Unit 3 Revision

All occupy the same volume 2g H2 32g O2 17g NH3 16g CH4 1 mole 1 mole 1 mole 1 mole All occupy the same volume Volume of 1 mole of a gas – MOLAR VOLUME

All occupy the same volume 1g H2 16g O2 8.5g NH3 8g CH4 0.5 mole 0.5 mole 0.5 mole 0.5 mole All occupy the same volume Volume of 1 mole of a gas – MOLAR VOLUME Volume of 0.5 mole of a gas – ½ x MOLAR VOLUME

2 mol 1 mol 0.25 mol 0.1 mol MOLAR VOLUME = 24 litres VOLUME 16g CH4 4g CH4 1.6g CH4 32g CH4 2 mol 1 mol 0.25 mol 0.1 mol MOLAR VOLUME = 24 litres VOLUME = 6 litres VOLUME = 2.4 litres VOLUME = 48 litres

100cm3 of propene burned completely with 900cm3 of oxygen 100cm3 of propene burned completely with 900cm3 of oxygen. What will be the volume and composition of the resulting gas mixture? C3H6 (g) + 4½O2(g)  3CO2(g) + 3H2O(l)

PERCENTAGE YIELD CALCULATIONS 5g of methanol reacts with excess ethanoic acid. What is the theoretical yield of ethyl methanoate CH3OH + CH3COOH CH3OOCCH3 1 mol 1mol 32g 74g 1g 2.3125g 5g 11.5625g

PERCENTAGE YIELD CALCULATIONS 5g of methanol reacts with excess ethanoic acid to produce 9.6g of methyl ethanoate. Calculate the percentage yield. CH3OH + CH3COOH CH3OOCCH3 1 mol 1mol 32g 74g 1g 2.3125g 5g 11.5625g % yield = 9.6 x 100 11.5625 83%

Reactants Products Reactants Products Equilibrium lies to the right Forward (reactants) Concentration Backward (products) Time Reactants Products Equilibrium lies to the left Forward (reactants) Concentration Backward (products) Time

Balance....Why add acid....

Would you need to add an indicator? Write the ion electron equation for the reduction reaction

Lab analysis: chromatography video Hydrolyse the protein using acid or alkali and then use chromatography Using known amino acids alongside the hydrolysed protein allows identification of the amino acids in the protein.

A, B, C, D and E - five known amino acids. P - hydrolysed protein. P contains four amino acids because 4 spots are present.

The hydrolysed protein also contains another unknown amino acid. This can be identified by running another chromatogram with different known samples of pure amino acids.

Results Volume of I2 (cm3) 1st 2nd 3rd 4th Initial 20.0 Final 30.9 30.2 30.1 Difference 10.9 10.2 10.1

Results Volume of I2 (cm3) 1st 2nd 3rd 4th Initial 20.0 Final 30.1 30.2 30.9 Difference 10.1 10.2 10.9

TARGET EQUATION: complete combustion of CO: CO + ½ O2  CO2 (a) X -½ = -½ (-216) (b) X 1 = -394 = > -286 kJ mol-1

(a) X -1 = +446 (b) X 1 = -394 (c) X 1 = -184 = > -132 kJ mol-1 (a)

Example 2 Calculate the enthalpy of formation of propane using bond enthalpies. C H C (s) H H C (s) + C (s) H H 3C (s) + 4H2 (g) C3H8 (g)

Example 2 Calculate the enthalpy of formation of propane using bond enthalpies. 3C (s) + 4H2 (g) C3H8 (g) Convert 3 moles of carbon solid into 3 moles of carbon gaseous atoms AND Convert 4 moles of hydrogen molecules into 8 moles of hydrogen atoms

Example 2 Calculate the enthalpy of formation of propane. 3C (s) + 4H2 (g) C3H8 (g) C H C (s) C (s) H H H C (s) + C (s) H H Bond Breaking 3 3 x C(s) C(g) = 3 x 716 = 2148 4 x 4 H H = 4 x 436 = 1744 Total put in = 3892

Example 2 Calculate the enthalpy of formation of propane. 3C (s) + 4H2 (g) C3H8 (g) C H C (s) H H C (s) + C (s) H H Bond Breaking 3 x C(s) C(g) = 3 x 716 = 2148 4 x H H = 4 x 436 = 1744 Total put in = 3892

Example 2 Calculate the enthalpy of formation of propane. 3C (s) + 4H2 (g) C3H8 (g) C H C H C H C H C H C (s) H H C (s) + C H C C C H C C H C H C H C (s) H H Bond Breaking Bond Making 3 x C(s) C(g) = 2148 C 2 x 2 = 2 x -348 = -696 4 x H H = 1744 8 C H 8 x = 8 x -412 = -3296 Total put in = 3892 Total given out = –3992

Example 2 Calculate the enthalpy of formation of propane. 3C (s) + 4H2 (g) C3H8 (g) C H C (s) H H C (s) + C (s) H H Bond Breaking Bond Making 3 x C(s) C(g) = 2148 C 2 x = 2 x -348 = -696 4 x H H = 1744 C H 8 x = 8 x -412 = -3296 Total put in = 3892 Total given out = –3992 ∆H = 3892 + (–3992) = 3892 – 3992 = – 100 kJ