Topic 5.1
Measuring 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 ml
water
• 100 C
100C
100ml of water
contains twice the
heat of 50 ml.
Enthalpy
H
• 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
vs
• 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)
Enthalpy (Heat) of a Reaction
• the amount of energy absorbed or given off
in a chemical reaction
• H = ∑  Hproducts − ∑  Hreactants
Exothermic reactions
• 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)
ENTHALPY
activation energy
energy necessary
to get the
reaction going
energy given
out, ∆H is
negative
reactants
products
REACTION CO-ORDINATE
H2 + Cl2  2HCl
Energy taken in
to break bonds.
H, H, Cl, Cl
(Atoms)
Energy given out when
bonds are made.
energy
H-H, Cl-Cl
Reactants
Overall energy
change, H
H-Cl, H-Cl
Products
H2 + Cl2  2HCl
Energy in
= +678kJ
H, H, Cl, Cl
(Atoms)
Energy out = -862kJ
energy
H-H, Cl-Cl
Reactants
Overall energy
change,
H = -184kJ
H-Cl, H-Cl
Products
Endothermic reactions
• 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
activation energy
energy necessary
to get the
reaction going
ENTHALPY
products
reactants
REACTION CO-ORDINATE
Energy taken in
and now stored
in the bonds.
∆H is positive
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
Calculation of enthalpy change
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)
Specific Heat (c) values for Some
Common Substances
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
or
kJ kg-1 K-1
if multiply
by 1000
21
Heat energy change/transfer
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
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 = mcT
only 4 sig.
figs.
= (402.50 g) (0.902 J g-1 oC-1)(181.0oC)
= 65,712.955 J
= 65,710 J with correct 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: q (cold) = q (hot) so… mCT = mCT
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 3
On 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 = mcT
• 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
hexane is C6H14
– Need to find heat energy / mole
= 10.5 kJ/ 0.0021 mol
= 5000 kJ mol -1 or 5.0 x 103 kJ mol -1