Thermodynamics

Calculate the ∆Hrxn for the following chemical reaction. N2 + 3H2 → 2NH3

-93 kJ/mol

Calculate the change in entropy of 3 moles of liquid water if you heat it from 5˚C to 95˚C. The molar heat capacity of liquid water is 75.38 J/molK.

63.4 J/K

True or False: If a process in non-spontaneous, then there is nothing we can do to make this process occur.

False

Entropy

a state function that is a measure of the dispersal of energy

Open system

a system that allows both energy (heat flow and work) and matter to freely exchange with surroundings

A spontaneous process is associated with

an positive change in the entropy of the universe

An airtight greenhouse would be a rough example of a(n) _______ system.

closed The greenhouse is airtight which suggests no matter can enter or exit but it does allow light energy into the building. This is a rough example of a closed system.

Isolated system

contains both energy and matter within the system. Nothing is exchanged with the surroundings

The first law of thermodynamics states that ________ is conserved.

energy The first law of thermodynamics is where we derive the statement that "energy is never created nor destroyed."

The more microstates available for a system, the

higher the entropy of the system

q=

mc∆T

The change in enthalpy for a process at constant __________ is exactly equal to the heat (flow of thermal energy) between the system and the surroundings for that process.

pressure One way to define a change in enthalpy is to say that it is equal to the amount of heat involved in a process that occurs at constant pressure.

∆U

q + w

As a scientist, we pick a specific part of the world that we are interested in and we call it "the _________."

system The system is what we define as our specific focus.

Calorimeters measure the change in the _____________ of the water we can quantify the __________ of the chemical reaction.

temperature heat

Molar heat capacity

the energy required to raise the temperature of one mole of a substance by one degree Celsius q= nCΔT

The definition of internal energy is ΔU = q + w Which of these three values are state functions? Select all of the correct answers.

ΔU ΔU is a state function meaning that internal energy only depends on the current state of the system, it is path independent. q and w (heat and work) are both path functions.

When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston with an external pressure of 2.00 atm, the nitrogen gas expands from 2.00 to 5.00 L. What is the change in internal energy of this system?

+ 1.39 kJ 1. Write down all units given. q= 2 kJ Pext= 2 atm Vi= 2 L Vf= 5 L ΔV= 3 L ΔU= x 2. Calculate work. w= -PΔV -2 x 3= -6 3. Convert to kilojoules using L/atm. -6 x 101.325= -607.95 J/1000= -0.60795 kJ 4. Calculate ΔU. ΔU= q + w 2 + (-0.60795)= 1.39 kJ

Calculate the enthalpy change for the following chemical equation. 2SO2(g) + O2(g) → 2SO3(g) Use the following thermochemical data to solve for the change in enthalpy. ΔHf for SO2(g) = -16.9 kJ/mol ΔHf for SO3(g) = -21.9 kJ/mol

-10 kj molrxn

Calculate the standard enthalpy change for the following chemical equation. 2HCl (g) + F2 (g) → 2HF (l) + Cl2 (g) Use the following thermochemical equations to solve for the change in enthalpy. 4HCl (g) + O2 (g) → 2H2O (l) + 2Cl2 (g) ΔH° = -202.4 kJ/mol rxn ½ H2 (g) + ½ F2 (g) → HF (l) ΔH° = -600.0 kJ/mol rxn H2 (g) + ½ O2 (g) → H2O (l) ΔH° = -285.8 kJ/mol rxn

-1015.4 kj/molrxn

Nitric acid can be manufactured in a multi-step process, during which nitric oxide is oxidized to create nitrogen dioxide. 2NO (g) + O2 (g) → 2NO2 (g) Calculate the standard reaction enthalpy for the above reaction using the following thermodynamic data. N2 (g) + O2 (g) → 2NO (g) ∆H˚1 = 180.5 kJ/molrxn N2 (g) + 2O2 (g) → 2NO2 (g) ∆H˚2 = 66.4 kJ/molrxnq

-114.1 kj/molrxn

A 0.10g piece of chocolate cake is combusted with oxygen in a bomb calorimeter. The temperature of 4000g of H2O in the calorimeter is raised by 0.32 K. (The specific heat of the water is 1.0 cal/gK and the heat of vaporization of water is 540 cal/g). What is ∆U in units of kcal/g for the combustion of chocolate cake? Assume no heat is absorbed by the hardware of the calorimeter.

-12.8 kcal/kg 1. Write down all units given. m= 0.10 g mH2O= 4000 g ΔT= 0.32 K C(H2O)= 1 cal/gK ΔHvap= 540 cal/g 2. Find the amount of heat responsible for the increase in water temp. q= MCΔT 4000 x 1 x 0.32= 1280 cal= amount of heat release by the reaction (exothermic) 3. Use amount of system given to convert calories to in kcal. -1280 cal/0.10 g Cake x 1000 cal= -12.8 kcal/g

Using the bond energy data provided, calculate ΔH for the following reaction: H2 (g) + Cl2 (g) → 2HCl (g)

-186 kJ/mol

The standard molar enthalpy of formation of NH3 (g) is -46.11 kJ/mol. What is the standard molar internal energy of formation of NH3 (g)?

-43.63 kJ/mol

A 1.00 g sample of n-hexane (C6H14) undergoes complete combustion with excess O2 in a bomb calorimeter. The temperature of the 1502 g of water surrounding the bomb rises from 22.64°C to 29.30°C. The heat capacity of the hardware component of the calorimeter (everything that is not water) is 4042 J/°C. What is ΔU for the combustion of n-C6H14? One mole of n-C6H14 is 86.1 g. The specific heat of water is 4.184 J/g·°C.

-5.9 x 10^3 kJ/mol 1. Write down all units. m= 1 g C6H14 mH2O= 1502 g ΔTH2O= 29.3- 22.64=6.6 Cs= 4042 ΔU= x mwC6H14= 86.1 g 2. Recognize: q= heat absorbed by the water + heat absorbed by the calorimeter q=mcΔT (1502 x 4.184 x 6.6) + (4042 x 6.6)= 68,153.2 J 3. Convert to figure out how much heat is released per mole. 68,153 J x 1 kJ/1000 J x 86.1 g /1 mol= (divided by 1000)= 5,867.99 or 5.9 x 10^3 4. Remember that the system lost heat, so ΔU must be negative.

When 0.485 g of compound X is burned completely in a bomb calorimeter containing 3000 g of water, a temperature rise of 0.285°C is observed. What is ΔU of the reaction for the combustion of compound X? The hardware component of the calorimeter has a heat capacity of 3.81 kJ/°C. The specific heat of water is 4.184 J/g·°C, and the MW of X is 56.0 g/mol.

-538 kJ/mol 1. Write down all units given. 2. Remember that q= mcΔT. Calculate the q of water. 3000 x 4.184 x 0.285= 3,577.32/1000= 3.57 kJ 3. Calculate the q of the calorimeter. 3.5 x 0.285= 0.9975 kj x 1000= 997.5 J 4. Calculate the total heat of the reaction. qrxn= - (qcalorimeter + q bomb) - (997.5 + 3577.32 kJ)= -4,574.82 kJ 5. Calculate moles of x. 0.485/56= 0.0086 mol 6. Divide qrxn by moles of x. -4574.82/0.0086/100= 531.9 kJ/mol

Calculate the standard enthalpy change for the following chemical equation. 4FeO (s) + O2 (g) → 2Fe2O3 (s) Use the following thermochemical equations to solve for the change in enthalpy. Fe (s) + ½ O2 (g) → FeO (s) ΔH = -269 kJ/mol 2Fe (s) + 3/2 O2 (g) → Fe2O3 (s) ΔH = -825 kJ/mol

-574 kj/mol

For a certain reaction at constant pressure, the change in internal energy is -52 kJ. In addition, the system does 46 kJ of expansion work. What is ΔH for this process?

-6 kJ 1. Write down all units. ΔU= -52 w= -46 ΔH= x 2. We know that ΔU= q + w -52= -46 + q -(-46) ΔH= -6 kj

What is the value of work when a piston of volume 0.2 L expands against an external pressure of 200 kPa to a volume of 3.4 L?

-640 J 1. Write down all units given. Vi= 0.2 L Vf= 3.4 L Pext= 200 kPa ΔV= 3.2 2. Remember w= -PΔV -200 x 3.2= -640 J

A CD player and its battery together do 500 kJ of work. In addition, the battery releases 250 kJ of energy as heat. Furthermore, the CD player releases 50 kJ as heat due to friction from spinning. What is the change in internal energy of the system, where the system is regarded as the battery and CD player together.

-800 kJ To calculate internal energy we must sum the heat and the work. Here the system does 500 kJ of work. This means that w = -500 kJ. The system releases a total of 300 kJ of heat (q = -300 kJ). Altogether this means that ∆U = -300 kJ + (-500 kJ) = -800 kJ. 1. Pay close attention to the vocabulary in the problem. The CD Player and Battery DO 500 kj of work. w= -500 kJ Battery RELEASES 250 kJ of energy as heat. q= -250 kJ CD player RELEASES 50 kJ as heat. q= -50 kJ 2. Add your heat released together. -250 kJ + -50 kJ= -300 kJ. 3. Use your equation for change in energy. Δsys= q + w -300 kJ + -500 kJ= -800 kJ

Calculate the standard reaction enthalpy of the following reaction: SO2 (g) + ½ O2 (g) → SO3 (g)

-99 kJ mol/rxn

w=

-PΔV

How much heat is required to heat 1 mol of a sample from 150 to 250 degrees Celsius?

.... in notes

What mass of liquid ethanol (C2H5OH) must be burned to supply 500 kJ of heat? The standard enthalpy of combustion of ethanol at 298 K is -1368 kJ/mol.

16.8 g 1. Write down all units. mwC2H5OH= 46.07 g ΔH= 500 kj ΔHrxn= -1368 kj/mol 2. Recognize what the equation is telling you. Ethanol is burned, ΔH= -500 3. Remember ΔH=nΔHrxn -500= -1368x /-1368= x= 0.36 mol C2H5OH 0.36 x 46.07 g/1 mol=16.8 g C2H5OH

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (-97.6˚C) and 1 atm? ∆H˚fus = 3.17 kJ/mol.

18 j/K

Calculate the standard reaction enthalpy for the following chemical equation. CH4 (g) + H2O (g) → CO (g) + 3H2 (g) Use the following thermochemical equations to solve for the change in enthalpy. 2H2 (g) + CO (g) → CH3OH (l) ΔH° = -128.3 kJ/mol 2CH4 (g) + O2 (g) → 2CH3OH (l) ΔH° = -328.1 kJ/mol 2H2 (g) + O2 (g) → 2H2O (g) ΔH° = -483.6 kJ/mol

206.1 kj/mol

Burning 1 mol of methane in oxygen to form CO2 (g) and H2O (g) produces 803 kJ of energy. How much energy is produced when 3 mol of methane is burned?

2409 kJ 1. Write down all units. ΔH= 803 kj 1 mol CH4 3 mol methane burned 2. Recognize what you're given. 1 mol= 803 kJ energy 3 mol x 803= 2409 kJ

When 4.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston at an external pressure of 3.00 atm, the nitrogen gas expands from 1.00 L to 4.00 L against this constant pressure. What is ΔU for the process? Note: 1 L*atm = 0.1013 kJ.

3.09 kJ 1. Write down all units given. q= 4 kJ Pext= 3 atm Vi= 1 L Vf= 4 L ΔV= 3 L 1 L/atm= 0.1013 2. Calculate w. w= -PΔV -3 x 3 x 101.325= -911.925/1000= -0.912= w 3. Calculate ΔU. ΔU= q + w 4 + (-0.912)= 3.088

monoatomic gas

3/2RT Monatomic gases just have 3 modes of freedom, Multiplying those modes of freedom with the molar internal energy of one mode of freedom (Um, single mode = (1/2)RT) gives you these answers.

A 100g sample of hot copper is placed in a coffee cup calorimeter containing 100 grams of water at room temperature. After some time the temperature of the water and the copper become a constant at 50˚C. Calculate the initial temperature of the copper piece. The specific heat of copper is 0.385 J/g˚C. The specific heat of liquid water is 4.184 J/g˚C.

322 degrees Celsius 1. Write down all units. M Cu= 100g M H2O= 100 g Tf= 50 degrees Celsius Ti= ? C(Cu)= 0.385 J C(H2O)= 4.184 J Room temp= 25 degrees celsius 2. Recognize that the heat lost by the system must be gained by the surroundings. Heat gained by the water= Heat lost by Cu 3. Calculate ΔT. 50- 25= 25 degrees celsius 4. Remember mCΔT. 100 x 0.385 x (50-25) = 100 x 4.184 x (Ti - 50) 5. Solve for Ti Ti- 50= 4.184 x 25/0.385 Ti= 322

Calculate the average S‒F bond energy in SF6 using the following ΔHf values: SF6 (g) = -1209 kJ/mol S (g) = 279 kJ/mol F (g) = 79 kJ/mol

327 kJ/mol bonds The S-F bond energy in SF6 is defined as the ΔH for the reaction in which one mole of S-F bonds in SF6 are broken to give gaseous atoms. Each molecule of SF6 has six S-F bonds. So, each mole of SF6 has six moles of S-F bonds. Thus, we need to break the S-F bonds in only 1/6 mol of SF6. So, we need the ΔH for the following reaction:

nonlinear gas molecules

3RT Nonlinear gas molecules have six, Multiplying those modes of freedom with the molar internal energy of one mode of freedom (Um, single mode = (1/2)RT) gives you these answers.

linear gas molecule

5/2RT Linear gas molecules have 5, Multiplying those modes of freedom with the molar internal energy of one mode of freedom (Um, single mode = (1/2)RT) gives you these answers.

Consider the following specific heat capacities: H2O (s) = 2.09 J/g·°C H2O (l) = 4.18 J/g·°C H2O (g) = 2.03 J/g·°C The heat of fusion for water is 334 J/g and its heat of vaporization is 2260 J/g. Calculate the amount of heat required to convert 93 g of ice at -36°C completely to liquid water at 35°C.

52 kJ 1. Write down all units CsH2O= 2.09 J ClH2O= 4.18 J CgH20= 2.03 J Hfus= 334 j/g Hvap= 2260 j/h q= x mass of ice= 93 g Ti= -36 Tf= 35 2. Recognize what the question is asking of you. Raise the temperature of the ice, Melt the ice, heat the liquid. Heat the ice= mCsΔT Melt the ice= q= mΔH Heat the liquid= mCsΔT Add all energies together. 3. Plug into equations. 93 x 2.09 x (0- (-36))=6,997.32/1000= 6.99 93 x 334=31062/1000=31.062 J 93 x 4.18 x 35=13605.9/1000=13.605 J 4. Add all the energies together. 31.062 + 6.99 + 13.6= 51.657 J

What mass of propane gas (C3H8) must be burned under constant pressure to supply 2775 kJ of heat? The standard enthalpy of combustion of propane at 298 K is -2220 kJ/mol.

55 g 1. Write down all units given. ∆H= 2775 kJ ∆Hrxn= -2220 kJ/mol mw C3H8= 44.1 g 2. Recognize what the problem is telling you. propane gas is BURNED, releasing energy, exothermic ∆H= -2775 kJ 3. Remember ∆H= n∆Hrxn -2775= (n)(-2220) 4. Solve for moles. -2775/-2220= 1.25 moles of C3H8 5. Convert moles to grams. 1.25 moles C3H8 x 44.1 g C3H8/1 mol C3H8= 55.125 g C3H8

Consider the following chemical equation: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l) ΔH = -2220 kJ/mol rxn How much thermal energy is given off when 11.0 g of propane gas (C3H8) is burned at constant pressure?

555 kJ 1. Write down all units. ΔH = -2220 kJ/mol rxn 1 mol C3H8 11 g C3H8 2. Convert mass of propane to moles of propane. 11 g x 1 mol/44.1 g= 0.25 mol 3. Remember ΔH = nΔHrxn 0.25 x -2220=-555 problem says given off, energy will be positive to the surroundings.

Calculate the energy required to heat 10.0 g of solid ice at 0.0˚C to liquid water at 80.0˚C. It may help to draw a heating curve and to use some of the following values: Cs,ice : 2.09 J/g˚C Cs,water : 4.184 J/g˚C Cs,steam : 2.03 J/g˚C ∆Hfus, water : 334 J/g ∆Hvap, water : 2260 J/g

6.7 J 1. Write down all your units. Cs Ice= 2.09 j/g Cs Water= 4.184 j/g Cs Steam= 2.03 J/g ∆Hfus= 334 J/g ∆HVap= 2260 j/g m= 10 ∆T= 80-0= 80 2. Figure out what the equation has told you to do. Heat 10 g of solid ice to liquid water at 80 degrees celsius. How do you do this? Melt the ice, then heat the water. 3. Find out what energy is needed to melt the ice. qfus = m∆H 10 x 334= 3340 kJ 4. Find out how much energy is needed to heat the water. q= mCH2O∆T 10 x 4.184 x (80-0)= 3,347.2 5. Find out total energy by adding energy from melting and heating together. 3340 + 3347.2= 6,687.2 kJ 6. Convert answer into joules. 6,687.2/1000= 6.7 joules

A 100 W electric heater (1 W = 1 J/s) operates for 11 min to heat the gas in a cylinder. At the same time, the gas expands from 1 L to 6 L against a constant atmospheric pressure of 3.527 atm. What is the change in internal energy of the gas?

64.21 J 1. Write down all units given. E= 100 j/s Time= 11 min Vi= 1 L Vf= 6 L ΔV= 6-1= 5 L Pext= 3.527 atm ΔU= x 2. Convert time to seconds. 11 x 60= 660 seconds 3. Calculate the heat transferred to the cylinder. q= 100 x 660= 66,000/1000= 66 J 4. Calculate work. w= -PΔV -3.527 x 5= -17.365 L/atm 5. Convert work to kilojoules. -17.365 l atm x 101.325 J/1 atm= -1,786.86 J/1000= -1.786 kJ 6. Calculate internal energy of the gas. ΔU= q + w 66 + (-1.786) =64.214 J

The specific heat for liquid argon and gaseous argon is 25.0 J/mol·°C and 20.8 J/mol·°C, respectively. The enthalpy of vaporization of argon is 6506 J/mol. How much energy is required to convert 1 mole of liquid Ar from 5°C below its boiling point to 1 mole of gaseous Ar at 5°C above its boiling point?

6735 j 1. Write down all units. CslAr= 25 j/mol CsgAr= 20.8 j/mol ΔHvap= 6506 J/mol ΔT= 5 2. Recognize what the question is asking. Raise the temperature to its boiling point. Convert from liquid to gas. Raise the temperature 5 degrees above boiling point. q= nCsΔT q= nΔH q=nCsΔT 3. Plug into equation. 1 x 25 x 5= 125 1 x 6506= 6506 1 x 20.8 x 5= 104 4. Add up all your values of q. 125 + 6506 + 104= 6,735 j

Carbon monoxide reacts with oxygen to form carbon dioxide by the following reaction: 2CO(g) + O2(g) → 2CO2(g) ΔH for this reaction is -135.28 kcal. How much heat would be released if 12.0 moles of carbon monoxide reacted with sufficient oxygen to produce carbon dioxide? Use only the information provided in this question.

812 kCal 1. Write down all units given 2 moles CO 1 mol O2 2 Moles CO2 ΔH= -135.28 n= 12 moles 2. Remember that elements in their standard states have ΔH of 0. ΔHO2= 0 3. Remember that Carbon monoxide reacts with oxygen to form carbon dioxide by the following reaction: ΔH= ΔHproducts + ΔHreactants -135.28= x + 0 +x2 x= -67.64 -67.64 x 12= -811.68

Calculate the change in temperature produced by the addition of 1 kcal of heat to 100 g of steel. Csteel= 0.118

84.7 degrees celsius 1. Write down all units. q= 1 m= 100 C= 0.118 2. Remember q=mcΔT 3. Calculate the amount needed to raise the 100g by 1 degree. 100 x 0.118= 11.8 c 4. Convert kcal to cal. 1 kcal= 1000 calories. 5. Divide calories given by calories needed. 1000/11.8= 84.75

A piece of metal with a mass of 22 g at 92 °C is placed in a calorimeter containing 53.7 g of water at 21 °C. The final temperature of the mixture is 55.3 °C. What is the specific heat capacity of the metal? Assume that there is no energy lost to the surroundings.

9.5 J 1. Write down all units given. n= 22 g T1metal = 92 T2water = 21 Tf=55.3 ΔTmetal= 92-55.3= 36.7 degrees celsius ΔTwater=55.3-21= 34.3 degrees celsisus mass of H2O= 53.7 2. Understand what the problem is telling you. The heat lost by the metal is gained by the water. q= mcΔT= mcΔT 3. Plug in values and solve for C. 22 x c x 36.7= 53.7 x 1 x 34.3 807.4c= 1,841.91/807.4= 2.28 c= 2.28 4. Using specific heat of water, convert specific heat of the metal to joules from calories. 1 cal = 4.184 J 2.28 x 4.184= 9.54 J

Which of the following is not a state function? A. Heat B. Volume C. Pressure D. Entropy E. Free energy

A. Heat Heat and work are both path dependent properties!

Which of the following statements concerning the first law of thermodynamics is/are true? Select all of the correct answers. A. Internal energy lost by a system is always gained by the surroundings. B. The universe is an isolated system. C. The internal energy of the universe is always increasing.

A. Internal energy lost by a system is always gained by the surroundings. B. The universe is an isolated system. Statement I is false; the first law states that the energy of the universe is conserved, in other words a constant value. Statement II and III are true; internal energy in the uni- verse is conserved, and thus energy lost by the system is always gained by the surroundings. The universe is the most obvious example of an isolated system in that energy and matter are conserved in the universe.

Which of the following is not a way of defining enthalpy or change in enthalpy? A. The heat transfer at constant volume B. ∆H = Hfinal - Hinitial C. A state function that is very nearly equivalent to the "energy" of the system D. ∆H = ∆U + P∆V

A. The heat transfer at constant volume

A measurement of the heat produced when a known amount of sugar burns in a constant volume calorimeter would enable us to first calculate the A. molar internal energy change for the combustion of sugar. B. work done per mole by the combustion of sugar. C. molar entropy change form the combustion of sugar. D. molar Gibbs' Free Energy change for the combustion of sugar. E. molar enthalpy change for the combustion of sugar.

A. molar internal energy change for the combustion of sugar

Which of the following properties of a substance does not change when the amount of the substance changes? A. temperature B. heat content C. mass D. volume

A. temperature The temperature of a substance does not depend on how much of that substance is present. Temperature is an intensive property that is not dependent on the amount of matter.

Water has a heat capacity that is approximately ten times larger than the heat capacity of copper metal. Assuming 100 J of energy is deposited into equal masses of the two materials as heat energy, what is true about the change in temperature? A. The answer cannot be determined without knowing the actual masses of the two materials. B. The temperature change for Cu will be 10-fold higher than that for water. C. The temperature change for water will be 10-fold higher than that for Cu. D. The temperature changes will be the same since each receives the same amount of heat. E. The answer cannot be determined without knowing the initial temperatures of the two materials.

B. The temperature change for Cu will be 10-fold higher than that for water. Heat is proportional to a change in temperature with the heat capacity as the proportionality constant. So if two objects absorb the same amount of heat, the temperature changes of the objects are dictated by their respective heat capacities. Copper has a heat capacity that is 10 times smaller than that of water, so it will experience 10 times greater a temperature change comparatively.

Which of the following substances have ΔHf° = 0? Select all of the correct answers. Na (s) HCl (g) HCl (aq) F2 (g) C (s, graphite) C (s, diamond)

C (s, graphite) ΔHf° = 0 for elements in their standard states. Carbon exists naturally as solid graphite. F2 (g) ΔHf° = 0 for elements in their standard states. Fluorine exists naturally as a diatomic gas. Na (s) ΔHf° = 0 for elements in their standard states. Sodium exists naturally as a monatomic solid. ΔHf° = 0 for elements in their standard states. Fluorine exists naturally as a diatomic gas. Sodium exists naturally as a monatomic solid. Carbon exists naturally as solid graphite.

Which of the following measurements is of an intensive property? A. 11.4 grams B. 80 milliliters C. 100 Kelvin D. 16 cubic feet

C. 100 kelvin Intensive properties are independent of the amount of matter present. The temperature indicating of a substance is the same regardless of the size (large or small) of the sample.

Which of these reactions below is a correctly written standard formation reaction? A. 2H2 (g) + O2 (g) → 2H2O (l) B. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) C. 2Fe (s) + 1.5 O2 (g) → Fe2O3 (s) D. C (diamond,s) + 2H2 (g) → CH4 (g)

C. 2Fe (s) + 1.5 O2 (g) → Fe2O3 (s)

Which of the following is true of a general thermodynamic state function? A. The change in the value of a state function is always negative for a reaction. B. The value of the state function remains constant. C. The change of the value of a state function is dependent on the path of a process. D. The change of the value of a state function is independent of a path of a process.

D. The change of the value of a state function is independent of a path of a process. A change in a state function describes a difference between the two states. It is independent of the process or pathway by which the change occurs.

An endothermic reaction corresponds to one in which A. ∆Hsystem is negative; heat is released B. ∆Hsystem is negative; heat is absorbed C. ∆Hsystem is positive; heat is released D. ∆Hsystem is positive; heat is absorbed

D. ∆Hsystem is positive; heat is absorbed In endothermic reactions, heat flows into a system, or heat is absorbed. Since the amount of energy in the system increases, ∆H is positive.

Which of the following will best help determine the direction of heat flow in a system? A. work B. pressure C.internal energy D. enthalpy E.temperature

E. temperature

What is the first law of thermodynamics?

Energy cannot be created or destroyed. Energy is conserved. Energy merely changes forms.

Suppose you have 5 moles of a molecule that has 3 possible orientations. How would you set up an equation to solve for the residual entropy of these 5 moles of this substance?

S = 5•R•ln(3)

A system had 150 kJ of work done on it and its internal energy increased by 60 kJ. How much energy did the system gain or lose as heat?

The system lost 90 kJ of energy as heat. 1. Write down all units given. w= 150 kJ ΔU= 60 q= x 2. Identify the language used in the problem. Work done on it, w is positive (150 kJ) ΔU increased. 3. Remember ΔU= q + w. ΔU is given and so is w, so solve for q. 60= q + 150 60-150= q -90= q

Enthalpy

The total heat content of a system at constant pressure

How are ∆Hvaporization and ∆Hcondensation related for a given substance?

They are equal in magnitude and opposite in sign with ∆Hvaporization generally being listed as positive in sign (endothermic) and ∆Hcondensation being negative in sign (exothermic). For a given substance the enthalpy of vaporization is equal and opposite to the enthalpy of condensation. Vaporization is an endothermic process (requires thermal energy) while condensation is an exothermic process (releases thermal energy).

When two objects have different temperatures, heat flows spontaneously from __________ _____________ to ____________ _____________. This result occurs because the change in entropy for the lower temperature object is _________ ____________ the change in entropy for the higher temperature object.

high temperature low temperature greater than

When a system "does work" it will impart organized molecular motion in the surroundings such that molecules are moving __________.

in the same direction

If the products of a reaction have higher energy than the reactants, then the reaction...

is endothermic.

Closed system

one that only allows for energy to exchange with the surroundings while keeping the matter contained

If a process is carried out at constant pressure and the volume of the system decreases, then ΔV is ___________ and the work is ___________.

positive negative

A coffee cup calorimeter is a constant ________ calorimeter.

pressure