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Lecture 14: Thermochemistry

Lecture 14: Thermochemistry. Ch 14 Suggested HW : 19 , 27, 39, 23, 42, 82. We depend on energy for our existence. For example, batteries produce electric current from redox reactions, which is used to run machines

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Lecture 14: Thermochemistry

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  1. Lecture 14: Thermochemistry Ch 14 Suggested HW: 19, 27, 39, 23, 42, 82

  2. We depend on energy for our existence. For example, batteries produce electric current from redox reactions, which is used to run machines • Combustion reactions release thermal energy from chemical compounds that are used to move vehicles, provide electricity, and heat our homes • Plants use solar energy in photosynthesis to convert CO2 into: • the oxygen we breathe • Sugars which the plant itself feeds upon • These plants are consumed by animals, and so on..

  3. Introduction • Thermodynamics is the study of transformations of energy from one form to another. • The first law of thermodynamics, introduced in this lecture, allows us to “keep track” of energy change. • Two fundamental concepts of thermodynamics are heat and work.

  4. Energy Basics • What is Energy? • Energy is defined as the capacity to perform “work” • How do we define work? • Work is defined as the product of a force times the distance over which that force is applied (F x d) • Ex. Pushing an object along a rough surface • The force can be exerted by a human, steam engine, electric motor, etc. • In SI units, force is in Newtons (N), distance is in meters. Thus, work (and therefore, energy) is expressed in Newton-meters (N•m) • We use the unit JOULE (J) to represent a Newton-meter (1 J = 1 N•m)

  5. What is Energy? Work? • You can not perform work without energy • Example: You need at least 100 J of energy to do 100 J of work • Lets use a gas cylinder equipped with a piston as an example. The cylinder is our system. Everything else is our surroundings. • The application of a force to the piston causes displacement of the system by a distance Δh (final-initial) • The work (in J) done is: hinitial Δh h final

  6. Work • Compression:Since the compression is caused by the surroundings, we say that work is done on the system by the surroundings. • Work done onthe system is ALWAYSPOSITIVE.Energy from the surroundings is transferred into the system, so the system gains that energy. • Expansion: If the gas expands against the piston, then work is doneby the system on the surroundings. • Work done bythe system is ALWAYS NEGATIVEbecause the system now has that much less energy. Compression hfinal Δh h initial Expansion F hinitial Δh h final

  7. About Gases • Gases are the most understood form of matter • Even though different gases have different chemical properties, they tend to exhibit similar physical properties • This situation arises because gas molecules expand to fill a given space, and are relatively far apart from one another • Thus, each molecule behaves as if the others are not there

  8. Pressure • The most readily measured properties of a gas are its temperature, volume, andpressure • Pressure describes the force that a gas exerts on an area, A (P = F/A) • The image below shows gas molecules inside of a cubic container. The gas molecules strike against the walls of the container. These collisions are the source of the pressure.

  9. Atmospheric Pressure • You and I are currently experiencing an attractive force that pulls us toward the center of the earth (gravity). • Gas molecules in the atmosphere also experience gravity. • Because of their small masses and thermal energies, gas molecules can somewhat counteract gravity, which is why gases don’t just sit on the surface • Nonetheless, gravity causes the gases in the atmosphere to “press down” on the surface. This is atmospheric pressure.

  10. Atmospheric Pressure • To calculate atmospheric pressure, we calculate the mass of a 1m2 column of air extending through the entire atmosphere. • That column would have an approximate mass of 104 kg. • The force exerted on that 1m2 area would be: • F= ma = (104kg)(9.8 ms-2) = 105 N • Then, P = 105N/1m2 = 105N/m2 = 105Pa • SI unit of pressure is the Pascal(Pa). Related units are bar, mmHg, and atmospheres.

  11. Units of Pressure 1 atm = 760 mm Hg

  12. Defining Work in Terms of Pressure and Volume • Pressure is defined as force/area. We can manipulate this expression: • Plugging this new expression of force into the work equation: • In chemistry, it is more useful to think of work in terms of volume, rather than distance since the change in volume of a system is much easier to quantify. • Remember, work must be in Joules. For a volume in liters and a pressure in atmospheres, the equation above would yield work in units of L•atm. Volume = area x height

  13. Examples • Calculate the work, in Joules, required to compress a gas from a volume of 5.0 L to 2.5 L with a constant pressure of 1.5 atm? • Calculate the work, in Joules, required to expand a gas from a volume of 2.5 L to 5.0 L against a constant pressure of 1.5 atm? ΔV = -2.5 L P = 1.5 atm The system is compressed, so the surroundings do work on the system. Work is positive ΔV = 2.5 L The system expands, so the system does work on the surroundings. Work is negative

  14. Conservation of Energy Energy is never created or destroyed, merely converted between forms and transferred from place to place. The total energy of the universe is finite.

  15. Conservation of Energy, contd. • Imagine we have a perfectly insulated box. Inside that box is trapped air, a light bulb, wires and a battery • When a connection is made between the battery and bulb, chemical energy is converted to electrical energy which transmits the wiring, yielding light • Light interacts with the air, generating heat • Once the battery is dead, the amount of heat energy in the box is exactly equal to the total chemical energy that was initially in the battery • Energy is conserved! • However, the chemical energy was a useful form for generating light. The heat is not as useful

  16. Heat • Energy that is not used to perform work or stored is lost as heat. With respect to mechanical systems, this is an undesirable but inevitable feature of nature. • For example, only 25% of the thermal energy released from the combustion of gasoline is actually used to perform work (move the car). • The rest is lost as heat to the surroundings (heating up the car and the air around the car). • Without this limitation, cars could easily exceed 100 mpg. • Heat, denoted with the symbol q, will always flow when there is a temperature difference between the system and the surroundings (from high temp to low temp)

  17. Heat • The sign convention for heat is the same as for work. • When heat is lost by the system, it must be transferred to the surroundings. In such a case, qsys is negative (qsurr is positive). A reaction in which qsys < 0 is called an exothermic reaction Fuel + O2 CO2 + H2O + heat H2O CaCl2(s) Ca2+(aq) + 2Cl-(aq) + heat

  18. Heat • When heat is absorbed by the system, it is taken from the surroundings, and qsys is positive (qsurr is negative). When qsys >0, the reaction is endothermic. Al(NO3)3(s) + heat ------> Al3+(aq) + 3NO3-(aq) H2O(s) + heat -----> H2O (L)

  19. First Law of Thermodynamics • The change in the internal energy of a system is given by the first law of thermodynamics: • This says that, in going from some initial state, to some final state, the total energy change must equal the total energy transferred as heat and work

  20. State Functions • In thermodynamics, we must distinguish between those terms that are state functions, and those that are not. • A state function is a property of the system that is based on the system’s condition, but NOT on the path taken to reach that condition. • Example: Imagine that our system is 50g of water at 25oC. We can attain this condition by cooling water hotter water to 25oC, or by heating cooler water to 25oC. The value of U is the same either way. So U is a state function

  21. Heat and Work are NOT State Functions • This may seem confusing because we said previously that U is a state function, and ΔU=q+w. • However, the amount of heat and work transferred will depend on the path, although their sum does not. • Take the system (battery) below. ΔU of the battery is the change in energy between the “fully-charged” state and the “dead” state. The value of ΔU doesn’t depend on HOW the energy is used, but q and w do. Fully Charged Heat Heat ΔU Battery is short circuited. Coil heats up. No work is done. All energy is lost as heat. Work Some energy is used to do work (turn fan). Dead

  22. Enthalpy • Enthalpy (H) is the heat absorbed or released during a reaction at constant pressure(ΔH = qP). Enthalpy is a state function. • Now, we see that a positive value of ΔH indicates an endothermic reaction, and negative ΔH indicates an exothermic reaction. • For a reaction at a constant pressure of 1 bar (.987 atm), we can define the standard enthalpy (denoted by ‘o’ symbol) as: • Enthalpy is a molar quantity, and is usually expressed in

  23. Calculating ΔHo • The enthalpy change associated with a reaction can be determined by several methods, depending on the information provided. These methods include: • Stoichiometric calculation, using a standard molar enthalpy • Hess’s Law, using standard reactions of known ΔHo • Heats of formation • Bond enthalpy calculations

  24. Stoichiometric Enthalpy Calculations Enthalpy Diagram • Lets take the combustion of hydrogen gas as an example of an exothermic reaction. 2H2(g) + O2(g) -----> 2H2O(g) ΔHorxn= -483.6 kJ/mol Enthalpy is an extensiveproperty. The magnitude of ΔH is proportional to the amount of reactant consumed. So, in excess O2, 2 mol of H2 will evolve -483.6 kJ. 4 mol of H2 will evolve double that. Enthalpy of the reverse reactionhas the opposite sign. The decomposition of 2 moles of H2O into 2 moles of H2 and 1 mole of O2 is +483.6 kJ. 2H2(g) + O2(g) ΔH < 0 exothermic 3. The state of the reactants matters. If liquid water formed instead of steam, ΔH would be different. 2H2O(g)

  25. Example • Calculate the heat released when 30.0 g of iron(III)oxide is combined with 15.0 g of Al(s) at 1 bar. • The negative sign of tells us that the products have less “heat content” than the reactants. Therefore, heat is transferred to the surroundings (exothermic). • To determine the enthalpy change of the reaction, we first determine the moles of the limiting reactant Fe2O3(s) + 2Al(s) -----> Al2O3(s) + 2 Fe(s) ΔHorxn= -851.5 kJ/mol Limiting Reactant

  26. Example, continued Fe2O3(s) + 2Al(s) -----> Al2O3(s) + 2 Fe(s) ΔHorxn= -851.5 kJ/mol 0.556 mol • Now, knowing the limiting reactant, we use the stoichiometry in the reaction to determine the heat released. • We see that there are -851.5 kJ of heat evolved for every 2 mol of Al consumed. We set up the calculation as shown below:

  27. Hess’s Law • For multi-step processes, enthalpy changes are additive. • Lets take the following reaction: • This reaction is two steps, and proceeds through an intermediate (short-lived species), as shown below.

  28. Hess’s Law, continued • We can add these reactions together, canceling intermediates on opposing sides of the arrow. • Since we know the enthalpy change in each step, we can determine the overall enthalpy change of the reaction in question. • This additive property is known as Hess’s Law.

  29. Hess’s Law • The usefulness of Hess’s Law is that it allows you to express the value of ΔHo of an unknown reaction as the sum of the ΔHo values of known standard reactions.

  30. Example • Given the following ΔHorxn values, calculate ΔHorxn for the reaction: 2SO2(g) + O2(g)  2SO3(g) (1) SO2(g) ---> S(s) + O2(g) ΔHorxn (1) = 296.8 kJ/mol (2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHorxn (2) = -791. 4 kJ/mol It is necessary to multiply reaction (1) by a factor of 2 in order to get 2 moles of SO2(g), as well as to cancel out S(s) and 2 moles of O2(g). (1) 2[SO2(g) ---> S(s) + O2(g) ]ΔHorxn (1) = 2 (296.8 kJ/mol) (2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHorxn (2) = -791. 4 kJ/mol (3) 2SO2(g) + O2(g) ---> 2SO3(g) ΔHorxn (3) = -197.8 kJ/mol

  31. Group Example • Given the following values of ΔHorxn: • Calculate the value of ΔHorxn for the following reaction = -157.3 kJ/mol = -33.2 kJ/mol

  32. = -157.3 kJ/mol = -33.2 kJ/mol • The final reaction has 4 mol of Cu(s) on the reactant side. Reaction (1) must be multiplied by 4. • The final reaction has 2 mol of NO2(g). This indicates that reaction (2) needs to be multiplied by 2. = -629.2 kJ/mol = -66.4 kJ/mol = -695.6 kJ/mol

  33. Group Example 2 • Given the following values of ΔHorxn: • Calculate the value of ΔHorxn for the following reaction: = -639.4 kJ/mol = -887.0 kJ/mol (3) • Reaction (1) must be reversed and multiplied by ½ . Reaction (2) must also be multiplied by ½ . This gives us 1 mole of PCl3 on the left and 1 mole of PCl5on the right. = +319.7 kJ/mol = -443.5 kJ/mol (3) = -123.8 kJ/mol

  34. Enthalpies of Formation • The enthalpy change associated with a the formation of a molecules is called the enthalpy change of formation, ΔHf (also called heat of formation). • For a given reaction, if you know the enthalpies of formation of the reactants and products:

  35. Enthalpies of Formation For pure elements, and for elements in their most stable form, ΔHof = 0 kJ/mol • This means pure metals, diatomic gases, and pure liquids like Hg(L) and Br2 (L),

  36. The enthalpies of formation are exactly known for many elements.

  37. Example • Calculate the enthalpy change for the combustion of benzene, C6H6 to form CO2 and H2O using standard heats of formation. Using the heats of formation of the reactants and the products, we can directly calculate

  38. Molar Bond Enthalpies • The enthalpy change for the complete dissociation of water: is = + 925 kJ/mol. • In a water molecule, there are 2 O-H bonds. The positive value of enthalpy indicates that heat (energy) must be absorbed by the system in order to break these bonds. • Therefore, we can estimate that the molar bond enthalpy, Hbond, of each O-H bond (energy needed to break the bond) is 463.5 kJ/mol

  39. Molar Bond Enthalpies • In general, if we consider a reaction in the most basic sense, then we would imagine that the formation of a compound proceeds as follows: Combination of H and O to form H2O. Dissociation of H2 and O2 into H and O energy released when product bonds form energy absorbed to break reactant bonds +

  40. Table of common bond enthalpies. These are the energies to break the bonds. When you formbonds, the sign of H will be negative.

  41. Example • Using molar bond enthalpies, estimate of the following: • As you can see, this reaction proceeds by breaking one C-H bond, and a Cl-Cl bond, followed by the formation of a C-Cl bond and an H-Cl bond H H Cl Cl C H Cl H H C Cl H H H Energy released. New bonds formed. Energy absorbed. Bonds broken + -331 kJ/mol + (-431 kJ/mol) 414 kJ/mol + 243 kJ/mol =-105 kJ/mol Energy released by new C-Cl and H-Cl bonds Energy to break bonds

  42. Example • Use molar bond enthalpies to calculate the enthalpy of formation of hydrazine, N2H4. Compare this with the known value, 95.4 kJ/mol. • To do this, we do not need to know the exact reaction leading to the formation of hydrazine. We need only to know the bonds involved, and we can make a good estimation from there. • Lets just assume the following balanced reaction: •• •• H H N N N H H N •• •• + 4(-390 kJ/mol) + (-159 kJ/mol) (945 kJ/mol) + 2(435 kJ/mol) Energy released by forming 4 N-H bonds and 1 N-N single bond Energy to break N-N triple bond and 2 H-H single bonds 96 kJ/mol

  43. Heat Capacity • The heat capacity of a substance, denoted as ‘c’, is defined as the heat energy required to raise the temperature of a sample one degree Kelvin at constant pressure, usually in J/oK. • As you may recall, at constant pressure, ΔH = qp. You can therefore interchange heat and enthalpy in the equation above. ΔT = TF - Ti • Example: When 4.219 kJ of heat is added to 36 grams of water at one bar, the temperature of the water increases from 10oC to 38.05oC. Find the heat capacity of 36 g of water.

  44. Molar Heat Capacity • The molar heat capacity, denoted as Cp, defines the energy needed to raise a moleof a substance one degree K. • We can easily find Cp of water from the previous example by converting the 36 g to moles, then dividing heat by moles. • Using our previous relationship for heat and heat capacity, we obtain: where n is the number of moles of substance in the system. Thus, using molar heat capacities, we can directly calculate heat absorbed.

  45. Example • Calculate the joules of heat required to raise the temperature of 150 kg of water from 18oC to 60oC, assuming no loss to the surroundings. • Note: A change of x oC is exactly equal to a change of x oK. Thus, the ΔT value could have also been reported as 42 oC. The units of Cp can also reflect this.

  46. Heat Transfer • Heat ALWAYSflows from hotter substances to cooler substancesuntil both substances are the same temperature. • Note: heat and temperature are not the same.Heat is energy. Temperature determines the flow of heat. • In an insulated system, no heat is transferred to the surroundings. Therefore, if we have two materials in contact, the heat lost by one material is exactly equal to the heat gained by the other. • Remember: Heat loss is negative. Heat gain is positive

  47. Example • 50g of copper metal at 80oC is placed in 100 mL of water at 10oC within an insulated system. What will the final temperature of the water and copper be? Use the molar heat capacities of Cu and H2O(L) as 24.4 J/moloK and 75.3 J/moloK,

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