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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?
3
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.