.4 Energetics Flashcards

1
Q

Draw graph for exothermic + endothermic and give basic info and examples

A
Endothermic
Heat absorbed 
Enthalpy of reactants < products 
ΔH = positive
Examples: Respiration
Combustion of fuels
Exothermic: 
Heat given out 
Enthalpy of reactants > products 
ΔH = negative
Examples: Photosynthesis
Thermal decomposition of calcium carbonate
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2
Q

Introduction to Energetics and 1st Law of Thermodynamics

A

All chemical reactions are accompanied with some form of energy change. Usually heat energy is either absorbed or given out*. The products and reactants in a chemical reaction have two main forms (or classes) of energy:
Chemical potential energy
Kinetic energy

As the 1st Law of Thermodynamics states ‘Energy can be neither created nor destroyed but it can be converted from one form to another’.

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3
Q

ΔH (enthalpy change)- why can’t measure directly, how to measure indirectly

A

We cannot measure enthalpy directly because we cannot measure absolute gravitational or kinetic energy. However we can calculate ΔH indirectly.

The symbol used to indicate change in enthalpy is ΔH. ΔH is the heat energy change at constant pressure when:

Reactants → Products

ΔH = Hproducts – Hreactants

(same as: ΔH = Hfinal – Hinitial)

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4
Q

Standard conditions and symbol

A

Enthalpy values vary according to the conditions.
Standard state conditions are said to be: 1. Pressure: 100 kPa (100 000 Pa) (1 atmosphere)
2. A stated temperature usually 298K (= 25oC)
3. Any solutions are of concentration 1 mol dm-3.

Standard state standard conditions are indicated by ΔHϴ298.

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5
Q

Key definitions of diff standard enthalpies

A

Standard enthalpy of combustion. Enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants and products in standard states under standard conditions.

Standard enthalpy of formation. Enthalpy change when 1 mole of a substance is formed from its constituent elements with all reactants and products in standard states under standard conditions.

Standard enthalpy of neutralisation. Enthalpy change when 1 mole of water is formed in a reaction between an acid and alkali under standard conditions

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6
Q

Key definitions of diff standard enthalpies and examples with CH4

A

Standard enthalpy of combustion. Enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants and products in standard states under standard conditions.
e.g. CH4 CH4(g) + 2O2(g) 🡪 CO2(g) + 2H2O(l)

Standard enthalpy of formation. Enthalpy change when 1 mole of a substance is formed from its constituent elements with all reactants and products in standard states under standard conditions.
e.g. CH4 C(s) + 2H2(g) 🡪 CH4(g)

Standard enthalpy of neutralisation. Enthalpy change when 1 mole of water is formed in a reaction between an acid and alkali under standard conditions
e.g. HCl + NaOH HCl(aq) + NaOH(aq) 🡪 NaCl(aq) + H2O(l)

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7
Q

Flame calorimeter

A

an improved version of the simple calorimeters used for measuring enthalpy changes of combustion. Itincorporates the following features that are designed to reduce heat loss even further:

  • the spiral chimney is made of copper
  • the flame is enclosed
  • the fuel burns in pure oxygen, rather than air.
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8
Q

Simple Calorimeter and draw apparatus

A

an apparatus for measuring the amount of heat involved in a chemical reaction or other process.
Follows the basic principle of heat being transferred to the water. As the specific heat capacity of water is known, we can work backwards to calculate how much heat was given out by the burning fuel.

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9
Q

Calorimetry- give examples for 1g of water hotter by 1C, 3g of water hotter by 10C

A

The science of measuring the heat of chemical reactions, i.e. calculating enthalpy change.

1 ºC hotter 1 g of water
Energy required = 4.18J

10 ºC hotter 3 g of water
Energy required = 3 x 10 x 4.18J = 125.4 J

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10
Q

Energy required equation, what they all stand for, units, how to use to work out ΔH

A

Energy required (q) = mass heated (m) x energy needed to make 1g of substance 1ºC hotter x temperature rise (ΔT)

q = mcΔT

Units: q = J; m = g; T = K

ΔH= Q/mol

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11
Q

Reactions in Solution (Cup Calorimeter)

A

The reaction is carried out in an insulated beaker and the temperature change measured. The reaction must be fast so that the maximum temperature is reached quickly (before it starts to cool). The specific heat capacity of the solution is taken as 4.18 Jg-1K-1 (the same as water) and we usually use the mass of the water (not the solution) in the calculation.

Biggest source of error= heat loss (for exothermic)

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12
Q

50 cm3 of 1.0 mol dm-3 hydrochloric acid was added to 50 cm3 of 1.0 mol dm-3 sodium hydroxide solution. The temperature rose by 6.8ºC. Calculate the standard enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3; the specific heat capacity of the solution is 4.18 Jg-1K-1.

A

q = mc∆T m = 100 c = 4.18 ∆T = 6.8 q

= 100 x 4.18 x 6.8 = 2842 J

∆H = q / mol

Mol HCl = conc x vol = 1.0 x 50/1000 = 0.050

(Mol NaOH = conc x vol = 1.0 x 50/1000 = 0.050)

HCl + NaOH → NaCl + H2O

∆H = –2.842 / 0.050 = -56.8 kJ mol-1 (3 sig fig)

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13
Q

Consider this reaction: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) ΔHθ298=-890kJmol-1 What does the minus sign indicate?

A

Reaction is EXOTHERMIC- products lower energy than reactants

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14
Q

Pros and Cons of Cup Calorimetry

A

PROS
1. Heat generated is from the solutions themselves so only has to be kept in calorimeter.

  1. Polystyrene cups are very good insulators (often you will use a cup within another cup).
  2. They have a low heat capacity, so absorb very little heat.

CONS
1. Inaccuracy in measuring mass and temperature

  1. Solution is assumed to be pure water (therefore c is assumed to be 4.18 Jg-1K-1)
  2. Some heat is absorbed by the cup.
  3. Heat is lost (or gained) from the surroundings. Not a completely closed system.
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15
Q

Bomb Calorimeter

A

The most accurate type of calorimeter.
A known mass of a fuel is placed inside a steel container (the “bomb”) and the container filled with oxygen under
pressure. The fuel is electrically ignited and heat evolved used to heat surrounding water. The heat capacity of the calorimeter is found by burning a substance with accurately known ΔHc.

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16
Q

Enthalpy change of combustion
200g of water were heated by burning ethanol in a spirit burner. The following mass measurements were recorded:
Mass of spirit burner and ethanol before burning = 58.25 g
Mass of spirit burner and ethanol after burning = 57.62 g

The initial temperature of the water was 20.7 oC and the highest temperature of the water was 41.0 oC. The specific heat capacity of water is 4.18 J K-1 g-1. Calculate a value for the standard enthalpy change of the combustion of ethanol in kJmol-1 to 3 sig figs.

A

-1240 kJmol-1

17
Q

What is Hess’ Law?

A

‘the enthalpy change for a chemical reaction depends only on the initial and final states and is independent of the path followed’.

Overall change in enthalpy is independent of the route taken

18
Q

Hess’s Law reaction diagram- what it tells us

A

If a reaction can occur by more than one route, the overall enthalpy change is independent of the route

Draw diagram

19
Q

Using Hess’s Law 1: Heat of Combustion
C(s) + 2H2(g) 🡪 CH4(g) ΔHθr = ? kJmol-1

Data given: ΔHθC C(s) = -394 kJmol-1 ΔHθC H2 (g) = -286 kJmol-1 ΔHθC CH4(g) = -890 kJmol-

A

+(-394) + (2 x -286) - (-890)= -76kJmol-1

20
Q

Using Hess’s Law 1: Heat of Combustion
C(s) + ½O2 (g) → CO(g)

Data given: ΔHθC C(s) = -394 kJmol-1 ΔHθC CO (g) = -283 kJmol-1

A

+(-3064) + (3 x -286) - (-32727)= -195kJmol-1

21
Q

Formation Q: Identify the constituent elements, in their standard states, used to form the following:
CH4

C6H12O6 + O2

CuCO3 + H2

HBr

H2O + CO2

A

Check

22
Q

Using Hess’ Law 2: Heat of Formation

If you formed all the products form their elements, you will need exactly the same elements to form the reactants.
If we know the enthalpy of formation for all reactants and products we can set up a thermochemical cycle as we did before:

CH4 (g) + 2O2 (g) → CO2(g) + 2H2O(l) ΔHθr= ?

Why value not written for ?

A

Value not written for oxygen bc it already exists in its standard state as an element

-891.1kJmol-1

23
Q

Enthalpy change of formation:
CH4(g) + Br2 (g) —–> CH3Br (g) + HBr(l)

Complete the thermochemical cycle for paths B and C
Calculate ΔHr º using the formation values below:
CH4(g) = -`74.9 kJmol-1
CH3Br(g) = -37.2 kJmol-1
HBr(l) = - 36.4 kJmol-1

A

-(-74.9) + (-37.2) + 36.4= 74.1 kJmol-1

24
Q

The enthalpy of formation is the enthalpy change when one mole of compound is formed from its constituent elements under standard conditions with all reactants and products in their standard states.

The enthalpies of all elements in their standard states are taken as ,,,

A

zero by definition.
So for example, hydrogen is found as H2(g) in standard conditions so any enthalpy calculations containing H2(g) will give no value for H2(g)

25
Q

1) Calculate the ΔHθ for the following reactions given the values of ΔHθf in the following table.

ZnCO3(s) -812
ZnO(s) -348
CO2(g) -393
CO(g) -111
H2O(l) -286
Fe2O3(s) -822
Al2O3(s) -166
C2H4(g) +52
a) ZnCO3(s)  →  ZnO(s)  +  CO2(g)
b) 2 CO(g)  +  O2(g)  →  2 CO2(g)
c) 2 Al(s)  +  Fe2O3(s)  →  2 Fe(s)  + Al2O3(s)
d) C2H4(g)  +  3 O2(g)  →  2 CO2(g)  +  2 H2O(l)
e) C2H4(g)  +  2 O2(g)  →  2 CO(g)  +  2 H2O(l)
A

a) ZnCO3(s) → ZnO(s) + CO2(g)

            Zn (s)   + C  (s)    +   3/2 O2 (g)

a = c – b = [-348 + -393] – [-812 ] = + 71 Kjmol-1

b)	2 CO(g)  +  O2(g)  →  2 CO2(g)
	= [2 x -393] – [2 x -111] = -564 Kjmol-1

c)	2 Al(s)  +  Fe2O3(s)  →  2 Fe(s)  + Al2O3(s)
	= [-166] – [-822] = 656 Kjmol-1

d)	C2H4(g)  +  3 O2(g)  →  2 CO2(g)  +  2 H2O(l)
	= [(2x -393) + (2 x -286)] – [52] = -1410 Kjmol-1 

e)	C2H4(g)  +  2 O2(g)  →  2 CO(g)  +  2 H2O(l)
	= [(2 x -111) + (2 x -286)] – [52] = -846 Kjmol-1
26
Q

Do questions 2-16 on AQA AS Chemistry Ch.7 Energetics sheet

HESS’S LAW 2- USING ENTHALPY CHANGE OF FORMATION

A

Check

27
Q
Do combustion reactions
CH4 + O2 🡪 
H2 + O2 🡪 
C6H12O6 + O2 🡪
Zn + O2 🡪
Cu + O2 🡪
CuCO3 + O2 🡪
A

CH4 + 2O2 —->2H2O + CO2

H2 + 1/2O2 —–> 2H2O

C6H12O6 + 602 —-> 6H2O + 6CO2

Zn + 1/2O2 —–> ZnO

Cu + 1/2O2 —–> CuO

28
Q

7) 25 cm3 of 1.00 mol dm-3 copper sulphate solution was put in a calorimeter and 6.0 g of zinc powder added. The temperature of the solution rose by 50.6°C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals.

A
q = 25 ( remeber m is mass of SOLUTION) x  4.18  x  50.6  =  5288 J
∆H = q / mol

Mol CuSO4 = conc x vol= 1.0 x 25/1000 = 0.025

Mol Zn = mass / Mr= 6.0 / 65.4 = 0.0917- XS so use 0.025

CuSO4 + Zn → ZnSO4 + Cu

∆H = –5.288 / 0.025= -212 kJ mol-1 (3 sig fig)

29
Q

What is ΔHθ for the change from methoxymethane (CH3OCH3) to ethanol? The ΔHθf of the two compounds are:
CH3OCH3 ΔHθf = -184kJmol-1
C2H5OH ΔHθf = -277kJmol-1

A

Draw diagram(s)- check in notes

-277= -184 + ΔHr
ΔHr= -93
30
Q

Bond enthalpy definition and important note

A

Bond Enthalpy is the standard enthalpy change associated with breaking one mole of bonds into one mole of gaseous atoms

Bond enthalpies are mean values as the actual strength of a bond depends completely upon the environment of a bond within a molecule. (C-H in methane and trichloromethane)

31
Q

Method 1: Hess’ Cycle

Calculate ΔH for the following bond enthalpies for the reaction: 2NCl3 🡪 N2 + 3Cl2

(N−Cl) = 200 kJ/mol 
(N≡N) = 941 kJ/mol 
(Cl−Cl) = 242 kJ/mol
A

Draw Hess’ cycle

ΔH = 6(200) − [941 + 3(242)]

ΔH = 1200 − (1667)

ΔH = − 467 kJmol-1

32
Q

Method 2: Bonds Broken – Bonds Formed

Calculate ΔH for the following bond enthalpies for the reaction: 2NCl3 🡪 N2 + 3Cl2

A

ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)

ΔH = 6(200) − [941 + 3(242)]

ΔH = 1200 − (1667)

ΔH = − 467 kJmol-1

33
Q

Method 3: Enthalpy Diagram
Steps

Example for
CH3COCH3(l) + H2(g) —-> CH3CH(OH)CH3(l)

CH3COCH3 (l) : ΔHθf (kJmol-1)= -248

H2- already gaseous (or bc standard state?), no value

CH3CH(OH)CH3 (l): ΔHθf (kJmol-1)= -318

A
  1. Draw an enthalpy energy axis
  2. Draw in the reactants
  3. Calculate energy input (to break bonds)
  4. Calculate the energy released (to make bonds)

Ans: -70