1. Unit 3 Revision

5. All occupy the same volume2g 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

6. All occupy the same volume1g 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

7. 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

10. 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) ½O2(g)  3CO2(g) H2O(l)

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

12. 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 mol
32g g
1g g
5g g
% yield = x 100
83%

17. 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

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

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

25. 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.

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

27. 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.

29. 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

30. 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

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

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

33. 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)

34. 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

35. 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

36. 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

37. 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

38. 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