ΔHEvery chemical reaction and change of physical

state releases or absorbs heat.Goal – to determine whether heat is absorbed or released during a chemical reaction, therefore

determine if exothermic or endothermic.

Thermochemistry

• Study of heat changes that accompany chemical reactions and phase changes

• Thermochemical equations are used to represent these reactions

• 4Fe(s) + 3 O2(g) → 2Fe2O3(s) + 1625kJ• 27kJ + NH4NO3(s) → NH4

+(aq) + NO3-(aq)

• First equation – exothermic- heat pack Second equation – endothermic – cold pack

Thermochemistry Terms

• System – specific part of the universe that contains reaction or process you want to study

• Surroundings – everything in universe other than system

• Universe = system + surroundings

Enthalpy

• Impossible to know the total heat content of a system because it depends upon many factors

• Chemists are interested in changes in energy during reactions

• For many reactions, energy lost or gained can be measured by calorimeter at constant pressure

• Enthalpy(H) – heat content of a system at constant pressure

• We measure change in H (heat content)

H0

H0

• Exothermic reaction a downhill change

• Endothermic reaction an uphill energy change

ΔH = H2 – H1

Products – reactants

Exothermic Reactions

reactants

productsEnt

halp

y (k

J)

• Is ΔH positive or negative?

• Why?

• Products - Reactants!

Endothermic ReactionsE

ntha

lpy

(kJ)

reactants

products

• ΔH is a positive #

Why is there a change in Enthalpy?

due to• energy required to break the

bonds in the reactants and • energy produced by forming the

chemical bonds in the products.

• A balanced equation can represent the energy absorbed or released

• Energy change for the reaction is called the • Enthalpy of Reaction

• and is represented by ΔHr

C(cr) + O2 → CO2 + energy (393.5kJ)∆Hr = -393.5kJ

Standard States

H = change in enthalpy• “ ° ” = Standard state enthalpy at

298.15K (25° C degrees) and 101.325 kilopascals (pressure)

• (No longer STP!)

• “f” = formation

ΔHºf

Elements in the Natural State

ELEMENTS ONLY

ΔHºf = 0

ALL OTHER COMPOUNDS

• Enthalpy of formation• found in standard tables• Table C-13 pg921

ΔHºf

Enthalpy of Formation

• Δ Hf represents the production of one mole from ...

• its free elements in their standard states. (units = kJ/mole)

Calculation of Enthalpy of Reaction

• ΔHr = ΣΔHºf (products) - ΣΔHºf

(reactants) • Σ (Sigma) is the symbol used to indicate

summation.• It means to ADD all the values of ΔH for all

the products and subtract ...• the sum of all the Δ H of the reactants

antsreactHproductsHH ffr tan

EXAMPLE 1

• Calculate the enthalpy change in the following chemical reaction

carbon monoxide + oxygen → carbon dioxide

• 1) Write a balanced equation

• 2CO (g) + O2 → 2CO2

2CO (g) + O2 → 2CO2

• 2) Calculate Δ Hf products

• 2 moles CO2 x (-393.5 kJ) = -787.0 kJ

• 3) Calculate Δ Hf reactants

• 2 moles CO + 1 mole O2 = 2(-110.5 kJ) + 1(0 kJ)

• =-221 kJ

•ΔHr =ΔHf (products) - ΔHf (reactants)

• Δ Hr = (-787.0 kJ) - (-221. kJ) = -566.0 kJ

• This means the 566.0 kJ are released in the reaction

tsreacproducts HHH tan tsreacproductsr HHH tan

Practice

• Compute ΔHºr for the following reactions:

2NO(g) + O2(g) → 2NO2(g)

-114.14kJ

__FeO(cr)+ O2(g) → __ Fe2O3(cr)

• -560.4kJ

4 2

• Solving for change in enthalpy we getΔHr = ΣΔHºf (products) - ΣΔHºf (reactants)

• Knowing the enthalpy of formation of each reactant and product, we can calculate the amount of energy produced or absorbed and predict whether a reaction will be exothermic or endothermic.

Exothermic Reactions

reactants

products

Ent

halp

y (k

J)

115 kJ

75 kJ

• Δ Hr= 75 kJ -115 kJ

• = - 40 kJ

• Negative Δ Hr for EXOTHERMIC

• Δ H = 247 kJ - 122 kJ• = 125 kJ

• Positive Δ Hr for ENDOTHERMIC

Endothermic ReactionsE

ntha

lpy

(kJ)

reactants

products 247 kJ

122 kJ

Spontaneous or Nonspontaneous That is the Question ?

• Suppose Δ H is negative Δ Hf products < Δ Hf reactants

• It’s EXOTHERMIC

• When Δ Hr is negative reaction tends to be

• SPONTANEOUS • Spontaneous means that it will occur without any

outside influence

• How about Δ H positive?

Hess’s Law

• the enthalpy change for a reaction is the sum of the enthalpy changes for a series of reactions that add up to the overall reaction.

Hess’s Law

N2 + 2O2 ↔ 2NO2 41kJ

N2O4 ↔ 2NO2 35kJ

N2 + 2O2 ↔ N2O4

Reverse the second reaction to get:

Hess’s Law

N2 + 2O2 ↔ 2NO2 ΔH= 41kJ

2NO2 ↔ N2O4 ΔH= 35kJ

N2 + 2O2 ↔ N2O4

Hess’s Law

N2 + 2O2 ↔ 2NO2 ΔH= 41kJ

2NO2 ↔ N2O4 ΔH= 35kJ

N2 + 2O2 ↔ N2O4 Reversing the second reaction reverses the sign of the enthalpy of the reaction

Hess’s Law

N2 + 2O2 ↔ 2NO2 ΔH= 41kJ

2NO2 ↔ N2O4 ΔH=-35kJ

N2 + 2O2 ↔ N2O4 Now add the two

Hess’s Law

N2 + 2O2 ↔ 2NO2 ΔH= 41kJ

2NO2 ↔ N2O4 ΔH=-35kJ

N2 + 2O2 ↔ N2O4 6kJ

End

Entropy

• (S) is the measure of the degree of disorder in a system.

• A spontaneous process is one that occurs in a system left to itself. No external action is needed to make it happen.

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• Increase in disorder;

• Decrease in disorder;

∆S > 0

∆S < 0

GIBBS FREE ENERGY

• Gibbs free energy indicates whether a reaction will occur or not.

G H TS

G H T S

T temp K

( .( ))

• Exergonic reactions (spontaneous)

• Endergonic reactions (nonspontaneous)

G0

G0

• Endothermic reactions occur spontaneously when T Δ S is large.

• The thermodynamic definition of a system in equilibrium is; when Δ H and Δ S have the same sign, and there is some temperature at which Δ H and T Δ S are numerically equal.

• A spontaneous reaction proceeds toward equilibrium.

• Chemical potential energy, G, is least at equilibrium.

• Enthalpy, entropy, and free energy depend on temperature. ( We will only work at 298.15 K and 100.00 kPa-standard states)

GIBBS FREE ENERGY CALCULATIONS

• Δ H values are relative; free elements are considered to have

• change in enthalpy is found by;

H f 0

H H Hr f products f reac ts ( ) ( tan )

• entropy changes and Gibbs free energy is computes as follows;

S H H

G G G

r f products f reac ts

r f products f reac ts

( ) ( tan )

( ) ( tan )

• 1. Organize the data you will use from the appropriate table.

• 2. Multiply each Δ Gf0 by the number of

moles from the balanced equation. (Always make sure the chemical equation is balanced). Substitute these values into the equation used to determine Δ Gf

0 .

• The Gibbs free energy decreases in a spontaneous reaction because the system is changing to a more stable state.

• To find

1. Multiply each by the number of moles from the balanced equation.

2. Substitute these values into the equation.

S

S