Energetics - · PDF fileOn complete combustion, 0.18g of hexane raised the temperature of...
Transcript of Energetics - · PDF fileOn complete combustion, 0.18g of hexane raised the temperature of...
EnergeticsTopic 5.1 – 5.2
Topic 5.1Exothermic and Endothermic
Reactions
? ?
• total energy of the universe is a constant
• if a system loses energy, it must be gained by the surroundings, and vice versa
• a measure of the heat content (not temp)
• you cannot measure the actual enthalpy of a substance
– but you can measure an enthalpy CHANGE because of energy it takes in or releases
• = Greek letter ‘delta’ meaning change
• H = heat.
• so, H means ‘heat change’.
HEnthalpy
• Why a standard?
– enthalpy values vary according to the conditions
• a substance under these conditions is said to
be in its standard state
– pressure: 100 kPa (1 atmosphere)
– temperature: usually 298K (25°C)
• if these were not standardized, then energy
could be leaving or entering the system
• modify the symbol from
Enthalpy Change Standard Enthalpy Change
(at 298K)
Enthalpy (Heat) of Reaction
• H = Hproducts − Hreactants
• lower energy is more stable
• heat energy is given out by the reaction hence the surroundings increase in temperature (feels hot)
• occurs when bonds are formed– new products are more stable and extra energy is
given off
• Hproducts < Hreactants
– H is negative
• examples– combustion of fuels
– respiration
– neutralization reactions (acid reacts with something)
Exothermic reactions
REACTION CO-ORDINATE
EN
TH
AL
PY
energy given out, ∆H is negativereactants
products
activation energyenergy necessary
to get the reaction going
H2 + Cl2 2HCl
energy
H-H, Cl-Cl
Reactants
H, H, Cl, Cl (Atoms)
H-Cl, H-ClProducts
Energy taken in to break bonds.
Energy given out when bonds are made.
Overall energy change, H
H2 + Cl2 2HCl
energy H-H, Cl-Cl
Reactants
H, H, Cl, Cl (Atoms)
H-Cl, H-ClProducts
Energy out = -862kJ
Overall energy change,H = -184kJ
Energy in= +678kJ
• heat energy is taken in by the reaction mixture hence the surroundings decrease in temperature (feels cold)
• occurs when bonds are broken– the reactants were more stable (bonds are
stronger)• therefore, took energy from the surrounding to break
bonds
• Hreactants < Hproduct
H is positive
• examples
Endothermic reactions
REACTION CO-ORDINATE
EN
TH
AL
PY
reactants
productsenergy taken
in, ∆H is positive
activation energyenergy necessary
to get the reaction going
14
Summary Table
Exothermic reactions
Endothermic reactions
Energy is given out to the surroundings
Energy is taken in from the surroundings
∆H is negative ∆H is positive
Products have less energy than reactants
Products have more energy than reactants
Topic 5.2Calculation of enthalpy change
Notice less sources of error here compared
to our lab…
• calorimetry– measurement of heat flow
• calorimeter– device used to measure heat flow
• heat– energy that is transferred from one object to
another due to a difference in temperature– measures total energy in a given substance
• temperature– a measure of the average kinetic energy of a
substance regardless how much is there
Temperature vs. Heat
• 50 ml water
• 100 C
100 ml water
100C
100ml of water contains twice the
heat of 50 ml.
Heat Capacity/Specific Heat• the amount of energy a substance absorbs depends on:
– mass of material– temperature– kind of material and its ability to absorb or retain heat.
• heat capacity– the amount of heat required to raise the temperature of
a substance 1 oC (or 1 Kelvin)• molar heat capacity
– the amount of heat required to raise the temperature of one mole 1 oC (or 1 Kelvin)
• specific heat – the amount of heat required to raise the temperature of
1 gram of a substance 1 oC (or 1 Kelvin)
19
Substance J g-1 K-1
Water (liquid) 4.184
Water (steam) 2.080
Water (ice) 2.050
Copper 0.385
Aluminum 0.897
Ethanol 2.44
Lead 0.127
Specific Heat (c) values for Some
Common Substances
20
orkJ kg-1 K-1
if multiply by 1000
Heat energy change
q = m c T
• q = change in heat (same as H if pressure held constant)
• m = mass in grams or kilograms
• c = specific heat in J g-1 K-1 or kJ kg-1 K-1
(or Celsius which has same increments as Kelvin)
• T = temperature change (final –initial)
Measuring the temperature change in a calorimetryexperiment can be difficult since the system is losing heat to the surroundings even as it is generating heat.
By plotting a graph of time vs. temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings.
Heat Transfer Problem 1Calculate the heat that would be required to heat an aluminum cooking pan whose mass is 402.5 grams, from 20.5oC to 201.5oC. The specific heat of aluminum is 0.902 J g-1 oC-1.
q = mcT
= (402.5 g) (0.902 J g-1 oC-1)(181.0oC)
= 65,712.955 J
= 65,710 J with correct sig. figs.
only 4 sig. figs.
Heat Transfer Problem 2What is the final temperature when 50.15 grams of water at 20.5oC is added to 80.65 grams water at 60.5oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 oC-1
Solution: q (Cold) = q (hot) so… mCT = mCT
Let T = final temperature
(50.15 g) (4.184 J g-1 oC-1)(T- 20.5oC)
= (80.65 g) (4.184 J g-1 oC-1)(60.5oC- T)
(50.15 g)(T- 20.5oC) = (80.65 g)(60.5oC- T)
50.15T -1030 = 4880 – 80.65T
130.80T = 5830
T = 44.6 oC
Heat Transfer Problem 3On complete combustion, 0.18g of hexane raised the
temperature of 100.5g water from 22.5°C to 47.5°C. Calculate its enthalpy of combustion.
• Heat absorbed by the water…
• q = mcT
• q = 100.5 (4.18) (25.0) = 10,500 J which is same as 10.5 kJ
• Moles of hexane burned = mass / molar mass
= 0.18 g / 86 g/mol
= 0.0021 moles of hexane
• Enthalpy change means find heat energy / mole
= 10.5 kJ/ 0.0021 mol
= 5000 kJ mol -1 or 5.0 x 103 kJ mol -1
hexane is C6H14