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