THERMODYNAMICS

IMPORTANT DEFINITIONS System: the object or collection of objects being studied. Surroundings: everything outside the system that can exchange energy with the system. Heat: the transfer of energy to a substance causing an increase in that substance’s average kinetic energy Temperature: a measure of the average kinetic energy of a material’s particles.

TEMPERATURE CURVE

1 ST LAW OF THERMODYNAMICS The total energy of the universe is constant Synonymous with law of conservation of energy

HEAT TRANSFER  Occurs from hot object to cold object  Continues until both objects are at same temperature (thermal equilibrium)  Heat lost by hotter object equals heat gained by cooler object.  Exothermic: heat transfer from system to surroundings  Endothermic: heat transfer from surroundings to system.

ENERGY UNITS Standard unit (SI) is the Joule 1 kJ = 1000 J 1 calorie (cal) = J 1000 cal = 1 Cal = kJ Cal = dietary calories

SPECIFIC HEAT CAPACITY q = mC p Δ T q = Heat energy (J) m = mass of substance (g) C p = specific heat capacity = amount of heat to raise temperature of 1 gram of a material 1 Kelvin Δ T = T f – T i (in Celsius)

SPECIFIC HEAT CAPACITY C p (ice) = 2.06 J/g*K C p (water) = J/g*K C p (steam) = 1.7 J/g*K

How much heat is required to raise the temperature of 43 grams of water from 22 degrees Celsius to 85 degrees Celsius?

A 47 gram chunk of hot copper is placed into 200 mL of water originally at 33 degrees Celsius. Afterwards, the water now has a temperature of 35.4 degrees Celsius. Assuming all heat lost by the copper was gained by the water, how hot was the copper before putting it into the water? The specific heat of copper is 0.38.

TEMPERATURE CHANGES AS HEAT IS ADDED

CHANGES OF STATE Heat of Fusion: Heat required to convert a substance from solid at its melting point to liquid  For water: 333 J/g or 6.00 kJ/mol Heat of Vaporization: Heat required to convert a substance from liquid at its boiling point to gas  For water: 2256 J/g or kJ/mol

˚C. How much heat will be required to convert this amount of water to steam at a temperature of 125 ˚C? I have 67 grams of ice at -10˚C. How much heat will be required to convert this amount of water to steam at a temperature of 125 ˚C?

ENTHALPY (H) The enthalpy of a substance is equal to its total energy plus a small added term that takes into account volume and pressure. It’s a measure of how much energy is stored in a substance.

ENTHALPY CHANGES ( Δ H) All reactions have a standard enthalpy change  Found at 1 atmosphere of pressure and 25˚C Δ H = H products – H reactants

WHAT DOES IT ALL MEAN? If Δ H is positive, it’s an endothermic reaction and heat is absorbed. If Δ H is negative, the reaction is exothermic and heat is released.

USING ENTHALPY How much heat is transferred when 9.22 g of glucose (C 6 H 12 O 6 ) in your body react with O 2 according to the following equation? C 6 H 12 O 6 + O 2  6CO 2 + 6H 2 O Δ H˚ = kJ

C + H 2 0  CO + H 2 How much heat energy is required to react with excess carbon and water to produce 45 L of hydrogen gas at STP?

HESS’S LAW In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

HESS’S LAW N 2 + O 2  2NO Δ H = 180 kJ 2NO + O 2  2NO 2 Δ H = -112 kJ Net: N 2 + 2O 2  2NO 2 Δ H = 68 kJ

S(s) + 3/2 O 2 (g)  SO 3 (g) Δ H = kJ 2SO 2 (g) + O 2 (g)  2SO 3 (g) Δ H = kJ What is the Δ H for the following net reaction? S(s) + O 2 (g)  SO 2 (g) Δ H = ? How many grams of sulfur dioxide gas can be created if 350 kJ of heat is transferred?

Calculate the standard enthalpy of combustion for benzene, C 6 H 6. C 6 H 6 (l) + 7 ½ O 2 (g)  6 CO 2 (g) + 3 H 2 O(l)

Nitroglycerin is a powerful explosive that forms four different gases when detonated: 2 C 3 H 5 (NO 3 ) 3 (l)  3 N 2 (g) + ½ O 2 (g) + 6 CO 2 (g) + 5 H 2 O(g) Calculate the enthalpy change when 10 grams of nitroglycerin is detonated. (The enthalpy of formation for nitroglycerin is -364 kJ/mol.)

1 ST LAW OF THERMODYNAMICS The energy of the universe is constant The energy change of the system must equal the heat lost or gained by the environment as well as the work done on or by the system from the environment. ∆E = q + W ∆E = change in internal energy content q = heat transferred to or from the system W = work transferred to or from the system

When the system is under a constant external pressure, the work done on or by the surroundings is equivalent to product of the pressure and the change in volume. W = -P∆V Joule (J) = 1 L*atm

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40 cm 3. If the combustion of this mixture releases 950 J of energy, to what volume will the gases expand against a constant pressure of 650 torr if all the energy of combustion is converted into work to push back the piston. (760 torr = 1atm)

SPONTANEITY AND MATTER Spontaneous changes occur without outside intervention (heat exchange, work, etc.)  Examples: Gas molecules spontaneously spread throughout a room…never the other way (congregating in one corner) A reaction at equilibrium will never spontaneously undergo a change that takes it away from equilibrium. Above 0 Celsius, ice will spontaneously melt.

HEAT AND SPONTANEITY Exothermic reactions appear to be spontaneous.  HCl + NaOH  NaCl and H2O.  Na and Cl2 react to make NaCl  Giving off heat is not a requirement for spontaneity.  Melting ice is a spontaneous endothermic reaction Heat (Enthalpy) alone does not determine the spontaneity of a reaction.

ENTROPY (S) Measurement of disorder or randomness within a system. 2 nd law of thermodynamics: the entropy of the universe always increases in a spontaneous reaction. Entropy is based on the concept that spontaneous processes result in dispersal of matter and energy.  Matter becomes more disorganized (goes through a change of state, becomes a solution, gases expand, etc.)  Energy is spread over a greater number of molecules

WHICH OF THE FOLLOWING PROCESSES WOULD HAVE AN INCREASE IN ENTROPY? Your room getting cluttered Iron rusts Salt dissolves in water A house is built A satellite falls back to earth A satellite is launched into orbit

STANDARD ENTROPY ∆S˚ The standard entropy of a substance is the entropy gained by converting it from a perfect crystal at 0 K to standard state conditions.  Measured in J/K*mol 3 rd law of thermodynamics: The entropy of a perfect crystal at 0K is 0 (no disorder).

GENERAL RULES Entropies of gases are much much larger than those for liquids, and liquids are greater than solids. Larger molecules tend to have greater entropies than smaller molecules. For a given substance, entropy increases as temperature increases. Large increases in entropy accompany changes in state.

ENTROPY CHANGES ∆S˚ sys = S˚ products - S˚ reactants Calculate the entropy for the oxidation of NO with O 2. 2 NO(g) + O 2 (g)  2 NO 2 (g) Does the result agree with predictions based on general rules of entropy?

CALCULATING ∆S˚ UNIV ∆S˚ univ = ∆S˚ surr + ∆S˚ sys ∆S˚ surr = q surr /T = -∆H˚ sys /T Positive Entropy change in the universe is a spontaneous reaction Negative Entropy change in the universe is nonspontaneous 0 Entropy change in the universe means reaction is at equilibrium

Determine if the following reaction is spontaneous under standard conditions. Na 2 CO 3 (s) + 2HCl(aq)  2 NaCl(aq) + H 2 O(l) + CO 2 (g)

GIBB’S FREE ENERGY Allows determination of spontaneity through only assessment of the system and with one term. As with entropy and enthalpy, typically we are more concerned only with changes in free energy “Free Energy” represents the maximum energy available to do useful work…free can be thought of as “available”.  The sum of the energies available from dispersal of energy (enthalpy) and dispersal of matter (entropy).

∆G˚ sys = ∆H˚ sys - T∆S˚ sys If ∆G˚ sys is negative, reaction is spontaneous If ∆G˚ sys = 0, reaction is at equilibrium If ∆G˚ sys is positive, reaction is not spontaneous

STANDARD FREE ENERGY OF FORMATION ∆G˚ rxn = ∆G˚ f, products - ∆G˚ f, reactants ∆G˚ f = free energy change when forming one mole of a compound from the component elements. Calculate the free energy change for the oxidation of 1 mol of SO 2 (g) to form SO 3 (g)

FREE ENERGY AND TEMPERATURE Some reactions are spontaneous at certain temperatures but not at others. This occurs when  Processes are entropy-favored but enthalpy-disfavored  Processes are entropy-disfavored but enthalpy-favored. Using free energy, we can determine the approximate temperature where the reaction would become spontaneous.  The free energy term is set to 0 and equation is solved for T.

Use thermodynamic parameters to estimate the boiling point of methanol (CH 3 OH).

FREE ENERGY AND EQUILIBRIUM Free energy decreases to a position of lower free energy until a minimum free energy is reached and the reaction is at equilbrium ∆G˚ rxn = -RT ln K  Large K values tend to be product favored and correspond to large negative values for ∆G˚ rxn  R = Gas constant = J/K*mol To determine if a reaction is at equilibrium or what direction it needs to go to get to equilibrium… ∆G rxn = ∆G˚ rxn + RT ln Q

SUMMARY ∆G˚ rxn = ∆G˚ f, products - ∆G˚ f, reactants = ∆H˚ sys - T∆S˚ sys =-RT ln K If ∆G˚ rxn is negative, reaction is spontaneous If ∆G˚ rxn = 0, reaction is at equilibrium If ∆G˚ rxn is positive, reaction is not spontaneous ∆G rxn = ∆G˚ rxn + RT ln Q ∆G rxn < 0… reaction proceeds spontaneously to products until equilibrium ∆G rxn > 0…reaction proceeds spontaneously toward reactants until equilibrium ∆G rxn = 0…reaction is at equilibrium

Determine the value of ∆G˚ rxn for the reaction C(s) + CO 2 (g)  2 CO(g) from thermodynamic data. Use this result to calculate the equilibrium constant. The equilibrium constant for the formation of Ag(NH 3 ) 2 + is 1.6 x Use this value to calculate the ∆G˚ for the reaction. Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq)