Chemistry 101 - Chapter 5
Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4) and look up the ΔH°f for this compound CCl4(l) ΔH°f = -139.3 kJ/mol
C(graphite) + 2 Cl2(g) --> CCl4 (l)
Calorimeter
Device used to measure heat from
Thermodynamics
The study of energy and itrs transformations
The standard enthalpy change for the reaction CaCO3(s) --> CaO(s) + CO2(g) ΔH°rxn = 178.1 kJ Calculate the ΔH°f of CaCO3(s)
The table says -1207.3 kJ/mol ΔH°f[CaO(s)] = -635.5 ΔH°f[CO2(g)] = -393.5 We know that the ΔH°rxn = 178.1 kJ 178.1 kJ = (-635.5 + -393.5) - (CaCO3) CaCO3 = (-635.5 + -393.5) -178.1 = -1207.1 kJ
Capacity of Calorimeter
The total heat capacity of the calorimeter Ccal = kJ/ΔT qrxn = - Ccal x ΔT
Fossil Fuels
coal, oil, and natural gas, which are presently our major sources of energy
What is the value of ΔE if Efinal equals Einitial?
ΔE will be 0
Fundamental principle of Thermochemistry
Energy is released when chemical bonds are formed;energy is consumed when chemical bonds are broken.
If we define the system as the reactants and products, what is the sign of work in this process, positive or negative?
Negative
Coal
Solid, contains hydrocarbons of high molecular weight as well as compounds containing sulfur, oxygen or nitrogen Most abundant fossil fuel
Thermochemistry
The relationships between chemical reactions and energy changes
When a reaction occurs in a flask, you notice that the flask gets colder. What is the sign of ΔH ?
This is Endothermic, therefore ΔH is positive (qp is positive)
ΔH°rxn = Sum of nΔH°f products - Sum of nΔH°f reactants
n = the stoichiometric coefficients of the relevant chemical equation
Enthalpies of Combustion
the amount of energy for combusting a substance in oxygen
Natural Gas
gaseous hydrocarbons, compounds of hydrogen and carbon
The complete combustion of ethanol. C2H5OH (molar mass = 46.0 g/mol), proceeds as follows: C2H5OH (l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(l) ΔH = -555 kJ What is the enthalpy change for combustion of 15.0 g ethanol? a) -12.1 kJ b) -555 kJ c) -181 kJ d) -422 kJ e) -1700 kJ
15.0 g ethanol x (1 mol ethanol/46.0 g) x (-555 kJ/1 mol ethanol) = -181 kJ Answer : C
Calculate ΔH for the reaction 2 C(s) + H2 (g) --> C2H2 (g) given the following chemical equations and their respective enthalpy changes: Check Image
2 C(s) + O2 (g) --> 2 CO2 (g) 2(-393.5) H2 (g) + 1/2 O2 (g) --> H2O (l) -285.8 2 CO2 (g) + H2O (l) --> C2H2 (g) + 5/2 O2 (g) +1299.6 (the 5/2 O2 is canceled out by the 2 O2 and then the 1/2 O2) 2 C(s) + H2 (g) --> C2H2 (g) +226.8 kJ
Calculate the ΔH for 2 NO (g) + O2 (g) --> N2O4 (g) using the following information: N2O4 (g) ---> 2 NO2 (g) ΔH = +57.9 kJ 2 NO (g) + O2 (g) --> NO2 (g) ΔH = -113.1 kJ a) -85.5 kJ b) -55.2 kJ c) -171.0 kJ d) 2.7 kJ e) -55.2 kJ
2 NO (g) + O2 (g) --> 2 NO2 (g) ΔH = -113.1 kJ 2 NO2 (g) --> N2O4 (g) -ΔH = -57.9 kJ 2 NO (g) + O2 (g) --> N2O4 (g) -113.1 + -57.9 kJ = -171 Answer : C
Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l) ---> 2 H2O (l) + O2 (g) ΔH = -196 kJ Calculate the quantity of heat released when 5.00 g of H2O2 (l) decomposes at constant pressure
5.00 g x (1 mol H2O2/ 34g) x (-196 kJ/2 mol H2O2) = -14.4 kJ
Use the average bond enthalpies in Table 5.4 to estimate ΔH for the "water splitting reaction": H2O(g) --> H2(g) + 1/2O2(g) a) 242 kJ b) 417 kJ c) 5 kJ d) −5 kJ e) −468 kJ
= [2D(H-O)] - [1D(H-H) + 1/2D(O=O)] = 2(463) - [436 + (1/2x495)] = 242.5 kJ Answer = A
State function
A property of a system that is determined by its state or condition and not by how it got to that state Its value is fixed when temperature, pressure, composition and physical form are specified (P, V, T, E and H are state functions)
Enthalpy
A quantity defined by the relationship H = E + PV; The enthalpy change, ΔH, for a reaction that occurs at constant pressure is the heat evolved or absorbed in the reaction H= E + PV ΔH = qp
A positively charged particle and a negatively charged particle are initially far apart. What happens to their electrostatic potential energy as they are brought closer together?
As the distance decreases, the overall Eel will get more negative
What is the bond enthalpy of C-H in CH4? CH4 (g) --> C(g) + 4H(g) ΔH = 1660 kJ
CH4 is D(C-H) = (1660 kJ/4 mol) = 415 kJ/mol Bond enthalpy of C-H in CH4 is 415 kJ/mol
A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 0.250 L, and the final volume is 0.980 L. If the piston expands against a constant pressure of 1.35 atm, how much work (in J) is done? 1 L-atm = 101.3 J
Delta V = 0.980 L - 0.250 L = 0.730 L P = 1.35 atm w = -P(DeltaV) = - (1.35 atm)(0.730L) -0.9855 L-atm x (101.3 /1 L-atm) = -99.8 J
Internal Energy
E The sum of ALL kinetic and potential energies of the components of the system The change of internal energy (Delta E) is defined as the heat, added to the system, plus the work, done on the system by its surroundings ΔE = q + w ΔE = E final - E initial
When H2 (g) and O2 (g) react to form H2O (l), heat is released to the surroundings. Consider the reverse reaction, namely, the formation of H2 (g) and O2 (g) from H2O (l): 2H2O (l) --> 2H2 (g) + O2 (g) Is this reaction exothermic or endothermic?
Endothermic
The First Law of Thermodynamics
Energy can be converted, but never created nor destroyed Energy is conserved in any process; one way to express teh law is that the change in internal energy, DeltaE, of a system in any process is equal to the heat, q, added to the system, plus work, w, done on the system Delta E = q + w
If the battery is defined as the system, what is the sign on w in part (b)?
Energy taken from the battery, w<0
Molten gold poured into a mold solidifies at atmospheric pressure. With the gold defined as the system, is the solidification an exothermic or endothermic process?
Exothermic
For which of these reactions at 25°C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose Δ H is an enthalpy of formation? a) 2 Na(s) + 1/2 O2(g) --> Na2O (s) b) 2 K(l) + Cl2(g) --> 2KCl (s) c) C6H12O6(S) --> C(diamond) + 6H2(g) + 3O2(g)
Note: a substance that is in its standard enthalpy of formation is one in which each reactant is an element in its standard state and the product is one mole of the compound. We need to examine each equation to determine (1) whether the reaction is one in which one mole of substance is formed from the elements, and (2) whether the reactant elements are in their standard states. a) Yes, this does represent a standard enthalpy off formation 1 mol of Na2O is formed from the elements sodium and oxygen in their proper states, solid Na and O2 gas. Therefore (a) does correspond to a standard enthalpy of formation b) No it is not, but K is given as a liquid which means it must be changed to solid form at room temperature. Furthermore, 2 mol KCl are formed, so the enthalpy change for the reaction as written is twice the standard enthalpy of formation of KCl(S). Reduce it to 1 mol by multiplying by 1/2 K(s) + 1/2 Cl2 (g) --> KCl (s) c) No, it does not form a substance from its elements. Instead, a substance decomposes to its elements, so this reaction must be reversed. Next the element carbon is given as diamond, whereas graphite is the standard state of carbon at room temperature and 1 atm pressure. The equation that correctly represents the enthalpy of formation of glucose from its elements is 6C(graphite) + 6 H2(g) + 3 O2 (g) --> C6H12O6 (s) Flip the equation and simplify for 1 mol
ΔHrxn and Bond Enthalpies
Our strategy for estimating reaction enthalpies is a straightforward application of Hess's law. We use the fact that breaking bonds is always endothermic and forming bonds is always exothermic. We therefore imagine that the reaction occurs in two steps: 1) We supply enough energy to break those bonds in the reactants that are not present in the products. The enthalpy of the system is increased by the sum of the bond enthalpies of the bonds that are broken. 2) We form the bonds in the products that were not present in the reactants. This step releases energy and therefore lowers the enthalpy of the system by the sum of the bond enthalpies of the bonds that are formed. ΔHrxn = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed) = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products) ΔHrxn > 0 Endothermic ΔHrxn < 0 Exothermic
Calorimetry
The experimental measurement of heat produced in chemical and physical processis
Change in energy and what does it mean?
Thermodynamic quantities such as ΔE have three parts: 1) a number 2) a unit 3) a sign (1) and (2) together give the magnitude of the change, (3) gives the direction Apositive value of ΔE results when E final > E initial, indicating that the system has gained energy from its surroundings. A negative value of ΔE results when E final < E initial, indicating that the system has lost energy to its surroundings Note = there is an inverse relation to how much energy is in the surroundings compared to the system. If the internal energy increases, then the surroundings decrease. If the internal energy of the system decreases then the surroundings increase.
a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6 (l), to CO2(g) and H2O(l) b) Compare the quantity of heat produced by consumption of 1.00 g propane with that produced by 1.00 g benzene
a) 2 C6H6 (l) + 15 O2(g) --> 12 CO2(g) + 6H2O(l) C(s) + O2(g) --> CO2(g) 2 H2(g) + O2(g) --> 2 H2O(l) C6H6 (l) --> 6 C(s) + 3 H2(g) = (12(-393.5) + 6(-285.83)) - (2(49.0+0) + 15(0)) =-6534.9 kJ/2 mol = -3267.45 kJ/mol b) 1.00 g benzene x (1 mol benzene/78.11 g) x (-3267.45kJ/mol) = -41.83 kJ/mol C3H8(g) + 5 O2(g) --> 3 CO2(gO + H2O(l) C3H8(g) --> 3C(s) + 4H2(g) ΔH = -ΔH°f[C3H8(g)] 3C(s) + 3 O2(G) --> 3CO2(g) ΔH = -ΔH°f[CO2(g)] 4 H2(g) + 2O2(g) --> 4 H2O (l) ΔH = -ΔH°f[H2O(l)] C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l) = -(-103.85)+3(-393.5)+4(-285.8) = -2220 kJ 1.00 g C3H8 x (1 mol C3H8/44.1 g C3H8) x (-2220 kJ /1 mol C3H8) = -50.34 kJ/mol difference is = -41.83 - (-50.34) = +8.51 kJ/mol
Indicate the sign of the enthalpy change, ΔH, in the following processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic: a) An ice cube melts; b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O
a) Endothermic, ΔH is positive b) Exothermic, ΔH is negative
Surroundings
everything that lies outside the system that we study; spectator ions or the container for the chemical reaction are examples of surroundings Anything that does not affect the chemical reaction
Qrxn
heat of reaction = -Cs x m x ΔT
The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2(g), CO2 (g) and H2O (l) 2 CH6N2 (l) + 5 O2 (g) --> 2 N2 (g) + 2 CO2 (g) + 6 H2O (l) When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 to 39.50 C. In a separate experiment the heat capacity of the calorimeter is measure to be 7.794 kJ/C. Calculate the heat of reaction for the combustion of a mole of CH6N2
qrxn = -Ccal x ΔT ΔT = 39.5 - 25 = 14.5 C Ccal = 7.794 kJ/C = -(7.794 kJ/C)(14.5 C) = -113 kJ there are 2 moles of CH6N2 in the reaction, so now we calculate for 1 mole -113 kJ/4.00g x (46.1 g CH6N2/1 mol CH6N2) = -1302.3 kJ CH6N2 or -1.30x10^3 kJ
Closed system
systems that can exchange energy but not matter with their surroundings ex: mixture of H2 gas and O2 gas in a cylinder fitted with a piston
Enthalpies of Fusion
the amount of energy for melting solids
What process corresponds to the −88 KJ enthalpy change?
the cooling/condensation of H2O (g) to (l)
Standard Enthalpy Change
the enthalpy change when all reactants and products are in their standard states ΔH° the "°" indicates standard-state conditions
Bond enthalpy
the enthalpy change, ΔH, for the breaking of a particular bond in 1 mole of a gaseous substance ex: Cl2(g) --> 2Cl(g) 242 kJ/mol D(Cl-CL) = 242kJ/mol The bond enthalpy is always a positive quantity because energy is required to break chemical bonds. Conversely, energy is always released when a bond forms between two gaseous atoms or molecular fragments. The greater the bond enthalpy, the stronger the bond.
System
the portion of the universe that we single out for study; A chemical reaction is an example of a system
Decomposition
the reverse of the formation reaction The enthalpy change for the reaction is the negative of the ΔH°f value
Hess' Law and ΔH°f values
we can use both to calculate the standard enthalpy change of any reaction if we know the ΔH°f values of all reactants and products Ex: C3H8(g) + 5 O2(g) --> 3 CO2(gO + H2O(l) write the three equations that are associated with the standard enthalpies of formation: C3H8(g) --> 3C(s) + 4H2(g) ΔH = -ΔH°f[C3H8(g)] 3C(s) + 3 O2(G) --> 3CO2(g) ΔH = -ΔH°f[CO2(g)] 4 H2(g) + 2O2(g) --> 4 H2O (l) ΔH = -ΔH°f[H2O(l)] C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l) ΔH°rxn = ΔH1 + ΔH2 + ΔH3 = -(-103.85)+3(-393.5)+4(-285.8) = -2220 kJ
Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4 and react to form a solid product C(s): A (g) + B (g) ---> C (s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system?
ΔE = -1150 J + 480 J = -670 J
Calculate the change in the internal energy for a process in which a system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.
ΔE = q + w = 140 J - 85 J = +55 J
Suppose a system receives a "deposit" of 50 J of work from the surroundings and loses a "withdrawal" of 85 J of heat to the surroundings. What is the magnitude and the sign of ΔE for this process?
ΔE = q + w = 50 - 85 = -35 J
A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a piston that makes a gas-tight seal so that the gases are a closed system. The cylinder is submerged in a large beaker of water whose temperature is 25∘C, and a spark is used to trigger a reaction in the cylinder. At the completion of the reaction, the piston has moved downward, and the temperature of the water bath has increased to 28∘C. If we define the system as the gases inside the cylinder, which of the following best describes the signs of q, w, and ΔE for this reaction?
ΔE = q + w q = 28C - 25C = 3 C a chemical process in which heat is released has a -q value work being done on the system is a +w The ΔE = q + w; because we know q is (-) and w is (+), we cannot predict the value of ΔE without more information
Calculate the enthalpy change for the reaction 2 H2O2(l) --> 2 H2O(l) + O2(g) using enthalpies of formation ΔH°f[H2O2(l)] = -187.8 kJ/mol ΔH°f[H2O(l)] = -285.8 kJ/mol a) -88.0 kJ b) -196.0 kJ c) +88.0 kJ d) +196.0 kJ e) more information needed
ΔH°rxn = Sum of nΔH°f products - Sum of nΔH°f reactants 2 H2O2 (l) --> 2 H2O (l) + O2(g) 2 H2O2(l) --> 2 H2(g) + 2 O2(g) = 2(-187.8) 2H2(g) + O2(g) --> 2H2O (l) = -285.8 H2O2(l) --> H2O(l) + 1/2O2(g) = (-285.8) - 2(-187.8) = 89.8 kJ/mol Answer: C
Exothermic Process
When the system loses heat to its surroundings
Bomb Calorimeter
A device for measuring the heat involved in the combustion of a substance under constant volume conditions
Given 2SO2(g) + O2(g) --> 2SO3(g) Which of the following equations is correct? Check Image
Answer: E
Is the reaction from Card 95 exothermic or endothermic?
Exothermic
Which releases the greatest amount of energy per gram when metabolized? Carbohydrates, proteins, or fats.
Fats
Which value would change most if this label were for skim milk instead of whole milk: grams of fat, grams of total carbohydrate, or grams of protein?
Grams of fat
Petroleum
Naturally occurring combustible liquid composed of hundreds of hydrocarbons and other organic compounds
Is a human being an isolated, closed, or open system?
Open because we exchange matter and energy with our surroundings
A stalk of celery has a caloric content (fuel value) of 9.0 kcal. If 1.0 kcal is provided by fat and there is very little protein, estimate the number of grams of carbohydrate and fat in the celery. a) 2 g carbohydrate and 0.1 g fat b) 2 g carbohydrate and 1 g fat c) 1 g carbohydrate and 2 g fat d) 32 g carbohydrate and 10 g fat
Protein = 17 kJ/g or 4 Cal/g Carbohydrate = 17 kJ/g or 4 Cal/g Fat = 38 kJ/g or 9 Cal/g 9.0 kCal = 1.0 kCal fat + 8.0 kCal carb 1.0 kcal x (4.18 kJ/ 1 kcal) x (1 g fat/38 kJ) = 0.11 g fat 8.0 kcal x (4.18 kJ/1 kcal) x (1 g carb/17 kJ) = 2 g carb Answer: A
C6H12O6(s) + 6 O2(g) --> 6CO2(g) + 6 H2O(l) ΔH° = -2803 kJ Based on the enthalpy of the above reaction, what is the fuel value of glucose on a per gram basis?
Reacting 1 mol of glucose releases 2803 kJ of energy. The molar mass of glucose is 180.6 g, so the amount of energy released when 1 g is reacted is 2803 kJ x (1 mol/180.6 g) = 15.56 kJ/g This value is close to the average value for carbohydrates (17 kJ/g) as expected
Endothermic Process
When the system absorbs heat from its surroundings
a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values of these substances, estimate the fuel value (caloric content) of this serving. b) A person of average weight uses about 100 Cal/mi when running or jogging. How many servings of this cereal provide the fuel value requirements to run 3 mi?
a) Protein = 17 kJ/g or 4 Cal/g Carbohydrate = 17 kJ/g or 4 Cal/g Fat = 38 kJ/g or 9 Cal/g 8 g protein x (4 Cal/ 1 g) = 32 Cal/g 26 g Carb x (4 Cal/g) = 104 Cal/g 2 g Fat c (9 Cal/g) = 18 Cal/g 32 + 104 + 18 = 154 Cal/g or 8 g protein x (17 kJ/ 1 g) = 136 kJ/g 26 g Carb x (17 kJ/g) = 442 kJ/g 2 g Fat c (38 kJ/g) = 76 kJ/g 136+442+76 = 654 kJ/g 654 kJ/g x (1kcal/4.18 kJ) = 160 kcal b) 3 miles x (100 Cal/mi) = 300 Cal 300/160 = servings or 300/154 = 2 servings
Trinitroglycerin, C3H5N3O9 (usually referred to simply as nitroglycerin), has been widely used as an explosive. Alfred Nobel used it to make dynamite in 1866. Rather surprisingly, it also is used as a medication, to relieve angina (chest pains resulting from partially blocked arteries to the heart) by dilating the blood vessels. At 1 atm pressure and 25°C, the enthalpy of decomposition of trinitroglycerin to form nitrogen gas, carbon dioxide gas, liquid water, and oxygen gas is −1541.4kJ/mol. a) Write a balanced chemical equation for the decomposition of trinitroglycerin. b) Calculate the standard heat of formation of trinitroglycerin. c) A standard dose of trinitroglycerin for relief of angina is 0.60 mg. If the sample is eventually oxidized in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released? d) One common form of trinitroglycerin melts at about 3°C. From this information and the formula for the substance, would you expect it to be a molecular or ionic compound? Explain. e) Describe the various conversions of forms of energy when trinitroglycerin is used as an explosive to break rock faces in highway construction.
a) 4C3H5N3O9 -->6N2(g) + 12CO2(g) + 10H2O(l) + O2(g) b) 4(−1541.4kJ/mol) = -6165.6 kJ/mol is enthalpy of decomp, -6165.6kJ/mol = [6(ΔH°f[N2(g)] + 12(ΔH°f[CO2(g)] + 10(ΔH°f[H2O(l)] + ΔH°f[O2(g)]] - 4ΔH°f[C3H5N3O9(l)] -6165.6 kj = [12(-393.5) + 10(-285.8 kJ)] - 4(C3H5N3O9(l)) C3H5N3O9 = -353.6 kJ/mol c) 0.60 mg C3H5N3O9(l) to moles 0.60x10^-3 g x (1 mol/227 g) x (1541.4 kJ/1mol) = 4.1x10^-3 kJ = 4.1 J d) Because trinitroglycerin melts below room temperature, we expect that it is a molecular compound. With few exceptions, ionic substances are generally hard, crystalline materials that melt at high temperatures. (find more in Sections 2.6 and 2.7) Also, the molecular formula suggests that it is a molecular substance because all of its constituent elements are nonmetals. e) The energy stored in trinitroglycerin is chemical potential energy. When the substance reacts explosively, it forms carbon dioxide, water, and nitrogen gas, which are of lower potential energy. In the course of the chemical transformation, energy is released in the form of heat; the gaseous reaction products are very hot. This high heat energy is transferred to the surroundings. Work is done as the gases expand against the surroundings, moving the solid materials and imparting kinetic energy to them. For example, a chunk of rock might be impelled upward. It has been given kinetic energy by transfer of energy from the hot, expanding gases. As the rock rises, its kinetic energy is transformed into potential energy. Eventually, it again acquires kinetic energy as it falls to Earth. When it strikes Earth, its kinetic energy is converted largely to thermal energy, though some work may be done on the surroundings as well.
a) Dry red beans contain 62% carbohydrate, 22% protein, and 1.5% fat. Estimate the fuel value of these beans. b) During a very light activity, such as reading or watching television, the average adult expends about 7 kJ/min. How many minutes of such activity can be sustained by the energy provided by a serving of chicken noodle soup containing 13 g protein, 15 g carbohydrate, and 5 g fat?
a) assume we are working with 1 g Carb = 0.62 g Protein = 0.22 g Fat = 0.015 g Protein = 17 kJ/g or 4 Cal/g Carbohydrate = 17 kJ/g or 4 Cal/g Fat = 38 kJ/g or 9 Cal/g 0.62 g carb x (17 kJ/1g) = 10.54 kJ 0.22 g Protein x (17 kJ/1g) = 3.74 kJ 0.015 g fat x (38 kJ/1g) = 0.57 kJ 10.54+3.74+0.57 = 14.85 kJ/g or 15 kJ/g b) 13 g protein x (17 kJ/1 g) = 221 kJ 15 g carb x (17 kJ/1 g) = 255 kJ 5 g fat x (38 kJ/ 1 g) = 190 kJ 221+255+190= 666 kJ 666 kJ x (1 min/7 kJ) = 95.1 min or 100 min
Enthalpy of Reaction
ΔH = H products - H reactants Enthalpy change associated with a chemical reaction "heat of reaction" or ΔHrxn
Enthalpy of Formation
"heat of formation" ΔHf f = that the substances has been formed from its constituent elements the enthalpy change that accompanies the formation of a substance from the stable forms of its components elements
Enthalpy diagrams
1) Enthalpy is an extensive property - the magnitude of ΔH is proportional to the amount of reactant consumed i.e if you double or halve the original reactants/products then the ΔH will double or halve as well 2) The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to ΔH for the reverse reaction i.e. If a reaction to make water was worth 180 kJ, then the decomposition of water would be -180 kJ 3) The enthalpy change for a reaction depends on the states of the reactants and products i.e. If the water were (g) instead of (l) then the ΔHrxn would be 182 kJ instead of 180 kJ
Use Table 5.4 to estimate ΔH for the following combustion reaction. Check image The combustion reaction is 2 molecules of C2H6 (gas) plus 7 O2 (gas) leads to 4 CO2 (gas) plus 6 H2O (gas).
2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(g) = [12D(C-H) + 2D(C-C) + 7D(O=O) - [8D(C=O) + 12D(H-O)] = [12(413) + 2(348) + 7(495)] - [8(799) + 12(463)] = -2831 kJ
Thermochemical Equations
Balanced chemical equations that show the associated enthalpy change ΔH = +/- ________ kJ
What is the ΔHrxn of Methane and Chlorine to produce Methyl Chloride (CH4) and Hydrogen Chloride, HCl: H--CH3(g) + Cl--Cl (g) --> Cl---CH3 (g) + H---Cl (g) ΔHrxn = ?
Bonds broken: 1 mol C—H, 1 mol Cl—Cl Bonds formed: 1 mole C—Cl, 1 mol H—Cl ΔHrxn = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed) = (413 + 242) - (328 + 431) = -104 kJ/mol
Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is -393.5 kJ/mol, and that of diamond is -395.4 kJ/mol: C(graphite) + O2 (g) --> CO2 (g) ΔH = -393.5 kJ C(diamond) + O2 (g) --> CO2 (g) ΔH = -395.4 kJ Calculate ΔH for the conversion of graphite to diamond: C(graphite) --> C (diamond) ΔH = ?
C(graphite) + O2 (g) --> CO2 (g) ΔH = -393.5 kJ CO2 (g) --> C(diamond) + O2 (g) -ΔH = +395.4 kJ C(graphite) --> C(diamond) ΔH = -393.5 + 395.4 = +1.9 kJ
Calculate the ΔH for the reaction C(s) + H2O(g) --> CO(g) + H2(g) give the following thermochemical equations: C(s) + O2(g) --> CO2(g) ΔH1 = -393.5 kJ 2 CO(g) + O2(g) --> 2 CO2 (g) ΔH2 = -566.0 kJ 2 H2(g) + O2(g) --> 2H2O (g) ΔH3 = -483.6 kJ a) -1443.1 kJ b) -918.3 kJ c) 131.3 kJ d) 262.6 kJ e) 656.1 kJ
C(s) + O2(g) --> CO2(g) ΔH1 = -393.5 kJ (1/2)2CO2(g) --> (1/2)2CO(g) + (1/2)O2 (g) -ΔH=+566.0/2 (1/2)2H2O(g) --> (1/2)2H2(g) + (1/2)O2(g) -ΔH=+483.6/2 -393.5 + 1/2(+566.) + 1/2(483.36) = +131.3 kJ Answer: C
Use the average bond enthalpies in Table 5.4 to estimate ΔH for the combustion of ethanol. C2H5OH
C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(g) = [5D(C-H) + 1D(C-C) + 1D(C-O) + 1D(O-H) + 6D(O=O)] - [4D(C=O) + 6D(H-O)] =[5(413) + 348 + 358 + 463 + 6(495)] - [4(799) + 6(463)] =[6204] - [5974] = 230 kJ
How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system?
CH4 (g) + O2 (g) --> CO2 (g) + H2O (l) ΔH = -890 kJ if mol of Ch4 burned = -890 kJ then we solve for moles then kJ 1 mol CH4/-890 kJ 4.50 g CH4 x (1 mol CH4/ 16.04 g CH4) x (-890 kJ/1 mol CH4) = -250 kJ
This figure shows the gas phase reaction between methane, CH4, and chlorine, Cl2 to produce methyl chloride, CH3Cl and hydrogen chloride, HCl. Using the average bond enthalpies in the table below, estimate the ∆Hrxn for the analogous gas phase reaction between methane, CH4, and bromine, Br2, to produce methyl bromide, CH3Br, and hydrogen bromide, HBr.
CH4 + Br2 --> CH3Br + HBr = (413 + 193) - (276 + 366) = -36 kJ/mol
The combustion of exactly 1.000 g of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat. If the combustion of 0.550 g of benzoic acid causes the temperature of the calorimeter to increase from 22.01°C to 24.27°C, calculate the heat capacity of the calorimeter. a) 32.7 kJ/°C b) 6.42 kJ/°C c) 21.2 kJ/°C d) 14.5 kJ/°C e) 0.660 kJ/°C
Cs = 26.38 kJ ΔT = 24.27 - 22.01 = 2.26 C qrxn = -Cs x m x ΔT qrxn = -Ccal x ΔT Ccal = -qrxn(ΔT) or kJ/ΔT Ccal = 26.38 kJ/2.26 C = 11.7 kJ/C at 1.00 g .550/1.00 x 100 = 55% 11.7 kJ/C x 55% = 6.43 kJ/C
Calculate the work, in J, if the volume of a system contracts from 1.55 to 0.85 L at a constant pressure of 0.985 atm.
Delta V = 0.85 L - 1.55 L = -0.7 L P = 0.985 w = -(0.985 atm)(-0.7 L) = 0.69 L-atm 0.69 L-atm x (101.3 J/1 L-atm) = 70 J
If a balloon is expanded from 0.055 to 1.403 L against an external pressure of 1.02 atm, how many L-atm of work is done? a) -0.056 L-atm b) -1.37 L-atm c) 1.43 L-atm d) 1.49 L-atm e) 139 L-atm
Delta V = 1.403 L - 0.055 L = 1.348 L P = -1.02 atm w = - (1.02 atm)(1.348 L) = -1.375 L-atm
A chemical reaction that gives off heat to its surroundings is said to be ___________________ and has a _________________ value of ΔH a) endothermic, positive b) endothermic, negative c) exothermic, positive d) exothermic, negative
Exothermic, negative Answer: D
Use Table 5.3 to calculate the enthalpy change for the combustion of 1 mol of ethanol: C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(l)
First we have to balance the equation: C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(l) ΔH°f[C2H5OH(l)] = -277.7 kJ/mol ΔH°f[CO2(g)] = -393.5 kJ/mol ΔH°f[H2O(l)] = -285.8 kJ/mol ΔH°f[O2] = 0 kJ/mol Then we follow the equation: ΔH°rxn = Σ(nΔH°f of products) - Σ(nΔH°f of reactants) = (2(-393.52) + 3(-285.8)) - ((-277.7) + 3 (0)) = -1644.4 - (-277.7) = -1366.7 kJ/mol so 1367 kJ
The enthalpy of reaction for the combustion of C to CO2 is -393.5 kJ/mol , and the enthalpy for combustion of CO to CO2 is -283.0 kJ/mol CO: 1. C(s) + O2(g) --> CO2 (g) ΔH = -393.5 kJ 2. CO(g) + 1/2 O2(g) --> CO2 (g) ΔH = -283.0 kJ Using these data, calculate the enthalpy for the combustion of C to CO: 3. C(s) + 1/2 O2(g) --> CO(g) ΔH = ?
First we want to arrange them so that C(s) is on the reactant side and CO is on the product side C(s) + O2(g) --> CO2 (g) ΔH = -393.5 kJ CO2 (g) --> CO (g) + 1/2 O2 (g) -ΔH = 283.0 kJ "add" these together C(s) + 1/2 O2 (g) --> CO (g) ΔH = -110.5 kJ When we add them together, CO2 appears on both sides and therefor cancel out. Likewise, 1/2 O2 is also eliminated because there is that much present on either side (simplify)
If the heat of formation of H2O(l) is -286 kJ/mol, which of the following thermochemical equations for the formation of H2O is correct? a) 2 H(g) + O(g) --> H2O (l) ΔH = -286 kJ b) 2 H2(g) + O2(g) --> 2 H2O (l) ΔH = -286 kJ c) H2(g) + O(g0 --> H2O(g) ΔH = -286 kJ d) H2O (l) --> H2(g) + 1/2O2(g) ΔH = -286 kJ e) H2(g) + 1/2O2(g) --> H2O(l) ΔH = -286 kJ
Firstly, per the table, H2O(l) = -286 kJ; which eliminates (c) and (d). H likes to exist as H2(g); an O likes to exist as O2(g). which eliminates (a). Standard enthalpy of enthalpy leads only to 1 mol of the compound; which eliminates (b) Answer: E
A 0.5865-g sample of lactic acid (HC3H5O3) reacts with oxygen in a calorimeter whose capacity is 4.812 kJ/C. The temperature increases from 23.10 to 24.95 C. Calculate the heat of combustion of lactic acid a) per gram b) per mole
HC3H5O3 + 3 O2 --> 3 CO2 + 3 H2O Ccal = 4.812 kJ/C ΔT = 24.95 - 23.10 = 1.85 C qrxn = -Ccal x ΔT = - (4.812 kJ/C)(1.85 C) = -8.9 kJ -8.9 kJ lactic acid/0.5865 g = -15.2 kJ/g 0.5865 g lactic acid x (1 mol/ 90.1 g) = 6.51 x 10^-3 mol -8.9 kJ/6.51x10^-3 mol = 1367.1 kJ/mol --> 1370 kJ/mol
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 to 27.5 C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/gK
HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) Delta T = 27.5 C - 21.0 C = 6.5 C --> 6.5 K qrxn = -Cs x m x ΔT total volume = 100 mL x (1 g/1 mL) = 100 g - 4.18J/gK x (100 g) x (6.5 K) = -2717 J because this occurs at a constant pressure ΔH = qp/mol Now to get moles of HCl = M = mol/L or mol = MxL = 1.0 M x 0.050 L = 0.050 mol HCl ΔH = -2.7 kJ/0.050 mol HCl = -54 kJ./mol
Hess's Law
If a reaction is carried out in a series of steps, ΔH for the overall reaction equals the sum of the enthalpy changes for the individual steps
Calculate ΔH for the reaction NO(g) + O(g) --> NO2(g) given the following information: NO(g) + O3(g) --> NO2(g) + O2(g) ΔH = -198.9 kJ O3(g) -- 3/2 O2(g) ΔH = -142.3 kJ O2(g) --> 2O(g) ΔH = 495.0 kJ
NO(g) + O3(g) --> NO2(g) + O2(g) ΔH = -198.9 kJ 3/2 O2(g) --> O3(g) -ΔH = +142.3 kJ (1/2) 2O(g) --> (1/2) O2(g) -ΔH = -495.0 kJ -198.9 kJ + 142.3 + 1/2(-495.0) = -304.1 kJ
If a system does not change its volume during the course of a process, does it do pressure-volume work?
No
Ozone, O3(g), is a form of elemental oxygen produced during electrical discharge. Is ΔH°f for O3 (g), necessarily 0?
No, because the most stable form of oxygen is O2(g). O3(g) has a ΔH°f of 142.3 kJ/mol
If the reaction to form water were written H2 (g) + 1/2O2 (g) --> H2O (g), would you expect the same value of ΔH as in 2 H2(g) + O2 (g) ---> 2 H2O (g) ΔH = -483.6 kJ Why or why not?
No, because there are half the amount of products and reactants involved in the reaction
Fuel Value
The energy released when 1 g of substance is combusted
Qsoln
The heat gained or lost by the solution and is therefore equal in magnitude but opposite in sign to the heat absorbed or released by the reaction qsoln = -qrxn qsoln = specific heat of solution x grams of solution x ΔT = -qrxn or = -Cs x m x ΔT specific heat of water = 4.18 J/gK
Enthalpy, what does it mean?
When ΔH is positive, ( that is, when qp is positive), the system has gained heat from the surroundings = ENDOTHERMIC When ΔH is negative, the system has released heat to the surroundings = EXOTHERMIC
Pressure-Volume (PV) work
Work performed by expansion of a gas against a resisting pressure w = -PΔV ΔV = Vfinal - Vinitial Pressure is usually given units of atm Volume is usually given units of L 1 L-atm = 101.3 J P = F/A F = force; A = area ΔV = A x Δh Magnitude of work = force x distance = F x Δh Magnitude of work = F x Δh = P x A x Δh = P x ΔV ΔE= qp - PΔV or qp = ΔE + PΔV THUS ΔH = qp
Suppose the overall reaction were modified to produce 2H2O(g) rather than 2H2O(l). Would any of the values of ΔH in the diagram stay the same? Check image
Yes, everything but the H2O (l)
You lose 5 pounds over a 30-day period. Which of the following quantities act like a state function: the amount of calories you consume, your weight, or the number of calories burned through exercise?
Your weight!
Which is more stable at room temperature, a Cl2 molecules or two separate Cl atoms? What is the energy difference between the Cl2 molecules and the separated atoms?
a Cl2 molecule is more stable because it lives naturally as a diatomic molecule (0 kj/mol). The energy difference is -242 kJm/ol because = 0 - 242 kJ = -242kJ/mol
Spontaneous Process
a process that is thermodynamically favored to happen; can be fast or slow; rates are not governed by thermodynamics
Standard State
a set of conditions at which most enthalpies are tabulated Pressure = 1 atm temperature = 298 K or 25 C
Open System
a system in which matter and energy can be exchanged with the surroundings ex: uncovered pot of boiling water
Isolated System
a system in which neither energy nor matter can be exchanged with the surroundings ex: insulated thermos containing hot coffee
If the ions in the picture were allowed to move freely, they would ______________ and the potential energy of the system would _____________ (check image)
a) attract b) decrease This is because, as long as the ions are not too far apart, they will fight to come together (highest point of potential energy); therefore as they do get closer, the potential energy decreases
What effect do these changes have on ΔH for a reaction: a) Reversing the reaction b) Multiplying the coefficients of the equation for the reaction by 2?
a) changes the sign from negative to positive, or vice versa b) doubles the ΔH for the reaction
Electrostatic Potential
arises from the interactions between charged particles Eel = k(Q1xQ2)/d k = 8.99x10^9 J-m/C^2 the charges of Q1 and Q2 are typically on the order of magnitude of the charge of an electron (1.60x10^-19 C) When Q1 and Q2 are of the same charge, they will repel, making Eel a positive number When Q1 and Q2 are opposite charges, they will attract, and Eel will be negative
When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, the following reaction occurs: Mg(s) + 2 HCl (aq) --> MgCl2 (aq) + H2 (g) If the temperature of the solution increases from 23.0 C to 34.1 C as a result of this reaction, calculate the ΔH in Kj/mol of Mg. Assume that the solution has specific heat of 4.18 J/gC a) -19.1 kJ/mol b) -91 kj/mol c) -111 kJ/mol d) -191 kJ/mol e) -464 kJ/mol
q rxn = -Cs x m x ΔT -Cs = -4.18 kJ/mol m = 0.243 g ΔT = 34.1 c - 23.0 C = 11.1 C q rxn = -4.18 J/gC x (100 g) x (11.1 C) = -4649.8 J to -4.64 kJ 0.243 g Mg x (1 mol Mg/24.31 g) = 0.01 mol Mg = -4.64 kJ/0.01 mol Mg = -464 kJ/mol
Enthalpy of Vaporization
the amount of energy needed to go from liquid to gas
When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 to 23.11 C. The temperature increase is caused by the following reaction: AgNO3 (aq) + HCl (aq) --> AgCl (s) + HNO3 (aq) Calculate the ΔH for this reaction in kJ/mol AgNO3. assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/gC
ΔH = kJ/mol qrxn = - Cs x m x ΔT Cs = 4.18 J/gC m = 100 ml or 100 g ΔT = 23.11 C - 22.30 C = 0.81 C qrxn = -(4.18 J/gC)(100 g)(0.81 C) = -340 J to -0.34 kJ M = mol/L mol= MxL mol AgNO3 = 0.100 M x 0.05 L = 0.005 mol AgNO3 ΔH = -0.34 kJ/0.005 mol AgNO3 = -68 kJ/mol
Enthalpy Change at a Constant Pressure
ΔH = ΔE + (PxΔV) ΔH = (qp + w) - w = qp The subscript P on q indicates that the process occurs at constant pressure The change in enthalpy equals the heat qp gained or constant pressure
Standard Enthalpy of Formation
ΔH°f the change in enthalpy for the reaction that forms one more of the compound from its elements with all substances in their standard states: elements in their standard state --> 1 mol of the compound in standard state ΔHrxn = ΔH°f always reported in kJ/mol the standard enthalpy of formation of the most stable form of any element is zero
Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s): CuO(s) + H2(g) --> Cu(s) + H2O(l) ΔH° = -129.7 kJ
ΔH°rxn = ΔH°f(products) - ΔH°f(reactants) ΔH°f[CuO] = ΔH°f(products) - ΔH°rxn Cu(s) = 0 kJ H2O(l) = -285.83 kJ = -285.83 - (-129.7) = -156.13 kJ
