Enthalpy, Calorimetry, Hess's Law

ObjectivesTo know how to calculate enthalpiesTo know how to solve problems on

calorimetryTo know how to solve problems on Hess’ Law

How to Calculate Enthalpies

How?

Let’s redefine some terms• Enthalpy (H) is the sum of the internal energy of the

system plus the product of the pressure of the gas in the system and its volume

Hsys = Esys + PV

If pressure is kept constant, we can arrive at:

H sys = q (at constant pressure)

Where: H = H final – H initialq --- is heat

Let’s redefine some terms

• Enthalpy of Reaction (H) is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants

H = nH products – mH reactants

where n and m are the coefficients of the products and the reactants in the balanced equation

Remember

• Enthalpies are usually computed at standard conditions (25oC = 298K)

• Note that enthalpies of formation of elements at standard conditions are equal to zero

Let’s Calculate• Example 1: Calculate the enthalpy of the oxidation

reaction of benzene (C6H6) given with the chemical equation: C6H6(l) + 4.5O2(g) = 6CO2(g) + 3H20(l)

1. Find the enthalpies of formation for all chemical components of the reaction using a Table of Heat Formation

H C6H6(g) = 48.85 KJ/molH O2(g) = 0H CO2(g) = -393.509 KJ/molH H2O (l) = -285.83 KJ/mol

C6H6(l) + 4.5O2(g) = 6CO2(g) + 3H20(l)

2. Multiply each enthalpy value on the corresponding reaction coefficient and sum up the enthalpies of formation

C6H6(l) + 4.5O2(g) H C6H6(g) = 48.85 KJ/molH O2(g) = 0H CO2(g) = -393.509 KJ/molH H2O (l) = -285.83 KJ/mol

H initial = 48.95 KJ/mol+ 4.5 x O

H initial = 48.95 KJ/mol

3. Do the same to the final reagents

6CO2(g) + 3H20(l)H final = 6 x (-393.509 KJ/mol) + 3 x (-285.83 KJ/mol)

H final = -3218.544 KJ/mol

H initial = 48.95 KJ/mol

4. Subtract the enthalpy of formation of the initial reagents from the final reagentH reaction = H final – H initial

H reaction = -3218.544 KJ/mol – 48.95 KJ/mol

H reaction = -3267.494 KJ/mol

H final = -3218.544 KJ/mol

Some Important Enthalpy Changes1. Enthalpy Change of Combustion

Eg. C (graphite) + ½ O2 (g) CO2 (g) C (graphite) + O2 (g) CO2 (g)

- the enthalpy change which occurs when one mole of the substance is completely burnt in oxygen under standard conditions

chemist’s shorthand:Hc,m [(graphite)] = -393.5 kJ/mol

Some Important Enthalpy Changes

2. Enthalpy Change of Formation- the enthalpy change when one mole of the compound is formed from its elements under standard conditions

- may also be called Heats of Formation.

Eg. The SECF of methane, CH4, refers to the change:

C (graphite) + 2 H2 (g) ---->CH4 (g) H = -74.8 kJ/mol

Hf,m [CH4(g)] = -74.8 kJ/mol

3. Enthalpy Change of Atomisation

The SMECA of an element is the enthalpy change when one mole of its atoms in the gaseous state is formed from the element under standard conditions .

* Atomisation is always endothermic.

Eg. C (graphite C (g) H = 716.7 kJ/mol

Hat,m [(graphite)] = 716.7 kJ/mol

Some Important Enthalpy Changes

4. Enthalpy Change of Fusion- The enthalpy change when 1 mole of solid is converted to one mole of liquid at its melting point at standard pressure

Hfus,m [(H20)] = 716.7 kJ/mol H = 6.01 kJ/mol

Some Important Enthalpy Changes

5. Enthalpy Change of Vaporisation- The enthalpy change when 1 mole of liquid is converted to one mole of gas at its boiling point at standard pressureHvap,m [(H20)] = 716.7 kJ/mol H = 41.09 kJ/mol

Calorimetry

Two types of calorimetry• 1. measurements based on constant pressure• 2. measurement based on constant volume

Calor (Latin) + metry (Greek) = Calorimetry

“heat” + “to measure” =Science of measuring

the amount of heat

What is it?

Other termsCalorimeter – the device used to measure heat of

reactionHeat capacity – the amount of heat required to raise

its temperature by a given amount– SI unit: J/KFormula: q= CT

where: q - heatC - heat capacityT - change in temperature

= Tf-Ti

Other terms

• Specific heat capacity – gives the specific heat capacity per unit mass of a particular substance- SI unit: J/kgKFormula: q= mcT

where: q - heatm - massc - specific heat capacityT - change in temperature

= Tf-Ti

Other terms

• Molar enthalpy of a substanceFormula: H= mcT

nwhere: H - enthalpy change

m - massc - specific heat capacityT - change in temperature

= Tf-Tin - moles of substance

Hess’s Law of Heat Summation

By Germain Henri Hess

What is it?• Hess Law of Heat Summation states that the heat

absorbed or released during a reaction is the same whether the reaction occurs in one or several steps

• Rules1. Make sure to rearrange the given equations so that the

reactants and products are on the appropriate sides of the arrows

2. If you reverse equations, you must also reverse the sign of H3. If you multiply/divide equations to obtain a correct coefficient,

you must also multiply/divide the H by this coefficient

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some brain exercises