units (ideal gas equation) Calculate the volume occupied by 1.15 mol of an ideal gas at 1.01 x 10 5 Nm -2 and 20 C. pv = nrt T (in SI units) = = 253 K V = nrt p Vol = 1.15 mol x JK -1 mol -1 x 253 K = m x 10 5 Nm -2 NOT = 23.9 dm 3 Vol = 1.15 mol x JK -1 mol -1 x 20 C = m x 10 5 Nm -2 = 1.9 dm 3

16 Units (in Thermodynamics) Consider the reaction: Mg(s) + 1/2 O 2 (g) MgO(s) Once ignited, Mg burns giving off heat & light Reaction is exothermic Decrease in entropy G = H T S For a spontaneous change: G < 0

17 Units (in Thermodynamics) Calculate the Gibbs free energy ( G) for the reaction Mg(s) + 1/2 O 2 (g) MgO(s) at 298K, where H = kj mol -1 & S = J mol -1 K -1 G = H T S J mol -1 or kj mol -1 J mol -1 or kj mol -1 J mol -1 K -1

18 Units (in Thermodynamics) Calculate the Gibbs free energy ( G) for the reaction Mg(s) + 1/2 O 2 (g) MgO(s) at 298K, where H = kj mol -1 & S = J mol -1 K -1 J mol -1 G = H T S J mol -1 K -1 G = ( 601.7x1000) 298 ( 108.3) = J mol -1 = kj mol -1 NOT kj mol -1 J mol -1 K -1 G = ( 108.3) = kj mol -1

19 Percentage Yield The percentage yield of a reaction measures the effectiveness of a synthetic procedure % yield = 100 x actual yield (in g) theoretical yield (in g)

20 Percentage Yield For the following reaction using 9g of Mg, 34g of MgCl 2 were made. Calculate the % yield. Mg(s) + 2 HCl(aq) MgCl 2 (aq) + H 2 (g) Amount of Mg = mass = 9 g = moles molar mass 24 g Theoretical yield = no. of moles x molar mass = moles x 95 g = g % yield = 100 x actual yield = 34 g = 95.4% theoretical yield g NOT 100 x mass of product = 34 g = 378% mass of reactant 9 g

21 Percentage Yield For the following reaction using 50g of CaCl 2, 17g of Ca 3 (PO 4 ) 2 were made. Calculate the % yield. 3 CaCl Na 3 PO 4 Ca 3 (PO 4 ) NaCl 3 mol of reactant 1 mol of product Amount of CaCl 2 = mass = 50 g = 0.45 moles molar mass 111 g Theoretical yield = no. of moles x molar mass = 0.45 moles x 310 g = 46.5 g 3 % yield = 100 x actual yield = 17 g = 36.5% theoretical yield 46.5 g

22 Plotting data & graphs Plotting two variables shows the trend but a straight line can give more quantitative information First order kinetics rate constant k Solid/gas adsorption surface area A

23 Arrhenius expression Describes the effect of temperature on the rate of reaction and the rate constant k = A e Ea / RT mol dm -3 s -1 mol dm -3 s -1 J mol -1 J mol -1 K -1 x K

24 Arrhenius expression k = A e Ea / RT

25 Arrhenius expression Describes the effect of temperature on the rate of reaction and the rate constant k = A e Ea / RT 2 HI(g) H 2 (g) + I 2 (g)

26 Arrhenius expression Describes the effect of temperature on the rate of reaction and the rate constant k = A e Ea / RT Equation of a straight line: y = mx + c Plot y vs x gives gradient = m and intercept = c ln k = ln (A e Ea / RT ) = ln A + ln(e Ea / RT ) ln k = ln A Ea / RT

27 Arrhenius expression Describes the effect of temperature on the rate of reaction and the rate constant ln k = ln A Ea / RT y = mx + c Graph of ln k vs 1/T to give a Gradient = Ea / R Intercept = ln A

28 Arrhenius expression Describes the effect of temperature on the rate of reaction and the rate constant ln k = ln A Ea / RT 2 HI(g) H 2 (g) + I 2 (g)

29 Arrhenius expression Describes the effect of temperature on the rate of reaction and the rate constant y x Gradient = y/ x = Ea / R = 2.35 x 10 4 K Ea = 2.35 x 10 4 x = J mol -1 = 195 kjmol -1

30 Real Gases At high pressures and low temperatures, real gases depart from ideal gas behaviour Deviation greatest for gases that liquify The critical point is a point of inflexion Differentiate once: dp dv = 0 Differentiate twice: d 2 P dv 2 = 0

31 Molecular speeds Maxwell-Boltzmann distribution shows the spread of molecular speeds (energies) at a temperature Differentiate to find most probable speed Integrate to find mean speed

32 Summary Mathematical tools give us access to vital chemical information from experiment organic, inorganic, physical & analytical Units are important! Be consistent, use standard units Does your answer make sense? Think about how you plot your graphs Straight line graphs are very useful! To minimise errors - show your working & all steps