What is chemical energy?

Enthalpy, H

                                          

Conservation of energy (First Law of thermodynamics)

Energy cannot be created or destroyed just transferred between the system and the surroundings

  Heat loss in a chemical system = Heat gain to the surroundings Temperature increases
  Heat gain in a chemical system = Heat loss to the surroundings Temperature decreases

Enthalpy change, DH:

                DH    =    Hproducts    -    Hreactants

Exothermic reactions:

                                Hproducts    <    Hreactants

  DH = Hproducts - Hreactants
  DH = Small - Large
  DH = Negative value    

Energy is released from the system to the surroundings

Endothermic reactions:

                                Hproducts    >    Hreactants

  DH = Hproducts - Hreactants
  DH = Large - Small
  DH = Positive value    

Energy is transferred to the system from surroundings

Practical

Questions 1 - 2  P 185

 

Exothermic and endothermic reactions

 

1)  Oxidation of fuels:

CH4(g) + O2(g) CO2(g) + H2O(g) DH = - 890 Kj mol-1

Environmental impact - Example:

C8H18(l) + 12.5O2(g) 8CO2(g) + 9H2O(g) DH = - 5470 Kj mol-1
200g                    
Moles = m / Mr                    
Moles = 200 / 114                    
Moles = 1.75     1:8              
Moles ~ 2     Moles = 16            
Energy used = 2  x  5470       Volume = 16  x  24            
        Volume = 384 dm3            

 

2)  Respiration:

C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l) DH = -2801 Kj mol-1

 

Endothermic reactions:

 

1)  Photosynthesis:

6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) DH = +2801 Kj mol-1

2)  Thermal decomposition of limestone:

CaCO3(s) CaCO(s) + CO2(g) DH = +178 Kj mol-1

Questions 1 - 2  P187

 

Enthalpy profile diagrams

 

Simple enthalpy profile diagrams:

Exothermic reactions:

Where Hproducts  <  Hreactants

DH = negative

Endothermic reactions:

Where Hreactants  >  Hproducts

DH = positive

Activation energy:  Explanation

Exothermic reactions:

Endothermic reactions:
  • Even though the products are lower in energy than the reactants, a small amount of energy is needed to break the reactant bonds, the activation energy.

  • We say the reaction has to overcome the energy barrier.

  • After that, the exothermic nature of the reaction is enough to break more reactant bonds.

  • A way of thinking of this is like a cyclist riding a bike up a small hill to gain a large amount of 'free wheeling' or energy.

  • The products are higher in energy than the reactants, a small amount of energy is still needed to break the reactant bonds, the activation energy.

  • The reaction has to overcome the energy barrier.

  • This time there is no excess energy to break more reactant bonds, a sustained amount of energy needs to be continually supplied to keep the reaction going.

  • This time the cyclist rides a bike up a large hill to gain a small amount of 'free wheeling' or energy.

 

 

Questions  1-2  P189

 

Standard enthalpy changes

 

Standards:

Standard conditions:

 

                      Pressure at 100 kPa (which is 1 atmosphere)

                      Temperature at 298K (25oC)

                      1 Mole or 1 Molar solutions

                      Normal physical states at standard temperature and pressure (above conditions)

 

                      A standard enthalpy change is shown by  -   DH q

 

                       H - Enthalpy

                       D - Change in

               q symbol represents standard conditions

Standard states:

Substance Chemical symbol and state Explanation
Magnesium Mg(s) Magnesium is a solid under standard conditions, (s)
Hydrogen H2(g) Hydrogen is a gas under standard conditions, (g)
Water H2O(l) Water is a liquid under standard conditions, (l)

Standard enthalpy changes:

1)  Standard enthalpy change of reaction, DHrq

H2(g) + 1/2 O2(g) H2O(l) DHrq = -286 KJMol-1

For the reaction:

         Fe2O3(s)   +    2Al(s)           2Fe(s)    +       Al2O3(s)             DHqr   =  -851.50 KJ Mol-1

   1/2Fe2O3(s)    +    Al(s)              Fe(s)       +       1/2Al2O3(s)      DHqr   =  -425.75 KJ Mol-1

2)  Standard enthalpy of combustion DHqc

Definition:  The enthalpy change that occurs when 1 mole of a substance reacts completely with oxygen under standard conditions.  All reactants and products are in their standard states.

DHqc for ethane:

C2H6(g) + 3.5 O2(g) 3H2O(l) + 2CO2(g) DHcq = -1560 KJMol-1

3)  Standard enthalpy change of formation, DHqf

Is the enthalpy change that takes place when 1 mole of a compound is formed from its constituent elements in their standard states under the standard conditions.

 

DHq for water:

 

   H2(g)            +          1/2O2(g)                    H2O(l)              DHqf  =  -286 KJ Mol-1

A problem:

Questions  1 - 3   P191

Determination of enthalpy changes

Remember:

  Heat loss in a chemical system = Heat gain to the surroundings Temperature increases
  Heat gain in a chemical system = Heat loss to the surroundings Temperature decreases

                  

i)                    The energy change - for which we use temperature change

ii)                   The amounts in moles of the limiting reagent that reacts

i)  The energy change:

                                                  Q = mcDT

                                                          1000

                                                  Q       quantity of energy exchanged J, the 1000 converts to kJ

                                                  m      mass of the water g (ie cm3 as  the density if water is 1 gcm-3)

                                                  c       specific heat capacity j g-1 K-1

                                                  DT    rise in temperature K, (Tinitial  -  Tfinal)

ii)  Calculate the number of moles used:

 

                                                  No Moles = Mass         or        c  x  V

                                                                           Mr                       

 

iii)  Calculate the amount of energy exchanged per mole , this is the enthalpy:

 

                                                              Enthalpy = Energy

                                                                                    Moles

                                            (-)ve for exothermic reactions

                                            (+)ve for endothermic reactions

Direct determination of enthalpy changes:

  • Most reactions involve adding one reactant to another and measuring the temperature rise.
  • This is called direct determination and is carried out in a calorimeter.
  • A calorimeter is an insulated reaction vessel which minimises heat exchange to the air as the surroundings we measure is the water / solvent.
  • The simplest calorimeter is a polystyrene cup but they can get quite complex.

Example:

Excess Mg is added to 100cm3 of 2.00 Mol dm-3 CuSO4, the temperature rose from 20.0oC to 65oC.

i)  The energy change:

Q = (m x c x DT)  /  1000  
Q = (100 x 4.18 x 45)  /  1000  
Q =  18.81 kJ Temp increases, it is exothermic therefore is negative  
Q = - 18.81 kJ    

 

ii)  Calculate the number of moles used:

 

                                                  No Moles = c  x  V

 

                                                  No Moles = 2  x  0.100

 

                                                  No Moles =  0.2

                                                                       

iii)  Calculate the amount of energy exchanged per mole , this is the enthalpy:

 

                                                              Enthalpy = Energy

                                                                                    Moles

 

                                                              Enthalpy = - 18.81

                                                                                       0.2

 

                                                              Enthalpy = - 94.05 Kj Mol-1

Finally write the equation with the enthalpy change:

Mg(s) + CuSO4(aq) MgSO4(aq) + Cu(s) DHrq = - 94.05 KjMol-1

Practical

Questions  1 - 2   P193

Enthalpy change of combustion

CH4(g) + 2O2(g) CO2(g) + H2O(l) DHcq = - 890 KjMol-1
H2(g) + 1/2O2(g) H2O(g)     DHcq = - 286 KjMol-1
Al(s) + 3/4O2(g) 1/2 Al2O3(s)     DHcq = - 1676 KjMol-1

Experimental determination of DHc

i)                    The energy change - for which we use temperature change when heating water

ii)                   The amounts in moles of fuel used

  • To calculate DHc, measure a known mass (volume) of water - for Q = MCDT.
  • Take the initial temperature of the water.
  • Weigh the burner before heating.
  • Heat to raise the temperature by about 10oC.
  • Take a final temperature.
  • Reweigh the burner to get a mass of fuel used.

Example:

1.5g of propan-1-ol heated 250cm3 of water by 45oC.

i)  The energy change:

Q = (m x c x DT)  /  1000  
Q = (250 x 4.18 x 45)  /  1000  
Q = 47.025 kJ Temp rises, it is exothermic therefore is negative
Q = - 47.025 kJ  

 

ii)  Calculate the number of moles used:

 

                                                  No Moles = Mass

                                                                           Mr

 

                                                  No Moles = 1.5

                                                                         60

 

                                                  No Moles =  0.025

                                                                       

iii)  Calculate the amount of energy exchanged per mole , this is the enthalpy:

 

                                                              Enthalpy = Energy

                                                                                    Moles

 

                                                              Enthalpy = - 47.025

                                                                                       0.025

 

                                                              Enthalpy = - 1881 Kj Mol-1

Practical

Comparison of experimental value with standard enthalpy change:

Standard enthalpy change of combustion of C3H7OH, DHcq - 2021 Kj Mol-1
Experimental enthalpy change of combustion of C3H7OH, DHc - 1881 Kj Mol-1

Errors:

Improvements:

  • Use a Bomb calorimeter (left).
  • This apparatus reduces heat loss s the water is insulated from the surroundings.
  • It is burnt in oxygen to ensure complete combustion.

Questions  1-2  P195

Bond enthalpies

Bond enthalpy:

    Is the enthalpy change that takes place when breaking by homolytic fission1 mole of a given bond in the molecules of a gaseous species

  Breaking bonds = Energy is put in to break bonds Endothermic process
  Forming bonds = Energy is released when bonds are formed Exothermic process
H - H(g) 2H(g)     DH = +436 KjMol-1
H - Cl(g) H(g)

+

Cl(g) DH = + 432 KjMol-1

Average bond enthalpy:

 

Bond Average bond enthalpy  /  KJ Mol-1
C - H +413
O = O +497
O - H +463
C = C +612
H - H +436

 

Breaking and making bonds:  Explanation

 

  Breaking strong bonds Forming weak bonds Endothermic process
  Breaking weak bonds Forming strong bonds Exothermic process

Using bond enthalpies to determine enthalpy changes:

  Energy required to break bonds = S  (Bond enthalpies of bonds broken)  
  Breaking weak bonds = S  (Bond enthalpies of bonds formed)  

Formula:

DH = S  (Bond enthalpies of bonds broken)

-

S  (Bond enthalpies of bonds formed)

Worked example:

S  (Bond enthalpies of bonds broken)       S  (Bond enthalpies of bonds formed)
         

Reactants

Products

Bond

Number

Bond energy

Total

Bond

Number

Bond energy

Total

 C - H

 4

413 

 1652

 C = O

 2

805 

1610 

 O = O

 2

 497

 994

 O - H

 4

 463

1852

S  (Bond enthalpies of bonds broken) =

2646 

S  (Bond enthalpies of bonds formed) =

3462

         
DH = S  (Bond enthalpies of bonds broken)

-

S  (Bond enthalpies of bonds formed)
DH = 2646

-

3462
DH = - 816 Kj Mol-1    

Questions 1-2  P197

Enthalpy changes from DHcq - Hess's law

Measuring enthalpy changes indirectly:

                High activation energy

                Slow rate of reaction

                More than one reaction occurring at the same time

Definition:  Hesss Law  Explanation

 

The total enthalpy change accompanying a chemical change is independent of the route by which the chemical change takes place provided the initial and final conditions are the same.

Using Hesss law:  Enthalpy cycles

1.      Enthalpy changes of combustion

2.      Enthalpy changes of formation

 

1  Calculating DHq  from enthalpy changes of combustion:  Explanation

                                  3C(s)                +          4H2(g)                     C3H8(g)

            Route 2 = - 2326  +  + 2219

            Route 2 = - 107 Kj Mol-1

Remember

            Route 1 = Route 2

            Route 1 = - 107 Kj Mol-1

Using the formula:

 

DH = S  DHcq (reactants)

-

S  DHcq (products)
DH = (3 x - 394)  +  (4 x - 286)

-

- 2219
DH =

- 107 Kj Mol-1

   

 

Question 1  P199

 

 

Enthalpy changes from DHqf - Hess's law

 

2)  Using enthalpy change of formation    Explanation

                                  2SO2(g)           +          O2(g)                     2SO3(g)

            Route 2 = + 594  +  - 882

            Route 2 = - 288 Kj Mol-1

Remember

            Route 1 = Route 2

            Route 1 = - 288 Kj Mol-1

Using the formula:

 

DH = S  DHfq (products)

-

S  DHfq (reactants)
DH = (2 x - 441)

-

(2 x - 297)
DH =

- 288 Kj Mol-1

   

 

Other enthalpy cycles:

 

Practical

Alternative approach using bond enthalpies: Explanation

 

Review:  Using the Formula:

DH = S  (Bond enthalpies of bonds broken)

-

S  (Bond enthalpies of bonds formed)

Worked example:

S  (Bond enthalpies of bonds broken)       S  (Bond enthalpies of bonds formed)
         

Reactants

Products

Bond

Number

Bond energy

Total

Bond

Number

Bond energy

Total

 C - H

 4

413 

 1652

 C = O

 2

805 

1610 

 O = O

 2

 497

 994

 O - H

 4

 463

1852

S  (Bond enthalpies of bonds broken) =

2646 

S  (Bond enthalpies of bonds formed) =

3462

         
DH = S  (Bond enthalpies of bonds broken)

-

S  (Bond enthalpies of bonds formed)
DH = 2646

-

3462
DH = - 816 Kj Mol-1    

Alternative approach:  Explanation

 

Questions 1-2  P197

 

 

Summary of enthalpy cycles

 

Step 1:-    Write a balanced chemical equation for the reaction.

 

Step 2:-    Construct the Enthalpy cycle.

 

Step 3:-    Decide on your routes and draw them on the cycle

 

Step 4:-    Write in the DHq  for each compound / element next to the arrows.

 

Step 5:-    Look up the values of each DHq  and write them in.  Add them up for each route.

 

Step 6:-    Write out Hesss law Route 1  =  Route 2

 

Step 7:-    Put in your numbers.

 

Step 8:-    Calculate DHq

 

Question 1-2    P201  /  1-7  P215  /  1,3-5  P217