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Calculating Heat Changes
First Law of Thermodynamics:
the internal energy of an isolated system is constant
Signs (+/-) will tell you if energy is entering or leaving a system
+ indicates energy enters a system
- indicates energy leaves a system
•A change in internal energy can be identified with the heat supplied at constant volume
ENTHALPY (H)
(comes from Greek for “heat inside”)
•
The heat supplied is equal to the change in another thermodynamic property called enthalpy (H)
i.e. H = q
• this relation is only valid at constant pressure
As most reactions in chemistry take place at constant pressure we can say that:
A change in enthalpy = heat supplied
Heat Transfer
Specific Heat (Cp) amount of energy
required to raise the temp. of 1 kg of material by 1 degree Kelvin
units: J/(kg·K)or J/(kg·°C)
Heat Transfer Which sample will take
longer to heat to 100°C?
50 g Al 50 g Cu
• Al - It has a higher specific heat.• Al will also take longer to cool down.
Heat Transfer
Q = m T Cp
Q: heat (J)m: mass (kg)T: change in temperature (K or °C)Cp: specific heat (J/kg·K)
T = Tf - Ti
– Q = heat loss+ Q = heat gain
Heat Transfer A 32-g silver spoon cools from 60°C to 20°C. A 32-g silver spoon cools from 60°C to 20°C.
How much heat is lost by the spoon?How much heat is lost by the spoon?
GIVEN:
m = 32 g
Ti = 60°C
Tf = 20°C
Q = ?
Cp = 235 J/kg·K
WORK:
Q = m·T·Cp
m = 32 g = 0.032 kg
T = 20°C - 60°C = – 40°C
Q = (0.032kg)(-40°C)(235J/kg·K)Q = – 301 J
Thermochemical Equations
A chemical equation that includes the heat change
Heat of reaction- heat change for the equation exactly as written, represented by ΔH, when pressure is constant.
Hess’s law Hess’s Law states that the heat of a whole reaction
is equivalent to the sum of it’s steps. For example: C + O2 CO2 (pg. 165)The book tells us that this can occur as 2 steps
C + ½O2 CO H = – 110.5 kJCO + ½O2 CO2 H = – 283.0 kJ
C + CO + O2 CO + CO2 H = – 393.5 kJ
I.e. C + O2 CO2 H = – 393.5 kJ Hess’s law allows us to add equations. We add all reactants, products, & H values. We can also show how these steps add together via
an “enthalpy diagram” …
Steps in drawing enthalpy diagrams1. Balance the equation(s).2. Sketch a rough draft based on H values.3. Draw the overall chemical reaction as an enthalpy
diagram (with the reactants on one line, and the products on the other line).
4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line.
5. Check arrows: Start: two leading awayFinish: two pointing to finishIntermediate: one to, one away
6. Look at equations to help complete balancing (all levels must have the same # of all atoms).
7. Add axes and H values.
C + O2 CO2 H = – 393.5 kJ
Reactants
Intermediate
Products
C + O2
CO2
CO
Ent
halp
y
Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.
H = – 110.5 kJ
H = – 283.0 kJ
H = – 393.5 kJ
+ ½ O2
C + ½ O2 CO H = – 110.5 kJCO + ½ O2 CO2 H = – 283.0 kJ
Bond StrengthsBond Strengths
Bond strengths measured by bond enthalpy HB (+ve values)
• bond breaking requires energy (+ve H)
• bond making releases energy (-ve H)
Lattice EnthalpyLattice Enthalpy
A measure of the attraction between ions (the enthalpy change when a solid is broken up into a gas of its ions)
• all lattice enthalpies are positive
• I.e. energy is required o break up solids
Another Example of Hess’s Law Another Example of Hess’s Law
Given:
C(s) + ½ O2(g) CO(g) H = -110.5 kJ
CO2(g) CO(g) + ½ O2(g) H = 283.0 kJ
Calculate H for: C(s) + O2(g) CO2(g)
• If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Ho
f .
• Standard conditions (standard state): Most stable form of the substance at 1 atm and 25 oC (298 K).
• Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.
• Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.
Enthalpies of FormationEnthalpies of Formation
• If there is more than one state for a substance under standard conditions, the more stable one is used.
• Standard enthalpy of formation of the most stable form of an element is zero.
Enthalpies of FormationEnthalpies of Formation
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
• For a reaction
• Note: n & m are stoichiometric coefficients.• Calculate heat of reaction for the combustion of propane
gas giving carbon dioxide and water.
Enthalpies of FormationEnthalpies of Formation
reactantsproductsrxn ff HmHnH
Foods• 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.• Energy in our bodies comes from carbohydrates and fats
(mostly).
• Intestines: carbohydrates converted into glucose:C6H12O6 + 6O2 6CO2 + 6H2O, H = -2816 kJ
• Fats break down as follows:2C57H110O6 + 163O2 114CO2 + 110H2O, H = -75,520 kJ
Fats contain more energy; are not water soluble, so are good for energy storage.
Foods and FuelsFoods and Fuels