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Thermodynamics 1st law (Cons of Energy)

– Deals with changes in energy

Energy in chemical systems– Total energy of an isolated system is constantTotal energy of an isolated system is constant

Total energy = Potential energy + kinetic energy– Ep – mgh– Ek = ½ mv2

– Energy is transferred between a system and its surroundings by heat and/or work

– Energy is defined as the capacity to do work– Work = opposing force X distance moved

Measured in Joules1 J = 1kg x m2 /s

Thermodynamics of Chemical Reactions

Heat of Reaction– The quantity of energy released or absorbed as

heat during a reaction Every reaction is accompanied by transfer of energy as

heat

Thermochemical Equation– Equation that includes the heat of reaction

ex: CH4 (g) + 2O2 (g) CO2(g) + 2H2O (l) H = -890 kJ

Enthalpy - H– The quantity of energy released or absorbed as

heat during a reaction at a constant pressure ∆H

– Change in enthalpy

Enthalpy change ∆H = (sum of the Hf reactants) – (sum of the

Hf products)

When ∆H is >0 the reaction is endothermic

When ∆H is <0 the reaction is exothermic

Endothermic equations show heat as a reactant

Exothermic equations show heat as a product

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Identifying reactions 2H2O(g) 2H2(g) +O2(g) ∆H = +483.6

kJ/mol

2H2O( ) + 483 6 kJ/mol 2H2( ) 2H2O(g) + 483.6 kJ/mol 2H2(g)

+O2(g)

2H2(g) +O2(g) 2H2O(g) ∆H = -483.6 kJ/mol

2H2(g) +O2(g) 2H2O(g) + 483.6 kJ/mol

How reactions proceed This shows how energy changes as a reaction proceeds from

reactants to products

The activated complex is formed when the reactant particles have collided Thisparticles have collided. This is when the bonds between the reactant atoms are broken and the bonds between the product particles are formed.

The activation energy is the energy the reactants need to form the activated complex

How do the energy of the reactants and products compare?

Exothermic reactions

How does the energy of the reactants compare with the energy of the products?

Is energy released or absorbed?

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Writing Balanced Thermochemical Equations

The coefficients represent the numbers of moles of reactants and products, never the numbers of molecules.– This allows us to write these at fractions rather than whole

b hnumbers when necessary

This reaction releases 241.8 kJ of heat when one mole of water vapor is produced. The heat of formation of water vapor is -241.8 kJ. (Why is this negative?)

Since heat is released, where should it be placed in the equation?– H2 (g) + 1/2 O2 (g) → H2O (g) + 241.8 kJ

Standard Enthalpy Values

∆Ho is measured under standard conditionsP = 1 atmosphere

T = usually 25oC

with all species in standard states

Standard state of an element is zero under standard conditions (e.g., C = graphite and O2 = gas)

Hess’s Law The overall reaction enthalpy is the sum of

the reaction enthalpies of the steps into which the reaction can be divided

Enthalpy is dependant on state Enthalpy is dependant on state

– Therefore the value of ∆H is independent of the path between initial and final states

The reaction enthalpy thus, can be calculated from any sequence of reactions that can be balanced to add up to the reaction of interest

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Consider the oxidation of solid carbon graphite to

carbon dioxide C(gr) + O2 (g) CO2 (g)

– This reaction can be thought of as the outcome of 2 steps – first the oxidation of carbon to carbon monoxidecarbon monoxide

C(gr) + ½ O2 (g) CO(g) ∆H = -110.5 kJ– Then the oxidation of carbon monoxide to crbon

dioxide

CO(g) + ½ O2 (g) CO2 (g) ∆H = -283.0 kJ– This 2 step process is an example of a reaction

sequence

The net outcome of the sequence is the sum of the

steps

C(gr) + ½ O2 (g) CO(g) ∆H = -110.5 kJCO(g) + ½ O2 (g) CO2 (g) ∆H = -283.0 kJC(gr) + O2 (g) CO2 (g) ∆H = -393.5 kJ

Step by Step Select one of the reactants in the overall reaction and choose

a chemical reaction in which it also is a reactant

Select one of the products in the overall reaction and choose a chemical reaction in which it also is a product

Add these reactions and cancel species that appear on both sides of the equation

Cancel unwanted species in the sum obtained by adding an equation that has the same substance(s) on the opposite side of the arrow

Once the sequence is complete, combine the standard reaction enthalpies

– In each step we may need to reverse the equation or multiply it by a factor – we must do the same to the reaction enthalpy

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Consider the synthesis of propane (C3H8)

Use these three equations:– (a) C3H8(g) + 5O2 (g) 3CO2(g) + 4H2O(l) ∆H = -2220 kJ

– (b) C(gr) + O2 (g) CO2 (g) ∆H = -394 kJ

– (c) H2(g) + ½ O2 (g) H2O (l) ∆H = -286 kJ(c) H2(g) ½ O2 (g) H2O (l) ∆H 286 kJ

To treat carbon as a reactant there must be 3 carbons– 3C(gr) + 3O2 (g) 3CO2 (g) ∆H =(3) -394 kJ = -1182 kJ

Reverse (a) – 3CO2(g) + 4H2O(l) C3H8(g) + 5O2 (g) ∆H = +2220 kJ

Add– 3CO2(g) + 4H2O(l) + 3C(gr) + 3O2 (g) C3H8(g) + 5O2 (g) + 3CO2 (g)

∆H = (-1182 kJ) + 2200 kJ = 1038 kJ

Simplify by cancelling those on both sides 4H2O(l) + 3C(gr) C3H8(g) + 2O2 (g) ∆H = 1038 kJ

To cancel unwanted water and oxygen, multiply (c) by 4 4H2(g) + 2O2 (g) 4H2O (l) ∆H = 4 x (-286 kJ)= -1144kJ

Add4H O(l) 3C( ) 4H ( ) 2O ( ) C H ( ) 2O ( ) 4H O( ) 4H2O(l) + 3C(gr) + 4H2(g) + 2O2 (g) C3H8(g) + 2O2 (g) 4H2O(g)

∆H = 1038 kJ + (-1144kJ) = -106 kJ

Simplify

4H2(g) + 2O2 (g) C3H8(g) ∆H = -106 kJ

Hf – heat of formation– The energy released or absorbed as heat when

one mole of a compound is formed by the bi ti f it l t \i th t d dcombination of its elements o\in the standard

state

∆Hf– Change in heat of formation

∆Hreaction = ∆Hf products - ∆Hf reactants

Natural trend is toward decreased enthalpy (negative ∆H)

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Standard Molar Enthalpy of Formation, ΔHo

f

Calculate the heat given off when one mole of B5H9 reacts with excess oxygen according to the following reaction:

2B H (g) + 12O (g) →

Compound Hf (kJ/mol-K)

B5H9(g) 73.2

B O (g) 1272 772B5H9(g) + 12O2(g) → 5B2O3(g) + 9H2O(g)

Note: The heat of formation of any element under standard conditions is zero.

B2O3(g) -1272.77

O2(g) 0

H2O(g) -241.82

Calculating Enthalpy Change

Is this reaction endothermic or exothermic?

Enthalpy Calculation #2 Problem: Given that the standard heat of formation , ∆H°f ,

of H2O and CO are - 242 kJ/mol and - 111 kJ/mol ti l l l t th l h f threspectively calculate enthalpy change for the

reaction:

H2O(g) + C(graphite) → H2(g) + CO(g)

Why aren’t we given ∆H°f for C and H2?

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Entropy Entropy is a measure of the chaos or

disorder of a system

Entropy is represented by S Entropy is represented by S

The greater the entropy, the more disorder

Nature trends toward greater disorder – 2nd law of thermodynamics

Ssolid < Sliquid < Sgas

Entropy Standard entropy is entropy of a system at 1.0 atm

and 25 ºC ∆S = heat transferred/absolute temperature at

which transfer took place– Thus unit for entropy is J/K

∆S = Sf – Si

Positive entropy indicates increasing disorder∆S > 0

Negative entropy indicates decreasing disorder∆S < 0

Entropy example This is an example of thermal disorder

– As the system is heated, the supply of energy increases the the thermal motion and thus the disorder

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Entropy Example The decomposition of N2O4 into NO2 is also

accompanied by an increase in randomness.(N2O4 → 2NO2)

Whenever molecules break apart randomness Whenever molecules break apart, randomness increases

– This is an example of an increase in positional disorder

Increasing particle number

Solution processWhen NaCl dissolves in water, the crystal

breaks up, and the ions are surrounded by water molecules.

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Salt dissolves in water to form Na+ & Cl -

Less entropy More entropy

Is ∆S positive or negative for the system below?

Entropy of Chemical Reactions

Standard entropy of reaction:

∆S= ∑nS°(products) –∑nS°(reactants)

∑ means “the sum of” n = moles given by the coefficent

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Calculating Entropy Predict the entropy change, and

then calculate the standard entropy change for the following reactions at 25°C.– 2 CO(g) + O2(g) → 2 CO2(g)– 3 O2(g) → 2 O3(g)– 2 NaHCO3(s) → Na2CO3(s) + H2O(l) +

CO2(g)

Examples of standard molar entropies

Reaction Spontaneity A reaction that does occur under specific

conditions is called a spontaneous reaction. A reaction that does not occur under

specific conditions is called a nonspecific conditions is called a non-spontaneous reaction.

If a reaction is spontaneous in one direction, it will be non-spontaneous in the opposite direction

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Gibbs Free energy Gibbs free energy tells us whether a reaction is

spontaneous or not spontaneous. It combines the concepts of entropy and enthalpy. All reactions have by nature enthalpy change and

entropy change. All reactions occur at a certain temperature.p

A negative ∆H would be a spontaneous reaction A positive ∆S would be a spontaneous reaction G = ∆H – T∆S (temperature must be in Kelvin). If G < 0 The reaction is spontaneous.

– Any compound with a spontaneous formation reaction is thermodynamically stable

If G > 0 The reaction is not spontaneous.– Any compound with a non-spontaneous formation reaction

is thermodynamically unstable

Signs of ∆G, ∆S, and ∆H

Gibbs Free Energy Example

Iron metal can be produced by reducing iron(III) oxide with hydrogen:

Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g)∆H°= +98.8 kJ; ∆S°= +141.5 J/K;Note: You have J and kJ given above – you

must convertIs this reaction spontaneous at 25°C?

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Example Problem

Is the following reaction spontaneous under

∆H°f

(kJ/mol)S°(J/mol·K)

KClO3 -397.7 143.1under standard conditions?

4KClO3(s) → 3KClO4(s) + KCl(s)

KClO4 -432.8 151.0

KCl -436.7 82.6

Solution: Calculating ΔH for the reaction

Calculating ∆H°rxn

4KClO3(s) → 3KClO4(s) + KCl(s)∆H° [3∆H° (KClO ) ∆H° (KCl)]∆H°rxn = [3∆H°f (KClO4) + ∆H°f (KCl)] − [4∆H°f (KClO3)] = [3(−432.8kJ) + (−436.7kJ)] − [4(397.7kJ)] =

−144kJ Is this reaction endothermic or

exothermic?

Calculating ΔS° for the reaction

Calculating∆S°rxn

∆S°rxn= [3S°(KClO4)] + [S°(KCl)] − [4S°(KClO3)] = [3(151.0JK)+(82.6JK)] − [ ( 3)] [ ( ) ( )][4(143.1JK)] = −36.8JK

Is entropy increasing or decreasing?

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Calculating ΔGCalculating ∆G°rxn

∆G°rxn = ∆Hrxn - T∆S°-144kJ – [298K (-38.6 J/K)(1kJ/1000J)]

= 133 kJ= -133 kJ

∆G°rxn< 0; therefore, reaction is spontaneous.

A negative ∆G°rxn is an exergonic reaction.

–A positive ∆G°rxn is an endergonic reaction.