# ENTHALPY, ENTROPY AND GIBBS FREE ENERGY

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ENTHALPY, ENTROPY AND GIBBS FREE ENERGY

The First Law of Thermodynamics
Energy can neither be created or destroyed The energy of the universe is constant, but it can change forms.

Energy book keeper First Law accounts for energy, but it does not tell us why a particular process occurs in a given direction

Spontaneity DOES NOT MEAN FAST!!!
Means that the process occurs without any outside intervention

Energy released or absorbed during a chemical reaction (heat of reaction) is equal to the difference between the potential energy of the products and the potential energy of the reactants. In a chemical reaction; reactants --> products ΔPE = PE products - PE reactants PE can be thought as heat energy (H) Therefore, ΔH (kJ) = H products - H reactants

When ΔH is negative H products < H reactants and the reaction is exothermic.

When ΔH is positive H products > H reactants and the reaction is endothermic.

Energy released or absorbed by a chemical reaction can be represented by a potential energy diagram.

Activation Energy The activation energy is the minimum energy required to start a chemical reaction by providing colliding molecules with enough energy for effective collisions to occur. The activated complex is the short-lived and unstable intermediate species located at the highest of the activation energy.

Catalysts A catalyst provides an alternate reaction pathway, which has a lower activation energy than an uncatalyzed reaction.

Look at Table A-6 Notice the following:
Substances are in alphabetical order ΔHf (enthalpy of formation) is in kJ/mole Free elements have a ΔHf = 0 (they are not compounds formed from elements) The enthalpy of reaction is equal to the sum of the enthalpies of formation for the products – the sum of the enthalpies of formation for the reactants Σ ΔHr = ΔHf products - Σ ΔHf reactants

ENTHALPY CALCULATIONS
2CO(g) O2(g)  2CO2(g) ΔHr = Σ ΔHf products - Σ ΔHf reactants =[2 mole( kJ/mole)] - [2mole( kJ/mole) + 1mole(0kJ/mole)] = [ kJ] - [ kJ] = kJ The reaction is exothermic

ENTROPY Entropy is the degree of disorder
Represented with the symbol S Matter changes from a more ordered to less ordered state 2H2O  2H O2 H2(l)  H2(g) A positive ∆S means an increase in entropy

States of matter Ssolid < Sliquid << Sgas

Which has more entropy? 1. Solid or gaseous phosphorus
2. CH4(g) or C3H8(g) 3. NaCl(s) or NaCl(aq)

Second Law of Thermodynamics
In any spontaneous process there is always an increase in the entropy of the universe The entropy of the universe is constantly increasing

ENTROPY CALCULATIONS If the reaction increases entropy, ∆S is positive and the reaction is said to be ENTROPY-FAVORED Calculate the entropy change(∆S) for the following reaction CH4(g) O2(g)  CO2(g) H2O(l)

Three S’s Ssys = system Ssurr = surroundings Ssys + Ssurr = Suniv

Suniv If it is +, the entropy of the universe is increasing
Process is spontaneous If it is negative, the process is not spontaneous

Change of state H2O(l)  H2O(g) What happens to the S of the water?
Ssys= +

Heat is flowing from the surroundings to the system Random motion of particles decreases Ssurr = -

Which S controls the situation? DEPENDS ON TEMP
Is it spontaneous? Need to look at Suniv Which S controls the situation? DEPENDS ON TEMP

Exothermic Process Always increases entropy of surroundings
But, its significance depends on the temp at which the process occurs Energy transfer will be more significant at lower temps

GIBBS FREE ENERGY CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Gibbs free energy (∆G) is a measure of the chemical reaction potential of a system If ∆G is negative, the reaction is spontaneous If ∆G is positive, the reaction is not spontaneous Calculate the change in free energy for the following reaction CH4(g) O2(g)  CO2(g) H2O(l)

Gibbs Free Energy ∆H ∆S ∆G
Enthalpy and Entropy can be combined to predict reaction spontaneity ∆G = ∆H - T∆S ∆H ∆S ∆G Comments on Reaction - + Always spontaneous + or - Spontaneous at high temperatures Spontaneous at low temperatures Never spontaneous

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