Thermochemistry

Mrs. StoopsChemistry

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Thermochemistry or Thermodynamics

• reactions involving heat.• Abbreviations to know –

– rxn = reaction– E = Energy = delta = change

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Kinetic Energy vs Potential Energy

• Potential – stored energy• Kinetic – movement energy

PE

KE

Units of Energy – Joule or J also KJcalories 1 cal = 4.184 J

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System vs Surroundings

We will focus on the system – the system changes depending on what we are looking at. Could be a beaker, or the room, or the Earth, or the solar system

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Work & Heat

• Do you work?• Work = force x distance moved

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Heat

• Heat is energy transferred. Heat is transferred from HOT to COLD.

• Energy is the capacity to do work or transfer heat.• NEW symbol = delta – means change T means change in temperature = Tfinal - Tinitial

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1st Law of Thermo

1st law of Thermodynamics = Energy is CONSERVED.

E = Ef - Ei

E is positive means Ef Ei. Also Energy has moved from surroundings into the system

E is negative means Ef Ei. Energy has moved from system out to surroundings.

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System vs Surroundings

Another way to solve for EE = q + w = heat + work. Sign direction issues!

System

q +Heat entering

q –Heat leaving

W –Do work

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Example problem:

Calculate the change in energy on the system when 140 J of heat are absorbed and 85 J of work are done.

E = (+140J) + (- 85J) = 55 J

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Exothermic & Endothermic

Cue words are important!!!Absorbs heat = goes INTO the system

Endothermic = q +Releases heat = EXITS the system

Exothermic = q –When something feels cold – the heat is being taken

away and used, leaving the object “feeling” cold.

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State Functions

• State functions – depend on the current conditions only, not the path taken.

E, average temperature

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Homework Day 1

Page 203: 25, 29

Enthalpy - H• Heat gained or lost **yeah, and q is too!• State function H = H products – H reactants

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Enthalpy - H

• Sign indicates the direction of heat transfer only!!!

• +H = gained heat (into system), Endothermic, REACTANT

• -H = lost heat (out of system), Exothermic, PRODUCT

Hrxn – heat of reaction - we will calculate this later, for now it is given.

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Thermochemical equation

• a reaction shown with the H valueEx:

• 2H2 + O2 → 2H2O H = -483.6 KJ

What this means:• Since H is negative, the enthalpy is a

_____________ (which side of rxn?)

• 2H2 + O2 → 2H2O + 483.6 KJ

*** the sign tells the side of reaction ONLY!

product

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Rules of using Enthalpy:

1. Enthalpy is an extensive property. The coefficients matter!

Ex: CH4 + 2 O2 → CO2 + 2 H2O H = -890 KJ

• This is for 1 mole CH4 = based on the stoichiometry

For 4 moles CH4 the H becomes -890 x 4 = -3560 KJ

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Rules of Enthalpy

2. The enthalpy sign changes when the reaction is reversed

CH4 + 2 O2 → CO2 + 2 H2O H = -890 KJ

But if we switch products for reactants

CO2 + 2 H2O → CH4 + 2 O2 H = +890 KJ

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Rules of Enthalpy

3. Enthalpy depends on the states of the reactants and the products.

The states have been listed in the reactions – you need to pay attention to.

(s) solid, (l) liquid, (g) gas, & (aq) aqueous – dissolved in water

H2 (g) + O2 (g) → H2O (g)

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Using these ideas for stoichiometry . . .

If you have a Thermochemical equation, you can be asked to solve a problem to find the heat transferred.

Ex: CH4 + 2 O2 → CO2 + 2 H2O H = -890 KJ

If you have 50 KJ of heat, how many grams of oxygen can be reacted?

50KJ x 2 mol O2 x 32 g O2 = 3.6 g O2

890 KJ 1 mol O2

*** would the answer be negative? What does the sign tell us?

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Calorimetry

• Measures heat flow by measuring the temperature changes

• Heat capacity = amount of heat needed to raise the temperature by 1 (degree) Kelvin

• Molar heat capacity = amount of heat needed to raise the temperature by 1 degree for 1 mole

• Specific heat = heat capacity for 1 gram

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To find specific heat or use specific heat

Heat transferred = mass (in grams) x specific heat x change in temperature

Equation form:• q = m c t• Values for c are on page 181 of text, given, or

you are asked to solve for them

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Homework Day 2

page 204: 37a, 41, 43, 51, 53

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Hess’s Law

If the reaction is carried out in a series of steps, the H for the overall reaction will be equal to the sum of the H for all the steps.

Generic example:Overall reaction A D H = ?

Steps:1 A B H = x2 B C H = y3 C D H = z

Overall A D H = x + y + z

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Example:Find the H for the reaction: 2 C (s) + O2 (g) 2 CO (g)

Steps:

1 C (s) + O2 (g) CO2 (g) H = -393.5 KJ

2 CO (g) + ½ O2 (g) CO2 (g) H = - 283.0 KJ

Work:

2 C (s) + O2 (g) 2 CO (g)

2C (s) + 2O2 (g) 2CO2 (g) H x 2 = -393.5 x 2 = -787 KJ

2CO2 (g) 2CO(g) + O2 (g) H x 2, R = +283.0 x 2 = 566 KJ

**Cancel**2 C(s) + O2 (g) 2 CO(g) -221 KJ

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Enthalpy of Formation

Hf

Depend on the state (s, l, g, aq)Standard enthalpy = HºStandard enthalpy of formation = Hfº = are listed

for 298 KUnits are KJ/molListed in the text book = appendix C (page 1019)***Element values are zero***

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Enthalpy of Formation

Hrxn = n Hfºproducts - m Hfºreactants

What are these? - sigma, Greek letter, means sum ofn & m – coefficients from balanced equation

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ExampleFind the Hrxn for C2H5OH(l) + 3 O2 (g) 2 CO2(g) + 3 H2O

(l)

Look up values:

C2H5OH(l) -277.7 KJ

O2 (g) 0 KJ

CO2(g) -393.5 KJ

H2O (l) -285.83 KJ

Plug into the formula:

[(2*-393.5 KJ) + (3*-285.83 KJ)] – [(1*-277.7 KJ) + (3*0)] =

H = -1366.79 KJ PRODUCTS - REACTANTS

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Homework Day 3

P 205: 52, 61, 63, 69, 71b d,