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Thermo I

Thermo I. Chapter 5. Energy (section 5.1). Energy is- the capacity to do work or transfer heat. So what are work and heat ? Work-energy used to cause an object with mass to move Heat-energy used to cause the temperature of an object to increase. Energy. Kinetic (E k )-energy of motion

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Thermo I

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  1. Thermo I Chapter 5

  2. Energy (section 5.1) • Energy is- the capacity to do work or transfer heat. • So what are work and heat? • Work-energy used to cause an object with mass to move • Heat-energy used to cause the temperature of an object to increase.

  3. Energy • Kinetic (Ek)-energy of motion • EK = ½mv2 • Magnitude depends on the object’s mass (m) and its velocity (v). • Potential-depends on the object’s position relative to other objects. In chemistry, it is expressed as electrostatic potential energy

  4. Potential Energy • Eel= (KQ1Q2) /d, where K is a constant, and Q is the electrical charge on the two objects, and d is the distance between them. • K = 8.99x109 J-m/C2 (C is in coulombs) • Q will usually be about the size of electron charge (1.6x10-19 C) • When both Q’s have the same sign, the charges will repel one another, etc. • When Eel is positive, what is the situation?

  5. Units • SI unit for energy is the Joule (J) 1 kg-m2/s2 • Joules are tiny, so kJ will be appropriate. • calorie=4.184J • 1000cal=1 Cal=1kcal • What is the kinetic energy in joules of a 45g golf ball moving at 61 m/s? • What is this energy in calories? • What happens to this energy when the ball lands in a sand trap?

  6. Work and Heat • Two players: the system (that which we are focused upon), and the surroundings (everything else). • Work (w)= F*d A force (F) moves an object a distance (d). • When we (surroundings) cause an object (system) to move, we are performing work on the system, or transferring energy to it. • Heat, q, can enter or leave the system.

  7. First Law of Thermodynamics section 5.2 • Energy(E) is neither created nor destroyed. • The internal energy of a system is the sum of the kinetic & potential energies of all the components of the system. (We can’t know these values, but we can measure their change) • ∆E=Efinal – Einitial • Positive ∆E means the system gained energy. • Negative ∆E means the system lost energy.

  8. First Law, cont. • Positive ∆E denotes an endothermic process. • Negative ∆E denotes an exothermic process. • Endothermic-heat flows into the system from the surroundings. • Exothermic-heat flows out of the system to the surroundings.

  9. State Functions • A property of a system that is determined by specifying the system’s condition, or state (in temp., pressure, location, etc.) • The value of a state function depends only on the present state of the system, not the path it took to get to that state. • ∆E is a state function, since it is derived from an initial and final state, not how it got from initial to final. It describes the amount of change only. • q (heat) and w (work) are NOT state functions. • However, ∆E = q + w

  10. Enthalpy H section 5.3 • In chem., much of our work is in open systems (like a beaker), so work, w, is not readily obvious. • In a closed system, with a piston, pressure is constant, and the piston can be moved via a rxn. That generates a gas. This work would be visible. • This would be called pressure-volume work. • H accounts for heat flow in processes where pressure is constant, and the only work done is pressure-volume work.

  11. H2 gas is generated and moves the piston. Pressure-volume work.

  12. Enthalpy cont. • So, H = E + PV (internal energy plus pressure-volume work done) • Therefore, when a change occurs at constant pressure, ∆H = ∆E + P∆V • If ∆E= q + w, and w = -P∆V, then: • ∆H = (qP + w) – w, or ∆H = qP • This simply means that in most cases, ∆H = q • Positive ∆H means the system has gained heat from the surroundings (endothermic). • Negative ∆H means the system has lost heat to the surroundings (exothermic).

  13. Enthalpies of Reaction section 5.4 • ∆H = Hproducts – Hreactants • This change is known as the heat of reaction. It is the heat change that acoompanies a reaction. • Example: • CH4 + 2O2→ CO2 + 2H2O ∆H = -890 kJ • The production of water from its elements is an exothermic reaction the gives off 483.6 kJ of heat. • This is a thermochemical equation.

  14. Enthalpies of reaction cont. • Enthalpy is an extensive property. If you double the amount reactants consumed, ∆H doubles also. Enthalpy is stoichiometric. • The enthalpy change for a reaction is equal in magnitude, but opposite in sign, for its reverse reaction. • The enthalpy change for a reaction depends on the state (phase) of the reactants and products. • ∆H of reaction is usually in the units of kJ/mol.

  15. The value of ∆H for the reaction below is -72 kJ. • __________ kJ of heat are released when 1.0 mol of HBr is formed in this reaction. • A) 144 • B) 72 • C) 0.44 • D) 36 • E) -72 • d

  16. The value of ∆H for the reaction below is -126 kJ. • The amount of heat that is released by the reaction of 25.0 g of Na2O2 with water is __________ kJ. • A) 20.2 • B) 40.4 • C) 67.5 • D) 80.8 • E) -126 • a

  17. The value of ∆H for the reaction below is -1107 kJ: • How many kJ of heat are released when 15.75 g of Ba(s) reacts completely with oxygen to form BaO(s)? • A) 20.8 • B) 63.5 • C) 114 • D) 70.3 • E) 35.1 • b

  18. Calorimetry section 5.5 • ∆H can be measured experimentally, and this measurement of heat flow is called calorimetry. • The temp. change a substance undergoes when it absorbs heat is its heat capacity. Every pure substance has its own. • Heat capacity, C, is the heat required to raise the temperature 1°C. Specific heat capacity, s, is the heat required to raise 1g of that substance 1°C. Molar heat capacity, Cmolar……1 mole of a substance 1°C.

  19. Constant-pressure calorimeter • Really just a styrofoam cup. Styrofoam is an excellent insulator, so it holds heat very well. • Constant pressure because it’s open to the atmosphere.

  20. Constant-volume calorimeter • Called constant-volume because it is a closed system, so pressure varies, but the volume of the system does not. • Used to measure the energy stored in samples of fuels and foods.

  21. Calorimetry cont. • Heat, q, = m*s*∆T • Heat lost by the system is gained by the surroundings, and vice versa. • As stated earlier, ∆H and q are the same thing at constant pressure, so when we measure heat, we are measuring ∆H directly. • When mixing solutions, qSOLN = (spec. heat of solution)*(g of solution)*∆T • For solutions, the specific heat will be that of water: 4.184 J/g-°C

  22. Calorimetry • When 50.0mL of 0.100M AgNO3 and 50.0mL of 0.100M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30°C to 23.11°C. The temperature increase is caused by the following reaction: • AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) • Calculate ∆H for the reaction, assuming 100.0g of solution, and a specific heat of 4.184 J/g-°C. • -68000 J/mol or -68 kJ/mol (q= 338.9 J) • Would the ∆H value be different if the molarity of your reactants was different? • Write a net ionic equation for the reaction, just for fun.

  23. Hess’s Law section 5.6 • Allows you to calculate ∆H for a reaction without measuring anything, using known ∆H values for other reactions. • Stated: if a reaction is carried out in a series of steps, ∆H for the overall reaction will equal the sum of the enthalpy changes for the individual steps. • Some examples:

  24. Enthalpy of Formation section 5.7 • Using Hess’s Law, the enthalpy change can be calculated for many different kinds of reactions: • ∆Hvap-heat of vaporization-converting liquid to gas • ∆Hfus-heat of fusion-melting a solid • ∆Hcomb-heat of combustion-combusting a substance in oxygen • ∆Hf-heat of formation-heat associated with the formation of a compound from its elements. This one is kind of important. • Since the amount of enthalpy change depends on temp., pressure, and state (phase), it helps to compare reactions at what is called standard state, which is defined as 1 atm pressure and 298K or 25°C. From this, we have developed tables of enthalpy data called standard enthalpy changes.

  25. Standard enthalpy change • Symbolized as ∆H°, standard enthalpy change is the enthalpy change that occurs for a reaction when all of the reactants and products are at their standard state. That is, whatever state they are in at 1 atm and 298K. • So, a reaction with Fe(l) or H2O(s) could not be occurring at standard state conditions. • Even more specific, ∆H°f, or standard enthalpy of formation, is the enthalpy change that occurs when 1 mole of a compound is formed from its elements at standard state conditions. • These are listed in Appendix C in the back of the textbook.

  26. ∆H°f Std. enthalpy of formation • Let’s try writing a few equations, with their ∆H°f component alongside (from Appendix C): • HBr • AgNO3 • Hg2Cl2 • C2H5OH

  27. NH3 • SO2 • RbClO3 • NH4NO3

  28. ∆H°fStd. enthalpy of formation • If an element exists in more than one form at standard conditions, the most stable form of the element is listed. For example Cgraphite would be listed instead of Cdiamond. • ∆H°f of elements is zero, because no energy is needed, the element already would exist in its standard state at standard conditions. • ∆H°f equationor not (and if not, why not?): • 2Na(s) + ½O2(g) → Na2O(s) • 2K(l) + Cl2(g) → 2KCl(s) • C6H12O6(s) → 6C(diamond) + 6H2(g) + 3O2(g)

  29. Using ∆H°f to find ∆Hrxn • C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆Hcomb = ? • We can “add” these three equations to find ∆Hcomb. • C3H8(g) → 3C(s) + 4H2(g) ∆H1 = -∆H°f[C3H8(g)] • 3C(s) + 3O2(g) → 3CO2(g) ∆H2 = 3∆H°f[CO2(g)] • 4H2 + 2O2(g) → 4H2O(l) ∆H3 = 4∆H°f[H2O(l)] • _______________________________________________ C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆Hrxn = ∆H1 + ∆H2 + ∆H3 • You can use Hess’s Law and ∆H°f equations from a table to calculate ∆Hrxn for any reaction, so long as you choose the necessary reactions to work with.

  30. The short version: To find the heat of reaction, ∆Hrxn,total all of the ∆H°f‘s for the products, total all of the ∆H°f’s for the reactants, and subtract the reactant total from the product total. • Or: • ∆H°rxn= Σn∆H°f(products) – Σm∆H°f(reactants) • Now try # 69 and 71 in the textbook.

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