THERMODYNAMICS. Thermodynamics: (From Greek: thermos = heat and dynamic = change)

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1 THERMODYNAMICS Thermodynamics: (From Greek: thermos = heat and dynamic = change) The science that deals with the physical and chemical changes of matter due to work and heat flow Matter: The material of which any physical object is composed "Thermodynamics is the only physical theory of universal content which, within the framework of the applicability of its basic concepts, I am convinced will never be overthrown." Albert Einstein The foundations of thermodynamics System: A thermodynamic system is that part of the universe that is under consideration Surroundings (environment): Everything outside of the system boundary System Surroundings The foundations of thermodynamics There are three kinds of systems: Each kind deals with the nature of the exchange of energy and matter across the boundary 1. Isolated : No exchange of heat, matter or work across the boundary. 2. Closed : exchange of energy (heat and work) but not matter across the boundary. A boundary that allows exchange of work but not heat is called adiabatic A boundary that allows exchange of heat but not work is typically called rigid 3. Open : exchange of energy (heat and work) and matter across the boundary. 1

2 Properties of a System Intensive : Independent of the amount of matter present (i.e. Pressure, Temperature, density even if you add more material, the value stays the same) Extensive : Dependent on the amount of matter present (mass, volume these are additive if you add more material, the value increases) The foundations of thermodynamics Energy: The capacity of a system to do work Internal Energy: Changes that take place within a system owing to interaction of the system and its surroundings Work: The result of a force acting through a displacement (push on something and it moves) When work is done on the system by the surroundings or on the surroundings by the system, the internal energy of the system changes (thermodynamics) Example: An expanding bubble does work on the surroundings by pushing (displacing) on the surroundings as the bubble expands. The bubble has experienced a change in internal energy The foundations of thermodynamics Matter can exist in three states: these are referred to as the states of matter 1. Solid: 2. Liquid: 3. Gas: matter exists as discrete phases Phase: A mechanically separable, physically homogeneous portion of a system with a definite boundary 2

3 The foundations of thermodynamics Matter may change from one state to another (example:?) Likewise, matter may change from one phase to another (example:?) Keep in mind that to undergo a phase change does not necessarily mean a change in state. (example: Ice has many discrete solid phases) The foundations of thermodynamics Each system may be described completely by a set of physical and chemical properties. When any of the properties undergo a change, the system is said to undergo a change in state. (natural systems tend toward states of minimum energy) **A change in the state of a system does not refer to a change in the state of matter (we can refer to the changes in the state of matter as phase changes from now on)** A state property is one that does not depend on how the state was achieved. Any equation that relates the state properties of a system to each other is called an equation of state. Example: The ideal gas law PV=nRT is an equation of state because it relates the state variables T, P, V and n to each other. The laws of thermodynamics 0. The zeroth law: Systems in thermal equilibrium are said to have the same temperature. A system in equilibrium does not change with time We can have stable equilibrium: equilibrium occurring at the lowest energy state (natural systems tend toward the lowest energy state) Or we can have meta-stable equilibrium: equilibrium at an intermediate energy state with no perceived changes in state (however, not the lowest energy state) 3

4 The laws of thermodynamics 1. The first law: Conservation of energy. Energy cannot be created or destroyed, only modified in form The heat flowing into a system equals the increase in internal energy of the system plus the work done by the system. U=Q-W where U is the internal energy, Q is heat flow and W is work The units of work and heat are typically given in joules, however, in the past, the units of work were given in joules while the units of heat were in calories 1 calorie = joules (the calorie is not the same as that of diet lingo) Enthalpy H= U+P V where H (enthalpy change) is the heat change (Q) at constant pressure. Heat of fusion (heat absorbed when melting a substance at constant P) Heat of vaporization (heat absorbed when liquid is converted to gas) Heat of reaction (heat absorbed during a chemical reaction) SiO 2 (quartz) + CaCO 3 (calcite) = CaSiO 3 (wollastonite) + CO 2 (gas) Since H is a state variable, we only need to know the initial state (quartz and calcite at 25C and 1 atm) and the final state (wollastonite and CO 2 gas at 25C and 1 atm) in order to determine H (heat of reaction). H is calculated by H(products) H(reactants). H r = ( )-( ) = cal/mol is absorbed during the reaction If H is positive, heat is absorbed If H is negative, heat is given off 4

5 Reference State of chemicals and chemical components Because we cannot measure the absolute enthalpy of a component or chemical reaction but only changes that occur, we need to establish a reference point. This allows us to establish a scale Standard State: 25 o C / 1 atm At the standard state, the elements (in their elemental state) are said to have an enthalpy equal to 0 Joules. The laws of thermodynamics 2. The second law: Heat cannot be converted to work with one hundred percent efficiency Introduction of Entropy (S): A measure of the order of a system. In a closed system S always increases (directionality). This term essentially allows us to understand how chemical reactions can occur without a decrease in the internal energy of a system. S=Qr/T or Qr=T S The Gibb s function: G= H-T S G, H and S are all state functions and can be calculated similar to the H example used earlier (products reactants) A reaction will occur if the Gibb s function decreases. In other words, if Gr is negative then the reaction will proceed as written. If it is positive then the reaction will go in reverse. Therefore, if we look at the Gibb s function G= H-T S, a reaction will occur if the enthalpy change is small relative to the change in entropy. Or decreasing H (releasing heat) or increasing S will cause the reaction to go Think back to the enthalpy example if H is positive then heat is absorbed. In the old days, all reactions were thought to proceed if heat is released (decrease in U). However, some reactions still occur even though they absorb heat. They occur because of the increase in S Example: NaCl(s) = Na+(aq) + Cl-(aq) at 25C and 1 atm H=( kj/mol kj/mol) kj/mol = 3.71 kj/mol (heat absorbed endothermic) S=(58.45 J/molK J/molK)-72.1 J/molK = 42.9 J/molK so TS=12.8 kj/mol G=( ) = or G= = woohoo the reaction will go!! 5

6 The laws of thermodynamics 3. The third law: every perfectly ordered, pure crystalline substance has the same entropy at absolute zero (0 K). (S=0 at 0K) Le Chatilier s principle If a change is made to a system in equilibrium, the system responds by attempting to absorb the force causing the change For example: The more obvious one If pressure is applied to a system, the system will respond by decreasing volume. (the system becomes compressed) The less obvious but equally important one If temperature of a system is increased, the entropy and enthalpy of the system will both increase. Example: aragonite + heat = calcite (remember that H was positive indicating the reaction consumed heat) We can see that an increase in heat will drive the reaction to the right (this is qualitative so be careful, changes in heat do not necessarily mean changes in temperature) Calcite to Aragonite conversion V. If we look up the data, we note that Varagonite < Vcalcite. Therefore, increasing pressure should favor the formation of aragonite. H and S: Calcite has a greater entropy and enthalpy. Raising the temperature should favor the formation of calcite (or decreasing the temperature should favor the formation of aragonite). From this information, we can deduce the relationship between calcite and aragonite in temperature/pressure space. We have already determined that calcite is more stable than aragonite at 25 o C and 1 atm. Therefore, the phase diagram might look like this P Aragonite Calcite 25 o C and 1 atm T 6

7 The phase diagram we have constructed is a kind of free energy map. The diagram displays a T-P region where aragonite is stable (Garagonite<Gcalcite) and one where calcite is stable (Gcalcite<Garagonite). The line that separates the two phases is where Gcalcite=Garagonite and is called the phase boundary. Note that the line represents the combinations of T and P that Calcite and aragonite can co-exist. If the T and P of the system is not on the line, then calcite or aragonite will be stable but not both. Also note that, on the line, Gcalcite- Garagonite = 0. Gr=0 when the system is in equilibrium (no energy available to do work). 7

8 Gr = Vr P - Sr T At equilibrium G = 0, such that Vr P = Sr T Clapeyron Equation. So P/ T= S/ V V H S G cm^3/mol kj/mol J/molK kj/mol calcite aragonite The Clapeyron equation indicates that in T-P space, the slope of the equilibrium line between calcite and aragonite is equivalent to the ratio of the entropy change to the volume change (must be written in bars/ o C) or 15 for the case of calcite and aragonite (see below). (It is important to note that this form of the Clapeyron equation is only valid for pure solid phases, not for liquids or gases) Sr= =-4.2J/K Vr= = cm^3 or J/bar So -4.2/ = 15 Gibb s phase rule: When we look at the phase diagrams (like the calcite-aragonite diagram) we see that if the System is at a T and P that places you in the middle of a mineral field, we can adjust T and P arbitrarily. This is because we have 2 degrees of freedom. If we are on a line separating two phases, then we can move either P or T and the other is fixed (1 degree of freedom). If we are at a triple point, we are fixed at the specific T and P (0 degrees of freedom). Gibb s derived an equation to calculate the dof: F=C-P+2, where F is the degrees of freedom, C is the number of chemical components in a system, and P is the number of phases in the system the 2 is for T and P. The number of components in a system is the minimum number of chemical species needed to describe all of the phases in the system. Example, for the CaCO3 system, we have one component (C=1), when we are in a single mineral field we have 1 phase (P=1) so F=1-1+2=2 we have two degrees of freedom (an area). If we are on the line P=2 and then we get F=1-2+2=1 degrees of freedom (a line). If there were another phase composed of CaCO3, then we could get a fixed point where F=0 The equilibrium constant We know how to predict the direction of a spontaneous reaction (using the sign of the Gibb s free energy change) but how do we know how much reaction will occur? For example, we have calculated the standard state Gibb s free energy of reaction for the dissolution of Halite (NaCl) in water NaCl(s) = Na+(aq) + Cl-(aq) G o r = kj/mol We know it will occur spontaneously at 25 o C and 1 atmosphere (negative Gibb s free energy change) but will table salt continuously dissolve in water to no end? Observation shows us that there is a finite amount of salt that will dissolve in water. How can we predict how much salt will dissolve? The answer is the equilibrium constant. 8

9 The equilibrium constant Let us consider the general reaction. aa + bb = cc + dd where a,b,c,d are stoichiometric coefficients and A,B,C,D are chemical species (minerals, liquids, gases) For this and any reaction we can write the mass action quotient (Q) c C D Q= a A B d b If we introduce A and B, a reaction will occur to produce C and D. At some point, A and B will stop reacting and no more C and D will be produced (salt stops dissolving). At this time, the value of Q is not changing. When Q is constant, we have the equilibrium constant K. (at equilibrium Q=K) For the dissolution of salt we get [ Na+ ][ Cl ] Q= [ NaCl( s)] The equilibrium constant [ Na+ ][ Cl ] Q= [ NaCl( s)] As NaCl dissolves in water, the amount of product (Na+ and Cl-) will increase which results in a change to the value of Q. This change will continue until no more salt dissolves and the activities of Na+ and Cl- stop changing. At this time Q is constant. When a system is non-changing, we may be at equilibrium, at which time we say that Q=K (K is the equilibrium constant). In other words, at equilibrium, there is a relationship between the amounts of all components in a reaction that define equilibrium. The values listed in the mass action expression are not concentrations but activities (thermodynamic concentrations). We will learn more about this later By convention, activities of pure phase are equivalent to 1. Therefore, for the following reaction: Calcite + Quartz = Wollastonite + CO2(g) The mass action expression is written as aco2( g)* awoll K = acal* aqtz Since the activities (thermodynamic concentrations) of pure phases are = 1, the expression becomes K = pco2( g) Gases are given a p for the partial pressure of the gas (the thermodynamic pressure is the fugacity) 9

10 Activies are to concentrations as fugacities are to partial pressures (activities and fugacities are the thermodynamic variables used in the mass action expressions) we will learn more about this later The total pressure of a system is equal to the sum of the partial pressures. Thus, we can have a pco2 (or fco2)that is not equivalent to the total pressure of the system. Therefore, we do not assign a value of 1. Liquid water often has an activity of 1 but dissolved material (salts) can cause the activity of liquid water to deviate from unity. For most applications involving dilute aqueous solutions, the activity of water is taken to be 1. OK, we can write a balanced reaction, calculate the standard state Gibb s free energy change of the reaction, determine a value for the equilibrium constant and write a mass action expression for the reaction and equilibrium constant 10

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