topic 5.1 measuring energy changes total energy of the universe is a constant law of conservation of...
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Topic 5.1Measuring Energy Changes
• total energy of the universe is a constant• law of conservation of energy
– if a system loses energy, it must be gained by the surroundings, and vice versa
• when examining energy changes in a chemical reactions, we divide “the universe” into two parts
Temperature vs. Heat• temperature
– a measure of the average kinetic energy of the particles regardless of the amount
• heat (q)– one of many forms of energy– the transfer from a warmer body to a cooler body
as a result of a temperature gradient– measures the total energy in a given substance
(amount does matter)
• 50 ml water
• 100 C
100 ml water
100C
100ml of water contains twice the
heat of 50 ml.
• the amount of heat used or released (change) in a system
• a state function- meaning a quantity that only depends on the state at the moment– i.e. temp, pressure, volume– the value is independent of changes “along the
way”– only the beginning and end matter
• you cannot measure the actual enthalpy of a
substance directly
– but you can measure an enthalpy CHANGE
because of energy it takes in or releases
HEnthalpy
• use the symbol• a substance under these conditions is said
to be in its standard state–pressure: 100 kPa (1 atmosphere)–temperature: 298K (25°C)
vs
Enthalpy (Heat) of a Reaction• the amount of energy absorbed or given off in a
chemical reaction
• H = ∑ Hproducts − ∑ Hreactants
• 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
en
erg
y
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
en
erg
y 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)• overall, took energy from the surroundings
• Hreactants < Hproduct
H is positive• examples
– photosynthesis
Endothermic reactions
REACTION CO-ORDINATE
EN
TH
AL
PY
reactants
productsEnergy taken in and now stored in the bonds.∆H is positive
activation energyenergy necessary
to get the reaction going
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
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)
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
21
orkJ kg-1 K-1
if multiply by 1000
Heat energy change/transfer
q = m c DT
• q = change in heat (same as DH 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)
• DT = temperature change
Measuring the temperature change in a calorimetry experiment 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 1 Calculate the heat that would be required to heat an aluminum
cooking pan whose mass is 402.50 grams, from 20.5oC to 201.5oC. The specific heat of aluminum is 0.902 J g-1 oC-1.
q = mcDT
= (402.50 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 2 What 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: Dq (cold) = Dq (hot) so… mCDT = mCDT
Let Tf = final temperature of the water
(50.15 g) (4.184 J g-1 oC-1)(Tf - 20.5oC)
= (80.65 g) (4.184 J g-1 oC-1)(60.5oC - Tf)
(50.15 g)(Tf - 20.5oC) = (80.65 g)(60.5oC- Tf) 50.15Tf – 1030 = 4880 – 80.65Tf
130.80Tf = 5910
Tf = 45.2 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 in kJ mole -1 of hexane.
• Heat absorbed by the water…• q = mcDT• 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
– Need to 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