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

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)• 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 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 402.5 grams, from 20.5oC to 201.5oC. The specific heat of aluminum is 0.902 J g-1 oC-1.

q = mcDT

= (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 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 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 = 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

• 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