Презентация на тему The ideal gas equation

of gas molecules occupies 24.0 dm3
Conditions are not always

pV = nRT, convert units to m3, K and

100 kPa = 100 × 103 Pa

220 cm3 = 220 × 10−6 cm3

Examples

4.0 dm3 = 4.0 × 10−3 m3

48 °C = 48 + 273 = 321 K

make one of the resulting products

Atom economy is a

make one of the resulting products

Atom economy is a

MOLECULAR MASS OF DESIRED PRODUCT x 100
SUM OF MOLECULAR MASSES OF ALL PRODUCTS

of 1,2-dichloroethane, C2H4Cl2

Example 1

of 1,2-dichloroethane, C2H4Cl2

Equation C2H4 +

An ATOM ECONOMY of 100% is typical of an ADDITION REACTION

Example 1

of nitrobenzene, C6H5NO2

Example 2

of nitrobenzene, C6H5NO2

Equation C6H6 + HNO3

An ATOM ECONOMY of 100% is not possible with a SUBSTITUTION REACTION

Example 2

of ammonia from the thermal decomposition of ammonium sulphate.

Example 3

of ammonia from the thermal decomposition of ammonium sulphate.

In industry a low ATOM ECONOMY isn’t necessarily that bad if you can use some of the other products. If this reaction was used industrially, which it isn’t, the sulphuric acid would be a very useful by-product.

Example 3

(the required product is shown in red)


• CH3COCl

(the required product is shown in red)


• CH3COCl

70.2%

(the required product is shown in red)


• CH3COCl

70.2%

55.8%

33.9%

substitution reactions will have less than 100% atom economy

•

a reaction
Percentage yield

all the REACTANTS are converted into products.
This will give

of products which would be obtained from the balanced

an excess of chlorine to produce 4.0g of sodium

(Ar reactants: Na=23 Cl=35.5 Mr product: NaCl= 58.5)

58.5 x 0.1 =

Theoretical yield of NaCl =

5.85g

What is the percentage yield?

% Yield = Actual yield x 100% Theoretical yield

% Yield = 4.0g x 100% = 5.85g

68%

2Na(s) + Cl2(g) ⇒ 2NaCl(s)

= 0.1 mol Na

Theoretically 0.1 mol Na should yield 0.1 mol NaCl

with an excess of oxygen to produce 0.8g of

What is the percentage yield?

% Yield = 0.8g x 100% = 2g

40%

2Mg(s) + O2(g) ⇒ 2MgO(s)

(Ar reactants: Mg=24 O=16 Mr product: MgO= 40)

= 0.05 mol Mg

Theoretically 0.05 mol Mg should yield 0.05 mol MgO

40 x 0.05 =

Theoretical yield of MgO =

2g

% Yield = Actual yield x 100% Theoretical yield