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Spontaneous Change_CompleteNotes.pdf

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Department
Chemistry
Course
CHEM102
Professor
Sai Yiu
Semester
Winter

Description
UNIVERSITY OF ALBERTA INTRODUCTORY UNIVERSITY CHEMISTRY II (CHEM 102) Spontaneous Change Introduction Many chemical reactions or processes are spontaneous (occurs by itself without outside intervention): • A ball rolls down a hill spontaneously but never rolls back up the hill • Heat flows from hot object to cooler one. The reverse process never occurs spontaneously • Iron rusts spontaneously as shown by the following chemical equation 4Fe (s) + 3O (g) → 2Fe O (s) 2 2 3 But iron oxide does not spontaneously change to iron metal and oxygen • Methane burns to form carbon dioxide and water but the reversed reaction is not spontaneous CH (4) + 2O (g2 → CO (g) 2 2H O (l)2 • Below 0 C, water spontaneously freezes and at temperatures above 0 C, ice melts spontaneously Limitations of the First Law of Thermodynamics The First Law of Thermodynamics is good for bookkeeping of energy in chemical reactions such as: • How much energy is involved in the chemical reaction • Does energy flow in or out of the system The First Law of Thermodynamics does not make sense as to the direction of a reaction. Spontaneous processes and entropy A process is said to be spontaneous if it occurs without outside intervention. A non-spontaneous process will not occur unless some external action is continuously applied. A spontaneous process does not mean instantaneous and has nothing to do with how long the process takes to occur. For example: CH (g) + 2O (g) → CO (g) + 2H O (l) 4 2 2 2 It won’t happen unless a spark is applied. 1 Thermodynamics Remember: How fast a Reactants reaction occurs is determined Energy Thermodynamics by the domain of kinetics – the reaction pathway. Kinetics Products Extent of a chemical reaction What is the law that predicts direction of a chemical reaction? Enthalpy change Since many spontaneous reactions are exothermic, it was thought that exothermicity was the criterion for spontaneity. o CH (4) + 2O (g)2→ CO (g) + 22 O (l) 2 ∆H = -802 kJ 2Fe (s) + 3/2O (g) → Fe O (s) ∆H = -826 kJ 2 2 3 Almost all exothermic reactions are spontaneous. o However, some endothermic reactions are also spontaneous at 25 C. o H 2 (s) → H O (2) ∆H = +6.02 kJ (melting of ice) o H 2 (l) → H O 2g) ∆H = +44.0 kJ (evaporation of water) Disorder and entropy In all spontaneous processes, the driving force is an increase of the entropy (S) of the universe. The entropy is viewed as a measure of randomness or disorder: The natural progression of things is from order to disorder (i.e. from lower entropy to higher entropy) Entropy is a thermodynamic state function that describes the number of arrangements (microstates) that are available to a system existing in a given state. Entropy is therefore closely associated with probability. Expansion of an ideal gas into a vacuum Fig 20.2, p885 2 The gas molecules fill the other bulb spontaneously because there is an increase of microstates available. A C B D BulbA BulbB If there are four molecules A, B, C and D are in bulb A, the number of possible microstates available after expanding into bulb B are: Bulb A Bulb B A, B, C, D B, C, D A A, C, D B A, B, D C A, B, C D A, B C, D A, C B, D A, D B, C There are a total of 8 x 2 = 16 microstates available for mixing 4 molecules. From the Boltzmann equation, S = k lnW where, S = entropy of a system k = Boltzmann’s constant = R / NA(gas constant / Avogadro’s number) W = number of microstates • A system with more microstates would be expected to have more disorder and higher entropy. • Entropy is a state function and so ∆S is dependent on the initial and final values. ∆S = S finalSinitial Which of the following changes would result in an increase of entropy (∆S > 0)? CO 2s) → CO (g2 N 2P = 1.0 x 10 atm) → N (P2= 1.0 atm) (NH 4 2r 2 7s) → N (2) +Cr O 2s3 + 4H O 2g) NaCl (s) + H2O (l) → NaCl (aq) 3 Third Law of Thermodynamics Unlike internal energy (U) and enthalpy (H), the absolute entropy value (S) can be determined. The Third Law of Thermodynamics states that a perfect crystal has zero entropy at a temperature of absolute zero. Ssys= 0 at 0 K o Standard molar entropy S of different compounds can be determined (Appendix D, A18 - A24) Standard molar o entropy at 25 C Entropy of vaporization > Entropy of fusion Rapid rise in entropy due to phase change – Entropy of fusion Fig 20.7 p890 -1 -1 o From the standard molar entropy values (unit = J mol K ), ∆S can be calculated for different reactions. o o o ∆S = ∑n products products ∑n reactantreactants Example: Calculate the standard entropy for the following reaction: N 2g) + 3H (g2 → 2NH (g) 3 o Absolute entropy values S are found from the table o o o o ∆S = 2 mol x (S NH ) – 33mol x (S H ) – 1 mo2 (S N ) 2 = 2 mol x 192.5 J mol K - 3 mol x 130.7 J mol K - 1 mol x 191.6 J mol K -1 -1 -1 = -198.7 J K . 4 The result is consistent with the equation with 4 gas molecules react to give 2 gas molecules. Entropy and the Second Law of Thermodynamics At -10oC, the freezing of water is spontaneous: H2O (l) → H 2 (s) o However, ∆S < 0 because ice is more order than liquid water. Therefore, entropy change of the system ∆S sys cannot be used as a criterion for spontaneity. The Second Law of the Thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe To be spontaneous: ∆S = ∆S + ∆S > 0 univ surr sys We therefore need to take the entropy change of the surroundings into consideration for spontaneity The mathematical expression for entropy change of a system is given by: q ∆S = T Where q = heat T = temperature, K At constant pressure, q = q = ∆H p ∆H surr And so ∆S = surr T But ∆H = - ∆H surr sys Therefore, ∆H surr - ∆Hsys ∆Ssurr = T T - ∆H sys ∆Suniv= + ∆Ssys T Multiplied both sides by T gives T∆S = - ∆H + T∆S univ sys sys 5 - T∆S univ= ∆H sys- T∆S sys ∆G = ∆H - T∆S sys sys ∆G = Gibbs Free Energy = - T∆S univ The Gibbs Free Energy is another important thermodynamic state function. Its absolute value cannot be determined. G = T x S = J K -1K = J Therefore G has a unit of J or kJ Since measurement of ∆S surris difficult, we use ∆G as a criterion for spontaneity. • ∆G < 0 (spontaneous) • ∆G > 0 (non-spontaneous) • ∆G = 0 (equilibrium) The three factors (∆H, ∆S and T) that affect spontaneity are summarized below: p 905 Free energy change and the work done of a system ▯ For a spontaneous process, the free energy is the maximum work obtainable from the system as the process takes place. ∆G = w (max) For example: Burning of gasoline o C 8 18) + 25/2 O (g)2 8CO 2g) + 9 H O2(g) ∆G = - 1596.2 kJ /mol 6 This amount of free energy is used mainly as the kinetic energy of the motor car. However, most of the free energy is wasted as heat increasing the entropy of the surroundings. ▯ For a non-spontaneous process, the free energy is the minimum work that must be done to the system to make the process take place. In reality, the energy supplied to a system to make a non-spontaneous reaction to occur is always more than the minimum because some of the free energy change is lost a
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