Chem Unit6

Stoichiometry

στοιχεῖον (stoicheion, meaning element]) & μέτρον (metron, meaning measure)

Balancing Chemical Equations

• 2 Mg + O2 -> 2 MgO


• 2 Mg + O2 -> 2 MgO

1.2 x 1024 + 6 x 1023 form 1.2 x 1024 formula units

atoms O2 molecules MgO

Mg


• 2 Mg + O2 -> 2 MgO

2 moles + 1 mole form 2 moles

Mg O2 MgO

Mole Ratio


• 2 Mg + O2 -> 2 MgO

48.6 g + 32 g form 80.6 grams

Mg O2 MgO

Mass

Balance

The total mass of the product(s) must equal the total mass of the reactants

Yield

• Building Bicycles

Yield• Building Bicycles1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels

Finding Product Quantities

• How much O2 gas is formed when

49.89 g KClO3 decomposes?

KClO3 -> KCl + O2

Balance the Equation



2 KClO3 -> 2 KCl + 3 O2




2 KClO3 -> 2 KCl + 3 O2

1 mole49.89 g KClO3 x 122.45 g = 0.41 moles KClO3

3 moles O2

2 moles KClO3 x 0.41 moles KClO3

= 0.615 moles O2




2 KClO3 -> 2 KCl + 3 O2

49.89 g KClO3 ?

0.615 moles O2 x 32 g/molee = 19.68 grams O2

Standard Temperature & Pressure

• STP

Temperature = 0 °C (273.15 K; 32.00 °F)

Pressure = 101.325 kPa (1 atm; 760 Torr)

For Gases: Volume = 22.4 Liters/mole




2 KClO3 -> 2 KCl + 3 O2

49.89 g KClO3 ?

0.615 moles O2 x 22.4 L/mole = 13.776 Liters O2

Yield• Building Bicycles1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels

= 1 bicycle

Yield

• Theoretical Yield



2 KClO3 -> 2 KCl + 3 O2

49.89 g KClO3 6.12 Liters O2

What if only 5.5 Liters of O2 are produced?

Yield

• Theoretical Yield

2 KClO3 -> 2 KCl + 3 O2

49.89 g KClO3 6.12 Liters O2

What if only 5.5 Liters of O2 are produced?

5.5 Liters

6.12 Liters x 100% = 89.9% actual yield

Gas Yield

• Nitroglycerin decomposes to

produce N2, O2, CO2 and H2O

Gas Yield

• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O

227 g how much gas results?

Gas Yield

• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O

1 mole

227 g x 227 g = 1 mole nitroglycerin

Gas Yield

• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O

1.5 moles N2

1 mole nitroglycerin -> 0.25 mole O2

3 moles CO2

0.25 mole H2O

Gas Yield

• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O

1.5 moles N2 x 22.4 L/ mole = 33.6 L

0.25 mole O2 x 22.4 L/mole = 5.6 L

3 moles CO2 x 22.4 L/mole = 67.2 L

0.25 mole H2O x 22.4 L/mole = 5.6 L

112 Liters !

Making Aspirin

• White willow tree bark = pain reliever

Making Aspirin

• Salicylic acid acetylsalicylic acid

+ acetic anhydride -> and acetic acid

C7H6O3 + C2H6O3 -> C9H8O4 + C2H4O2

each aspirin tablet 325 mg

How much salicylic acid would be needed if we assume 90% yield?

Making Aspirin

C7H6O3 + C2H6O3 -> C9H8O4 + C2H4O2

0.325 g x 1 mole/180 g = 0.0018 moles of aspirin

90 % of 0.0018 = 0.9 x 0.0018 = 0.00162 moles

0.00162 moles x 138 g/mole = 0.224 grams of salicylic acid

Rocket Fuel

N2H4(l) + N2O4(l) -> N2(g) + H2O(g)

Is the reaction balanced?

Rocket Fuel

2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g)

What is the mole ratio

of N2H4 to N2?

Rocket Fuel

2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g)

What is the mole ratio

of N2H4 to N2?

3 moles N2

2 moles N2H4

Rocket Fuel

2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g)

How many Liters of N2 will be

produced from 16000 g of N2H4?

16000 g x 1 mole = 500 moles N2H4

32g

3 moles N2

500 moles N2H4 x 2 moles N2H4

= 750.0 moles N2

= 16800 Liters of N2

Rocket Fuel

2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g) ENERGY

How many Liters of N2 will be

produced from 16000 g of N2H4?

16000 g x 1 mole = 500 moles N2H4

32g

3 moles N2

500 moles N2H4 x 2 moles N2H4

= 750.0 moles N2

= 16800 Liters of N2

Chemical Change

• Energy is absorbed or released

Exothermic vs. Endothermic

Enthalpy ΔH

• Total energy of a system:Internal energy + (pressure) x (Volume)

Enthalpy ΔH

• Only way to measure ΔH is when heat is added or removed:

Molar Enthalpy Change: ΔH = C ΔT

(Recall Specific Heat Capacity: Q = m C ΔT)

Molar Enthalpy Change

• ΔH = CΔT

What is ΔH if an aluminum can is cooled from 25° C to 4° C?

Temperature Scales

Molar Enthalpy Change

• ΔH = CΔT

What is ΔH if an aluminum can is cooled from 25° C to 4° C? (C for Al = 24.2 J/K mole)

25° C = 298 K

4° C = 277 K

ΔH = CΔT = 24.2 J/K mole (21 K)

= 508.2 J/mole

Enthalpy Change

• Negative = Exothermic

C(s) + 1/2 O2(g) -> CO(g) ∆H = - 110.5 kJ

• Positive = Endothermic

C2H6(g) -> 2C(s) + 3H2(g) ∆H = + 4.83 kJ

Enthalpy Change (Endothermic)

A + B ----> C + D

Enthalpy Change (Exothermic)

W + X --> Y + Z

Enthalpy in Chemical Changes

Stoichiometric Enthalpy

• ∆H = ? When 4.8 g C reacts with O2?

• C + O2 -> CO2 ∆H = -393.5 kJ/mole

4.8 g x 1mole/12g = 0.4 moles

0.4 moles x (-393.5 kJ/mole)

= 157.4 kJ

Hess’ Law

• Total ∆H = Sum of ∆H for all steps

• Example:

H2S(g) + 2O2(g) ->SO3(g) + H2O(l)

∆H = ???????

Hess’ Law

H2S(g) + 2O2(g) -> H2SO4(l) ∆H= -628KJ

H2SO4(l) -> SO3(g) + H2O(g) ∆H= +164KJ

H2O(g) -> H2O(l) ∆H= -88KJ

H2S(g) + 2O2(g) ->SO3(g) + H2O(l) ∆H = ?

Hess’ Law

Germain Henri Hess

• Enthalpy change (∆H) depends only on the initial and final states of the reaction

(not on the intermediate stages)

Hess’ cycle

Hess’ Law

• CH4 + 2 O2 ----> CO2 + 2 H2O ∆H = ??

CH4 + 2 O2 -----> CO + 2 H2O + ½ O2

∆H = -607 kJ

CO + 2 H2O + ½ O2 ------> CO2 + 2 H2O ∆H = -283 kJ

Standard Enthalpy of Formation

• Formation from the composing elements:

• Glucose:

• 6 C + 6 H2 + 3 O2 --> C6H12O6

∆H = –1273.3 kJ

Hess’ Law

• CH4 + 2 O2 ----> CO2 + 2 H2O ∆H = -890 kJ

CH4 + 2 O2 -----> CO + 2 H2O + ½ O2

∆H = -607 kJ

CO + 2 H2O + ½ O2 ------> CO2 + 2 H2O ∆H = -283 kJ

Hess’ Cycle

Unit 6 Recap

• Stoichiometry Mole Ratios

Yield = actual/theoretical

Limiting and extra reactants

1 mole = 22.4 Liters at STP

• Enthalpy Change ∆Hendothermic vs. exothermic

∆H = C∆THess’ Law

Chem Unit6

Documents

Transcript of Chem Unit6