Thermochemistry [Thermochemical Equations, Enthalpy Change and Standard Enthalpy of Formation]

Enthalpy and Calorimetry

Reactions Based on Energy Profiles

1. Endothermic Reaction2. Exothermic Reaction

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Prepared: Kemikal Drills by Rea Abuan 2014

Endothermic Reaction

1. Reaction which absorbs or requires an amount of energy to proceed.

2. The total energy of the products are higher than the total energy of the reactants.

Exothermic Reaction

1. Reaction which releases an amount of energy.

2. The total energy of the products are lower than the total energy of the reactants.

Enthalpy

• Less familiar property of a system• Defined as

H = enthalpy, P = pressuresys

E = internal energy, V = volumesyswww.reaabuan.com/

blog/Prepared: Kemikal Drills by Rea Abuan 2014

Enthalpy is an example of a state

function

www.reaabuan.com/blog/Prepared: Kemikal

Drills by Rea Abuan 2014

So let’s define what is a state function first

State Functions

• These are properties of the system that are dependent ONLY on the state of the system. • They are not dependent on the

WAY the system came to be in that given state (pathway)

State Functions

• Expressed as capital letterse.g. Enthalpy, internal energy, Gibbs Free Energy, entropy, temperature, pressure, volume

State Functions

Enthalpy ΔH, Internal energy ΔE,

Gibbs Free Energy ΔG, Entropy ΔS,

State Functions

Temperature (Tf – Ti),

Pressure (Pf – Pi),

Volume (Vf –Vi)

Think of it this way

You and your friend agreed to meet at a coffee shop near your school after an hour.

You’re both coming from your house.

You went directly into the coffee shop after an hour. Your friend

took a detour to her house to pick up something before going to the

coffee shop.

Your final destination is the same. Your initial position is the

same as well. Think of state functions that way. Their values are independent of the pathway.

So going back to enthalpy

Enthalpy change, ΔH

• is defined as the quantity of heat transferred to or away from the system during a chemical reaction (physical change)

NOTE: system must have constant pressure and mole.

enthalpy of the reaction

Type of Reaction Enthalpy Change

EndothermicΔH > 0

(positive)

ExothermicΔH < 0

(negative)

Take note that these changes occur when P =

constant.

Now, remember that we cannot measure heat

directly.

Experimentally, heat associated with

chemical reactions are measured using

calorimetry

So let’s discuss calorimetry.

Calorimetry

is the science of measuring the change in the temperature of the system when it absorbs or

releases energy as heat.

Ok, one more time.

Calorimetry

1. Measures the temperature change of a body (note: you should know its specific heat)

2. The temperature change happens due to either absorption or release of energy from the body.

The calorimeter!

Before we continue, take

note of the requirement for

calorimetry

Remember? There should be a

contant pressure.

Constant-Pressure Calorimetry

• used for measuring qp (heat at constant temperature) for solution reactions

Constant-Pressure Calorimetry

• used for measuring ΔHrxn as well, at constant pressure.

NOTE: ΔHrxn ≠ qp

You want to use a cat in chemical

calculation?

Calorimetry Equations Based on

1. Heat transfer involving pure substances (pure water only)

2. Heat transfer involving mixed system (the whole calorimeter)

For Pure Substances

where q = heat absorbed /releasedm=mass (g)c = specific heat capacity (J/g⁰C)ΔT = change in temperature

For Mixed System

where q = heat absorbed /releasedC = heat capacity (J/⁰C)ΔT = change in temperature

Let’s check an example

A 3.0 g copper was heated from 20°C to 80°C. How much energy was used to heat Cu? (Specific heat capacity of Cu is 24

J/mol °C)

Given Values:

Mass = 3.0 gramsTinitial =20⁰C

Tfinal = 80 ⁰C

c = 24 J / mol ⁰C

Solution:

1. Find the value of ΔTΔT = Tfinal – Tinitial

ΔT = 80 ⁰C – 20⁰C ΔT = 60 ⁰C

Solution:

2. Solve for qq = mc ΔT q = (3.0 g) (24 J /mol ⁰C) (20⁰C)q =1440 J or 1.44 kJ

Now, we can calculate ΔH rxn

using qp

Ahm, where’s the cat? meow

Just remember that in

calorimetry, no heat flows out of the calorimeter

The entire system is adiabatic.qsystem = 0

Remember this qcal + qH2O = qrxn

Writing down each equation

Oh crap, there are 4 equations!

Nosebleed

Chill down a bit. We’ll go through examples for you.

When 5.35 kJ of heat is added to a calorimeter containing 23.00 g of water the temperature rises from 14.00oC to 29.55oC. Calculate the heat capacity of the calorimeter in J/oC. The specific heat of water is

4.184 J/g oC.

Given Values are..

qrxn = 5.35 kJmass water = 23.0 gTi = 14.00 ⁰ CTf = 29.55 ⁰ Cc water = 4.184 J/g oC

Solve for the change in temperature

ΔT = Tfinal – Tinitial

ΔT = 29.55 ⁰C – 14.00 ⁰C ΔT = 15.55 ⁰C

Find qwater

qwater = mc ΔT

qwater = (23.0 g) (4.184 J/g oC) (15.55⁰C)

qwater =1496 J or 1.496 kJ

Recall this qcal + qH2O = qrxn

1.496 kJ 5.35 kJ

Solving for qcal

qcal = qrxn - qH2O

qcal = 5.35 kJ - 1.496 kJ

qcal = 3.854 kJ

Recall again

where q = heat absorbed /releasedC = heat capacity (J/⁰C)ΔT = change in temperature

Calculating the heat capacity

Rearranging,

C = 3.854 kJ / 15.55 ⁰C C = 0.2478 kJ / ⁰C

Let’s end here. Check out the exercises on my blog.

Also, check the tutorial videos of problem solving on my blog.

Until the next post!

Thermochemistry [Thermochemical Equations, Enthalpy Change and Standard Enthalpy of Formation]

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