Thermochemistry

1. Thermochemistry Things are “heating up” now!

2. Nature of Energy • Energy – the capacity to do work (or to produce heat) • Work is a force acting over a distance • Moving an object • Heat is a form of energy • Chemicals may store potential energy in their bonds • Can be released as heat energy

3. First Law of Thermochemistry • Law of conservation of energy – energy is neither created nor destroyed • Can only be converted from one form to another • Potential energy – energy due to position • Kinetic energy – energy due to the motion of an object • KE = ½ mv2 • m = mass • v = velocity • Units are Joules

4. Heat & Temperature • Heat (J) • Measure of energy content • What is transferred during a temperature change • Temperature (°C) • Reflects random motion of particles in a substance • Indicates the direction in which heat energy will flow • Higher  lower

5. State Functions • A property of a system that depends only on its present state • Do not depend on what has happened in the system, or what might happen in the future • Independent of the pathway taken to get to that state • Eg: 1.0 L of water behind a dam has the same PE for work regardless of whether it flowed downhill to the dam or was taken uphill to the dam in a bucket. • PE is a state function dependent only on the current position of the water

6. Chemical Energy • Exothermic Reactions • Reactions that give off energy as the progress • Some of the PE stored in the chemical bonds is converted to thermal energy (random KE) through heat • Products are generally more stable (stronger bonds) than reactants • Endothermic Reactions • Reactions in which energy is absorbed from surroundings • Energy flows into the system to increase the PE of the system • Products are generally less stable (weaker bonds) than the reactants • Energy is needed to break bonds; released when formed

7. Thermodynamics • System Energy ΔE = q + w • q = heat • Positive in endothermic reactions • Negative in exothermic reactions • w = work • Negative if the system does work • Positive if work is done on the system

8. Example • Calculate the ΔE for a system undergoing an endothermic process in which 18.7 kJ of heat flows and where 4.3 kJ of work is done on the system. • ΔE = q + w • q = 18.7 kJ (endothermic – gaining heat) • w = 4.3 kJ (positive – work is done on the system) • ΔE = 18.7 kJ + 4.3 kJ = 23.0 kJ • System has gained 23.0 kJ of energy

9. Thermodynamics - Gases • Work done by gases w = -PΔV • By a gas (through expansion) • ΔV is positive • w is negative • To a gas (by compression) • ΔV is negative • w is positive • P is the P of surroundings

10. Example • Calculate the work associated with the expansion of a gas from 32 L to 75 L at a constant external pressure of 15 atm. • w = -PΔV • P = 15 atm • ΔV = 75 L – 32 L = 43 L • w = -(15 atm)(43 L) • w = -645 L · atm • -w because the gas expands so work is done to the surroundings • Energy flows out of the system

11. Enthalpy & Calorimetry • Enthalpy – measure of the energy released/absorbed by a substance when bonds are broken/formed during a reaction • Way to calculate heat flow in or out of a system • H = E + PV • In systems at constant pressure, where the only work is PV, the change in enthalpy is due only to energy flow as heat (ΔH = heat of reaction) • ΔH = Hproducts – Hreactants • Exothermic reactions: -ΔH • Endothermic reactions: +ΔH

12. Calorimetry • Calorimetry – the science of measuring heat • Heat capacity (c) • Ratio of heat absorbed to increase in temperature • c = heat absorbed temperature increase • Specific heat capacity – energy required to raise the temperature of 1 gram of a substance by 1°C • Molar heat capacity – energy required to raise the temperature of 1 mole of a substance by 1°C

13. q = mcΔT • q = heat • m = mass • c = specific heat • ΔT = change in T

14. Constant Pressure Calorimetry (soln) • Calculating Heat of Reaction (ΔH) • ΔH = (s)(m) (ΔT) • ΔH = (specific heat capacity) x (mass of soln) x (T change) • Heat of reaction is an extensive property • Dependent on the amount of substance • ΔH α moles of reactant

15. Constant Volume Calorimetry • Volume of bomb calorimeter cannot change • No work is done • Bomb calorimetry = constant volume • The heat capacity of the calorimeter must be known • kJ/°C • Bomb calorimetry – weighed reactants are placed inside a steel container and ignited • Used by industry to determine number of food calories that we consume • “Calories” on food labels are really kilocalories

16. ΔE = q + w, where w = 0  ΔE = q

17. Example – Bomb Calorimetry • Heat capacity of calorimeter = 11.3 kJ/°C • Compare heat of combustion for: • 1.50 g methane T increase = 7.3°C • 1.15 g hydrogen 14.3°C • 0.5269 g octane 2.25°C • Hcombustion CH4 = 54.0 kJ/g • Hcombustion H2 = 140.5 kJ/g • Hcombustion C8H18 = 48.3 kJ/g

18. Hess’s Law

19. Hess’s Law • Hess’s Law - In going from a particular set of reactants to a particular set of products, the change in enthalpy (ΔH) is the same whether the reaction takes place in one step or a series of steps

20. Hess’s Law • One step: N2(g) + 2 O2(g) 2 NO2(g) ΔH = 68 kJ • Two step: N2(g) + O2(g)2 NO(g) ΔH1 = 180 kJ 2 NO(g) + O2(g) 2 NO2(g) ΔH2 = -112 kJ N2(g) + 2 O2(g) 2 NO2(g) ΔH1 + ΔH2= 68 kJ

21. Characteristics of Enthalpy Changes • If a reaction is reversed, the sign on ΔH is reversed N2(g) + 2 O2(g) 2 NO2(g) ΔH = 68 kJ 2 NO2(g)  N2(g) + 2 O2(g) ΔH = -68 kJ • Magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. • If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer

22. Using Hess’s Law • Work backward from the final reaction • Reverse reactions as needed • Also reverse ΔH • Identical substances found on both sides of the summed equation cancel each other

23. Example • Given the following data • NH3(g)  ½ N2(g) + 3/2 H2(g)ΔH = 46 kJ • 2 H2O(g)  2 H2(g) + O2(g) ΔH = 484 kJ • Calculate ΔH for the reaction 2N2(g)+ 6 H2O(g)  3 O2(g) + 4 NH3(g)

24. Standard Enthalpies of Formation ΔHf°

25. Standard Enthalpy of Formation • The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state. • ΔHf° - Degree sign indicates that the process was carried out under standard conditions • Can measure enthalpy changes only by performing heat-flow experiments

26. Standard State • For a compound • Gaseous state • Pressure of 1 atm • Pure liquid or solid • Standard state is the pure liquid or solid • Substance in solution • Concentration of 1 M • For an element • The form in which the element exists at 1 atm and 25°C

27. ΔHf° always given per mole of product with the product in its standard state ½ N2(g) + O2(g)  NO2(g) ΔHf° = 34 kJ/mol C(s) + 2 H2(g) + ½ O2(g)  CH3OH(l) ΔHf° = -239 kJ/mol

28. Calculating Enthalpy Change • Elements in their standard states are not included • ΔHf° = 0 • The change in enthalpy for a reaction can be calculated from the enthalpies of formation of the reactants and products • ΔH°reaction = npΔH°f(products) - nrΔH°f(reactants)

29. Calculating ΔHf° • CH4(g) + O2(g)  CO2(g) + 2 H2O(l) • Using Hess’s Law – break apart the reactants into their respective elements • CH4(g) C(s) + 2 H2(g) C(s) + 2 H2(g)CH4(g) ΔHf° = -75 kJ/mol • O2(g)  O2(g) ΔHf° = 0 kJ/mol • C(s) + O2(g) CO2(g) ΔHf° = -394 kJ/mol • H2(g) + ½ O2(g)  H2O(l) ΔHf° = -286 kJ/mol • 2 H2O required: (-286 kJ/mol)(2) = ΔHf° = -572 kJ

30. Calculating ΔHf° • ΔHf° = (-ΔHf° CH4(g)) + (ΔHf° O2(g)) + (ΔHf° CO2(g)) + 2(ΔHf° H2O(l)) • ΔHf° = -(-75 kJ) + (0) + (-394 kJ) + (-572 kJ) • ΔHf° = -891 kJ

31. ΔHf° • If ΔHf° for a compound is negative, energy is released when the compound is formed from pure elements and the product is more stable that its constituent elements • Exothermic process • If ΔHf° for a compound is positive, energy is absorbed when the compound is formed from pure elements and the product is less stable than its constituent elements • Endothermic process