Chap 5 ThermoChemsitry
Endothermic
(of a chemical reaction or compound) occurring or formed with absorption of heat ex. gold bar problem
Open system
A system that can exchange both matter and energy with its surroundings.
Closed system
A system that can exchange energy with the surroundings, but not matter.
Isolated system
A system that can exchange neither energy nor matter with its surroundings.
Assume that the following reaction occurs at constant pressure:2Al(s)+3Cl2(g)⟶2AlCl3(s)
At constant pressure, ΔE=ΔH−PΔV. For an ideal gas at constant pressure and temperature, PΔV=RTΔn Because the value of Δ n for the given reaction is negative, the PΔV is negative. So, ΔEΔE must be larger the ΔHΔH.
Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions: a. Surroundings get colder and the system decreases in volume. b. Surroundings get hotter and the system expands in volume. c. Surroundings get hotter and the system decreases in volume. d. Surroundings get hotter and the system does not change in volume. Also assume that the magnitude of the volume and temperature changes are similar among the reactions.
B, D, C, A A temperature increase for the surroundings means that q is negative for the system. If the surroundings gain heat energy, then the system (the reaction) released heat energy. Change in temperature of the system relates to the heat transfer. Change in volume relates to work. Since energy is the sum of heat and work, ΔE = ΔU= q+w you will be able to compare the reactions after you determine the signs of heat and work for each case. Keep in mind that the reaction that releases the most energy will have the most negative ΔE At constant pressure, q=ΔH. When the surroundings get hotter, what is the sign of q? negative At constant pressure, w=−PΔV. When the system expands, what is the sign of w? negative
The complete combustion of ethanol, C2H5OH(l), to form H2O(g) and CO2(g) at constant pressure releases 1235 kJ of heat per mole of C2H5OH.
Balanced equation: C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(g)
Exothermic
Chemical Reaction in which energy is primarily given off in the form of heat ex. balloon problem
Classify the following by the sign of ΔE for the system.Drag the appropriate items to their respective bins. If no definitive classification can be made, drag the item into the bin labeled "Not enough data."
If q and w are both positive, then ΔE is positive because ΔE=q+w. Negative -the system expands and the surroundings get hotter Positive -the system contracts and the surrounding get colder Not enough data -the system contracts and the surroundings get hotter - the system expands and the surroundings get colder
An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 5.9 to 2.95 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.95 to 2.36 L . In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 5.90 to 2.36 L in one step If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules? Express your answer with the appropriate units.
Part 1 a. P1*ΔV1 + P2*ΔV2 b. 2.00(5.9-2.95) L*atm + 2.50 (2.95-2.36)L*atm = 7.375 L*atm c. Use conversion factor: d. 7.375 L*atm * 101.325J/L*atm = 747.27 J Part 2 e. q for the one-step process = PΔV = 2.50(5.90-2.36)L*atm * 101.325 J/L*atm = 896.73 J Hence q for the one-step process is greater by (896.73 J - 747.27J)= 149.46 J The difference of 149 J between the two-step and one-step process demonstrate that q is not a state function, but a path function that depends on path taken to get from the initial to final state.
State functions versus path functions
The change in internal energy, ΔE, is a state function because it depends only on the initial and final states of the system, and not on the path of change. In contrast, q and w are path functions because they depend on the path of change and not just the initial and final states of the system.
What is meant by the internal energy of a system?
The total internal energy (E) of a system is the sum of all the kinetic and potential energies of the system components.
Assume that the following reaction occurs at constant pressure:2Al(s)+3Cl2(g)⟶2AlCl3(s) If you are given ΔH for the reaction, what additional information do you need to determine ΔE for the process? Which quantity is larger for this reaction?
a. P and ΔV because ΔE=ΔH-PΔV b. ΔE>ΔH
By what means can the internal energy of a closed system increase? a. When the system increases it's general velocity. b. When work is done on the system. c. When heat is transferred to the system. d. When the flow of the hot matter is directed to the system.
b and c
According to the first law of thermodynamics, what quantity is conserved?
energy Energy can neither be created nor destroyed, it can only be transformed from one form to another. ΔE = q + w, where ΔU is change in internal energy of a system, q is heat transferred, and w is work done.
Which cannot leave or enter an isolated system?
heat, matter, or work
Which of the following cannot leave or enter a closed system: heat, work, or matter?
matter. We know that heat and work are forms of energy, so they can enter or leave the closed system.
Types of systems
open, closed, isolated
What do we call the part of the universe that is not part of the system?
surroundings
A piston has an external pressure of 15.0 atm. How much work has been done in joules if the cylinder goes from a volume of 0.120 liters to 0.560 liters?
w= -P(Dv) work = negative pressure times change in volume! Use the conversion factor that Mastering Chem gives you of the 1 L x atm = 101.3 J Change in volume is found by subtracting 0.560 and 0.120 which = 0.44 Plug what you have into that equation at top: w= -15.00(0.44) So that means work = -6.6 If it wants the answers in Joules than you have to multiply using the conversion factor given: 101.3 x -6.6 = -668.58 J So your answer is -669 J
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 40.0 atm and releases 61.2 kJ of heat. Before the reaction, the volume of the system was 6.80 L . After the reaction, the volume of the system was 3.00 L . Calculate the total internal energy change, ΔE, in kilojoules.
ΔE = -45.8 kJ a. w = 40.0 atm(6.80L - 3.00L) = 152 atm*L b. Given conversion factor: 1 L atm = 101.325 J c. 152x101.325 J = 15401.4J d. Convert to kJ 1 J = 10^-3 kJ = 15.4014 kJ e. 15.4014 kJ - 61.2 kJ = -45.8 kJ of internal energy change Although the system absorbed some energy in the form of work, the significant release of heat caused the total change to be negative.
A mole of X reacts at a constant pressure of 43.0 atm via the reaction. X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.
ΔE =-62.0 kJ Relationship between ΔH, ΔE and work done is given by first law of thermodynamics. ΔE = ΔH - PΔV Given, ΔH = -75.0 kJ = -75000 J. P = 43.0 atm. a. ΔV = Final volume - initial volume = (2.00 - 5.00) = -3.00 L b. PΔV = 43.0 × (-3.00) = -129 L atm c. given conversion factor: 1 L atm = 101.325 J d. -129 L atm = 129 × 101.325 = -13071 J Part2 e. ΔE = ΔH - PΔV = -75000 + 13071 = 61929 J f. Conversion from J to kilojoule 1 J = 10^-3 kJ g. Total energy change, ΔE = 61.929 kJ
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.653 kJ of heat. It shrinks on cooling, and the atmosphere does 384 J of work on the balloon.
ΔE=q+w a. The cooling removes energy from the balloon. Convert 0.653 kJ to -653J. b. The work done on the balloon increases the energy of the balloon. -653J + 384J = -269J is the change in internal energy The gain of energy from work performed on the balloon is less than the loss of energy through heat, so there is a net loss of internal energy in the balloon and the system is exothermic.
A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When 0.480 kJ of heat is added to the gas, it expands and does 216 J of work on the surroundings. What is the value of ΔH for this process? What is the value of ΔE for this process?
ΔH = 0.480kJ For a system under constant pressure, the change in enthalpy is equal to the heat, and the work energy (w) can be ignored because it effectively cancels. The relationship relating the change in enthalpy (ΔH) to the change in internal energy (ΔE), heat (qp, under constant pressure), and work (=−PΔV) is ΔH_sys=(ΔE)+PΔV =(qp+w_sys)−w_sys = qp ΔE = 0.264kJ 0.480kJ -.216kJ The change in internal energy is the sum of heat and work. Heat is gained by the system (gas), thus q = 0.480 kJ. Additionally, the system (gas) performs work, so w=−0.216 kJ. Summing the two terms provides the value for the change in internal energy.
A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When the gas undergoes a particular chemical reaction, it absorbs 826 J of heat from its surroundings and has 0.64 kJ of P−V work done on it by its surroundings. What is the value of ΔH for this process? What is the value of ΔE for this process?
ΔH = 0.826kJ ΔE = 1.47kJ b/c 0.64=0.826
Limestone stalactites and stalagmites are formed in caves by the following reaction: Ca2+(aq) + 2HCO3-(aq) --> CaCO3(s) + CO2(g) + H2O(l) If one mol of CaCO3 forms at 298 K under 1 atm pressure, the reaction performs 2.49 kJ of P - V work, pushing back the atmosphere as the gaseous CO2 forms. At the same time, 38.95 kJ of heat is absorbed from the environment. A.) What is the value of ΔH for this reaction? B.) What is the value of ΔE for this reaction?
ΔH = 38.95 kJ ΔE = 36.46 kJ 38.95-2.49
Calculate the change in internal energy of the following system: A 100.0-g bar of gold is heated from 25∘C to 50∘C during which it absorbs 322 J of heat. Assume the volume of the gold bar remains constant
ΔU = 322J + 0 = increase of 322J of internal energy Gold has a specific heat capacity of 0.129 J/(g⋅∘C). The amount of heat absorbed could be calculated by using this property, mass, and the change of temperature, but the value for heat is already provided. At a constant volume, no work is performed, and the change in internal energy is simply equal to the heat transferred to the system. The heated bar of gold performs no work, so its internal energy only increases and the system is endothermic.
