Chem 25 Chapter 9
What is the final temperature (in °C) of 150.1 g of water (specific heat = 4.184 J/g･°C) at 24.20°C that absorbed 950.0 J of heat?
25.71 °C Explanation: The change in temperature can be calculated using the heat, the mass, and the specific heat capacity. q=mcΔT 950.0J=(150.1g)×(4.184Jg⋅∘C)×(ΔT) ΔT=(Tf-Ti) Input into numeric solver
If a system has 65.0 J of heat added to it, and the system does 25.0 J of work on its surroundings, what is the change in the internal energy of the system, in J?
40 J (Delta U)=(65J)+(-25J)=40J
How many kcal are in 21.5 kJ?
5.14 kcal 21.5kJ(1kcal/4.184kJ)=5.14kcal
Convert 23.4 kJ to calories.
5590 cal 23.4kJ(1000J/1kJ)(1cal/4.184J)=5590 cal
Which one of the following reactions would produce the largest amount of heat per mole of oxygen? A) 2 Ca (s) + O₂ (g) → 2 CaO (s) ∆H° = -635 kJ/mol B) CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g) ∆H° = -802 kJ/mol C) 2 CH₃OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 4 H₂O(g) ∆H° = -1280 kJ/mol D) 2 Pb(s) + 3 O₂ (g) → 2 SO₂ (g) + 2 PbO (s) ∆H° = -828 kJ/mol
A) 2 Ca (s) + O₂ (g) → 2 CaO (s) ∆H° = -635 kJ/mol Explanation: Calculate the enthalpy per mole of oxygen for each option. The reaction enthalpy is per mole of reaction. -635 kJ/1 mol O2=-635 kJ per mole O2
A heated gas expands and presses on a piston. In this example, work is done: A) By the gas on the piston. B) On the gas by the heat source. C) By the piston on the gas. D) By the surroundings on the system.
A) By the gas on the piston. Explanation: Pressure-volume work is work done by the expansion of a gas against a constant pressure, such as that provided by a moveable piston.
The First Law of Thermodynamics tells us A) that the energy of the universe is always increasing. B) that the energy of the universe is a constant. C) that the energy for a chemical reaction is a constant. D) that a chemical reaction will always release energy.
B) that the energy of the universe is a constant.
A hot air ballon is expanding and its temperature is increasing. Which of the following is increasing. Which of the following is true? A. q>0; w>0 B. q>0; w<0 C. q<0; w<0 D. q<0; w>0
B. q>0; w<0 If Q>0, heat goes into the system If W<0, the system does work on the surroundings
Which one of the following processes is exothermic? A) Making coffee. B) Baking a cake. C) Burning wood in a campfire. D) Photosynthesis.
C) Burning wood in a campfire. An exothermic process releases heat. Making coffee, baking a cake, and photosynthesis require energy. Burning wood in a campfire releases heat that can be felt radiating from the campfire.
Calculate the work for the expansion of an ideal gas from 2.0 L to 5.0 L against a pressure of 2.0 atm at constant temperature. A. -6.0L atm B. 3.0 L atm C. 0 L atm D. -3.0 L atm E. 6.0 L atm
E. 6.0 L atm
How many kJ are in 25.4 kcal?
106kJ 25.4kcal(4.184kJ/1kcal)=106kJ
What quantity of heat (in kJ) would be required to convert 15.1 g of ice to water at 0.00°C? (∆Hfus = 6.01 kJ/mol for water)
5.04kJ q=n∆Hfus 15.1g H2O(1 mol H20/18.02g H20)=0.838 mol H2O 0.838 mol H2O(6.01kJ/1 mol H2O)=5.04 kJ 0 Celcius is constant from when water turns from ice to liquid
The calorie is defined based on the specific heat capacity of what substance? A) H₂O B) Al C) Pb D) NH₃ E) Hg
A) H₂O Explanation: One calorie (c) is defined as the amount of heat required to raise the temperature of 1 gram of water by 1°C. Water has a high specific heat capacity for a small molecule due to its intermolecular hydrogen bonding.
If a system does 84.0 kJ of work on its surroundings and releases 105 kJ of heat, what is the change in the internal energy of the system, in kJ?
-189 kJ Explanation: Change in internal energy (delta E)=q+w When work is done it has a negative sign and when heat is released, it is also a negative sign. (Delta E)=(-105kJ)+(-84.0kJ)=-189kJ
For the chemical reaction H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol. What is the molar enthalpy (in kJ/mol) for the reaction 3 H₂ (g) + 3 F₂ (g) → 6 HF (g)
-238 kJ/mol Since hydrofluoric acid (HF) is product the direction of the reaction can be kept as is. Then, multiply the equation by 3 to obtain the same mole ratio necessary for the complete balanced equation. 3×−79.2kJ/mol=−238kJ/mol
How many moles of KClO3 solid must be reacted according to the following balanced chemical equation to transfer -34.2kJ of heat? 2KClO3(s)--> 2KCl(s)+3O2(g) ∆H=-89.4
0.765 mol KClO3 -34.2kJ(2 mol KClO3/-89.4kJ)=0.765 mol KClO3
For the chemical reaction Ca (s) + ½ O₂ (g) → CaO (s) ∆H° = -635 kJ/mol. What is the molar enthalpy (in kJ/mol) for the reaction 2 CaO (s) → 2 Ca (s) + O₂ (g)?
1270 kJ/mol Explanation: Since calcium oxide (CaO) is a reactant and not a product reverse the equation. This also reverses the sign. Then, multiply the equation by 2 to obtain the same mole ratio necessary for the complete balanced equation. 2×635kJ/mol=1270kJ/mol
Convert 475 cal to joules.
1990 J Explanation: 475 cal(4.184J/1 cal)=1990J
If a reaction is exothermic, what is true of its reverse reaction? A) ∆H > 0 B) ∆H < 0 C) ∆H = 0 D) Not enough info
A) ∆H > 0 Explanation: The reverse reaction would be endothermic, which would have a positive ∆H.
A state function is best described as A) A function that depends on what route is taken between start and finish. B) A function that is independent of the amount of substance. C) A function that depends only on the starting point and ending point of a process. D) A function that only depends on the end point of a process.
C) A function that depends only on the starting point and ending point of a process. Explanation: A state function is best described as a function that depends only on the starting point and ending point of a process.
What type of property is heat? A) intensive physical property B) intensive chemical property C) extensive chemical property D) extensive physical property
D) extensive physical property Explanation: Unlike temperature, which measures average kinetic energy, heat measures internal energy and depends on the amount of substance present. Therefore, it is an extensive physical property.
Which of the following units of energy is defined as the heat it takes to raise the temperature of 1 gram of water by 1° C? A) kilojoule B) joule C) watt D) kilocalorie E) calorie
E) calorie Explanation: One calorie (c) is defined as the amount of heat required to raise the temperature of 1 gram of water by 1°C. The Calories (kcal) listed on nutritional labels are 1,000 of the heat calories.
Using the provided table and the equation below, determine the heat of formation (in kJ/mol) for KClO₂. 2 KClO₃ (s) → 2 KClO₂ (s) + O₂ (g) ∆H° = 296.2 kJ/mol *See Chem 101
-243.1 kJ/mol Explanation: The enthalpy for the reaction can be found by subtracting heats of formation of the reactants from that of the products. Be sure to include the coefficients from the balanced chemical equation. In this case, we can use the overall enthalpy to determine the unknown heat of formation by letting x = the heat of formation of KClO₂. *See Notebook/Chem 101 https://www.wyzant.com/resources/answers/727534/using-the-provided-table-and-the-equation-below-determine-the-heat-of-forma
Using the provided table, determine the enthalpy (in kJ/mol) for the reaction 2 K(s) + 2 H₂O(l) → 2 KOH (aq) + H₂(g) *See Chem 101
-393.3 kJ/mol https://www.wyzant.com/resources/answers/796012/using-the-provided-table-determine-the-enthalpy-for-the-reaction-2-k-s-2-h-
If a system has 35.5 J of heat added to it, and the system does 75.5 J of work on its surroundings, what is the change in the internal energy of the system, in J?
-40 J (Delta E)=(35.5J)+(-75.5J)=-40J
Determine the enthalpy of reaction for P₄O₁₀(s) + 6PCl₅(g) → 10POCl₃(g) P₄(s) + 6Cl₂(g) → 4PCl₃(g) ∆H° = -1225.6 kJ/mol P₄(s) + 5O₂(g) → P₄O₁₀(s) ∆H° = -2967.3 kJ/mol PCl₃(g) + Cl₂(g) → PCl₅(g) ∆H° = -84.2 kJ/mol PCl₃(g) + ½O₂(g) → POCl₃(g) ∆H° = -285.72 kJ/mol
-610.3 kJ/mol Explanation: Use Hess's law to determine the enthalpy of reaction for: P4O10(s)+6PCl5(g)⟶10POCl3(g)PX4OX10(s)+6PClX5(g)⟶10POClX3(g) Keep the first reaction the same. P₄(s)+6Cl₂(g)⟶4PCl₃(g)ΔH∘P₄(s)+6Cl₂(g)⟶4PCl₃(g)ΔH∘ = -1225.6 kJ/mol kJ/mol kJ/mol Flip the second reaction so that P4O10PX4OX10 is on the left side. P4O10(s)⟶P4(s)+5O2(g)ΔH∘PX4OX10(s)⟶PX4(s)+5OX2(g)ΔH∘ = +2967.3 kJ/mol kJ/mol Flip the third reaction so that PCl5PClX5 is on the right side. In addition, multiply by 6.6PCl5(g)⟶6PCl3(g)+6Cl2(g)ΔH∘6PClX5(g)⟶6PClX3(g)+6ClX2(g)ΔH∘ = +505.2 kJ/molkJ/mol Multiply the fourth reaction by 10: 10PCl3(g)+5O2(g)⟶10POCl3(g)ΔH∘10PClX3(g)+5OX2(g)⟶10POClX3(g)ΔH∘ = −2857.2 kJ/mol Finally, add these values together, keeping track of the signs to find the enthalpy of reaction. *See Notebook/Chem 101
If a system produces 111.0 kJ of heat and does 45.0 kJ of work on the surroundings, what is the ∆E for the surroundings, in kJ?
156 kJ ∆E=q+w ∆E=111.0 kJ+45.0kJ=156kJ Since heat is released from the system to the surroundings, qsurroundings is positive. Likewise, since work is done to the surroundings, wsurroundings has a positive sign.
In the following reaction, how many moles of CH₃OH are required to give off 1501 kJ of heat? 2 CH₃OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 4 H₂O(g) ∆H° = -1280. kJ
2.359 mol (-1501 kJ)(2 mol CH3OH/1280kJ)=2.345 or 2.359 mol (depending on your calculator)
Determine the enthalpy of reaction for 2CH₄(g) → C₂H₄(g) + 2H₂(g) using the equations: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ∆H° = −3120.8 kJ/mol CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ∆H° = −890.3 kJ/mol C₂H₄(g) + H₂(g) → C₂H₆(g) ∆H° = −136.3 kJ/mol H₂ + ½O₂(g) → H₂O(l) ∆H° = −285.8 kJ/mol
201.9 kJ/mol Use Hess's law to determine the enthalpy of reaction for: 2CH₄(g)⟶C₂H₄(g)+2H₂(g)2CH₄(g)⟶C₂H₄(g)+2H₂(g) Flip the first reaction: 4CO₂(g)+6H₂O(l)⟶2C₂H₆(g)+7O₂(g)ΔH∘4CO₂(g)+6H₂O(l)⟶2C₂H₆(g)+7O₂(g)ΔH∘ = +3120.8 kJ/molkJ/mol Multiply the second reaction ×4×4 to cancel the 4CO₂ from the first reaction:4CO₂ from the first reaction: 4CH₄(g)+8O₂(g)⟶4CO₂(g)+8H₂O(l)ΔH∘4CH₄(g)+8O₂(g)⟶4CO₂(g)+8H₂O(l)ΔH∘ = −3561.2 kJ/molkJ/mol Flip the third reaction to put C2H4CX2HX4 on product side and then multiply by 2: 2C₂H6(g)⟶2C₂H₄(g)+2H₂(g)ΔH∘2C₂HX6(g)⟶2C₂H₄(g)+2H₂(g)ΔH∘ = +272.6 kJ/molkJ/mol Flip the fourth reaction and then multiply by 2 to get a total of 8H2OHX2O on the reactant side: 2H₂O(l)⟶2H₂+O₂ΔH∘2H₂O(l)⟶2H₂+O₂ΔH∘ = +571.6 kJ/mol Finally, add these values together, keeping track of the signs to find the enthalpy of reaction. *See Notebook/Chem101
In the following reaction, what is the quantity of heat (in kJ) released when 3.21 moles of CH₄ are burned? CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g) ∆H° = -802 kJ/mol
2570kJ 3.21 mol CH4(802 kJ/1 mol CH4)=2570 kJ Take the absolute value of the delta H for quantity of heat.
Using the equations H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol Determine the molar enthalpy (in kJ/mol) for the reaction C (s) + 4 HF (g) → CF₄ (g) + 2 H₂ (g).
299.7 kJ/mol Explanation: Since HF is a reactant in the reaction and not a product, reverse the first equation for the formation of HF. This also reverses the sign on ∆H°. Next, multiply the equation by 2 to obtain the same mole ratio necessary for the complete equation. Finally, add the equation values together, keeping track of the signs to find the enthalpy of reaction. *See Chem 101/Notebook
Using the equations 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol Determine the molar enthalpy (in kJ/mol) for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s)
320.3 kJ/mol Explanation: For the formation of FeCl₃, from the balanced equation multiply the coefficients and ∆H° by 2 to obtain the same mole ratio necessary for the complete equation. Since SiCl₄ is a reactant in the reaction and not a product, reverse the first equation for the formation of SiCl₄ (s). This also reverses the sign on ∆H°. Next, multiply the equation by 3 to obtain the same mole ratio necessary for the complete equation. Finally, add the equation values together, keeping track of the signs to find the enthalpy of reaction. *See Chem 101/Notebook
What is the change in enthalpy in joules when 5.44 x 10^-4 mol of AgCl solid dissolves in water according to the following chemical equation: AgCl(s)--> Ag+(aq)+Cl-(aq) ∆H=65.5 kJ
35.6J 5.44 x 10^-4 mol AgCl(65.5kJ/1 mol AgCl)(1000J/1kJ)=35.6J
Determine the enthalpy of reaction for HCl(g) + NaNO₂(s) → HNO₂(l) + NaCl(s) 2NaCl(s) + H₂O(l) → 2HCl(g) + Na₂O(s) ∆H° = -507.1 kJ/mol NO(g) + NO₂(g) + Na₂O(s) → 2NaNO₂(s) ∆H° = -427.0 kJ/mol NO(g) + NO₂(g) → N₂O(g) + O₂(g) ∆H° = -43.01 kJ/mol 2HNO₂(l) → N₂O(g) + O₂(g) + H₂O(l) ∆H° = +34.02 kJ/mol
428.6 kJ/mol Explanation: Use Hess's law to determine the enthalpy of reaction for: HCl(g)+NaNO2(s)⟶HNO2(l)+NaCl(s)HCl(g)+NaNOX2(s)⟶HNOX2(l)+NaCl(s) Flip the first reaction and divide by 2: HCl(g)+½Na2O(s)⟶NaCl(s)+½H2O(ℓ)ΔH∘HCl(g)+½NaX2O(s)⟶NaCl(s)+½HX2O(ℓ)ΔH∘ = +253.6 kJ/molkJ/mol Flip the second reaction and divide by 2: NaNO2(s)⟶½NO(g)+½NO2(g)+½Na2O(s)ΔH∘NaNOX2(s)⟶½NO(g)+½NOX2(g)+½NaX2O(s)ΔH∘ = +213.5 kJ/molkJ/mol Divide the third reaction by 2: ½NO(g)+½NO2(g)⟶½N2O(g)+½O2(g)ΔH∘½NO(g)+½NOX2(g)⟶½NX2O(g)+½OX2(g)ΔH∘ = -21.51 kJ/molkJ/mol Flip the fourth reaction and divide by 2: ½N2O(g)+½O2(g)+½H2O(ℓ)⟶HNO2(ℓ)ΔH∘½NX2O(g)+½OX2(g)+½HX2O(ℓ)⟶HNOX2(ℓ)ΔH∘ = -17.01 kJ/mol Finally, add these values together, keeping track of the signs to find the enthalpy of reaction. *See Notebook/Chem101
How many grams of C6H6 liquid must decompose according to the following chemical equation to transfer 430 kJ of heat? C6H6(l)--> 3C2H2(a) ∆H=630kJ
53.3g C6H6 430kJ(3 mol C2H2/630 kJ)(1 mol C6H6/3 mol C2H2)(78.12g C6H6/1 mol C6H6)=53.3g C6H6
How many Joules are there in 2001 calories? (1 cal = 4.184 J)
8372J 2001cal(4.184J/1cal)=8372J
A plot of pressure-volume isotherms for an ideal gas at two different temperatures is shown below. In which of these processes would w < 0? A) X → Y B) Y → Z C) Z → X D) X → Y and Y → Z E) Y → Z and Z → X (See image on Chem 101)
A) X → Y Explanation: X → Y As the pressure decreases in this isothermal process, the volume of the gas increases. The w < 0 when an ideal gas expands.
Which of the following has the greatest amount of potential energy? A) a ball at the top of a hill. B) A ball rolling down a hill. C) A ball at the bottom of a hill. D) A ball thrown up and moving into the air.
A) a ball at the top of a hill. Explanation: Potential energy is the energy that an object possesses when not in movement. Potential energy is higher for objects that are higher off the ground. Therefore the ball at the top of the hill would have the greatest amount of potential energy.
An exothermic reaction causes the surroundings to A) warm up B) become acidic C) condense D) decrease in temperature E) release CO₂
A) warm up Explanation: An exothermic reaction transfers energy from the system to the surroundings. That tells us the surroundings will get warmer.
Calculate the work for the expansion of an ideal gas from 2.0L to 5.0L against a pressure of 2.0 atm at constant temperature. A. -6.0 L atm B. 3.0 L atm C. 0 L atm D. -3.0 L atm E. 6.0 L atm
A. -6.0 L atm w=-p∆V w=-(2.0atm)(5.0L-2.0L) w=-6.0Latm Explanation: Positive work is done on the gas when the gas is compressed; negative work is done on the gas when the gas expands. zero work is done on the gas when the gas volume is fixed.
Which one of the following processes is endothermic? A) Boiling a liquid. B) Freezing a solid. C) Condensation of a gas into a liquid. D) Condensation of a gas into a solid.
A. Boiling water To condense a gas, heat must be removed from the gas, which is an exothermic process. Freezing is an exothermic reaction Deposition as well
Which of the following statements is true? A. Change in enthalpy is a state function B. A chemist takes the surrounding point of view when determining the sign for work or heat C. Heat of reaction and change in enthalpy can always be used interchangeably D. In exothermic reactions, the reactants are lower in potential energy than the products
A. Change in enthalpy is a state function
Which of the following statements are true? A. Change in enthalpy is a state function B. A chemist takes the surroundings point of view when determining the sign for work or heat C. Heat of reaction and change in enthalpy can always be used interchangeably D. In exothermic reactions, the reactants are lower in potential energy than the products
A. Change in enthalpy is a state function H=E+PV Since internal energy, pressure, and volume are all state functions, enthalpy is also a state function.
Two solutions are mixed in a coffee cup and the calorimeter and the temperature goes up. A. Exothermic B. Endothermic C. Thermoneutral D. Neutralization
A. Exothermic
What is the identity of a 100. g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0 °C? A) Mg B) Al C) Ti D) Fe
Al Explanation: The specific heat capacity can be calculated using the mass, heat, and the change in temperature. Then, it can be compared to the table of different metals' specific heat capacity. This is equal to the specific heat capacity for aluminum. Q=mCAT 4680J=(100g)x(c)x(52C) C=0.90J-g-C
In which one of the following processes is the system endothermic? A) A process in which the air around the reaction becomes warmer. B) A process taking place in solution where the solution temperature decreases. C) The combustion of natural gas. D) The formation of frost on a window.
B) A process taking place in solution where the solution temperature decreases. Explanation: An endothermic process requires heat. The solution temperature decreases because the solvent is part of the surroundings and the chemical system is pulling heat from the surroundings.
For which one of the following is the enthalpy of the reaction the same as the enthalpy of formation? A) H₂ (g) + F₂ (g) → 2 HF (g) B) Ca (s) + ½ O₂ (g) → CaO (s) C) 2 CH₃OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 4 H₂O(g) D) SrO (s) + CO₂ (g) → SrCO₃ (s)
B) Ca (s) + ½ O₂ (g) → CaO (s) Explanation: Determine which enthalpy reaction has the elements in their natural states and is a synthesis reaction. Then, remember that for an enthalpy of formation, you must make 1 mole of the compound.
Which one of the equations below is an exothermic reaction? A) CO₂ (g) → C (s) + O₂ (g) ∆H° = 394 kJ/mol B) CaO (s) + H₂O (l) → Ca(OH)₂ (aq) ∆H° = -64 kJ/mol C) C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol D) N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol
B) CaO (s) + H₂O (l) → Ca(OH)₂ (aq) ∆H° = -64 kJ/mol Explanation: An exothermic reaction has a negative ∆H. The negative sign indicates that the energy is leaving the chemical system. The reaction enthalpy is per mole of reaction.
Which one of the following processes is exothermic? A) Boiling a liquid. B) Condensation of a gas into a liquid. C) Melting a solid. D) Converting solid into gas.
B) Condensation of a gas into a liquid. Explanation: An exothermic process releases heat. Boiling a liquid, melting a solid and converting a solid into a gas all require energy be put into the system. Gases have more energy than liquids and condensation is the change in phase from gas to liquid. This would result in the release of heat.
Cl₂ is a stable diatomic molecule. It can be decomposed to form two Cl atoms as shown below. Cl₂ → 2Cl(g) Predict the change in enthalpy (∆H) of this reaction. A) ∆H is (-); endothermic B) ∆H is (+); endothermic C) ∆H is (-); exothermic D) ∆H is (+); exothermic E) ∆H = 0; thermoneutral
B) ∆H is (+); endothermic Explanation: ∆H is (+); endothermic. Cl₂ is a stable molecule so bond-breaking will be endothermic requiring an input of heat. Bond-breaking is always endothermic.
Which of the following is not a state function? A. Energy B. Heat C. Temperature D. Potential Energy
B. Heat
To calculate the reaction enthalpy for the combustion of methane, which heat of formation can be omitted? A. Methane B. Oxygen C. Carbon dioxide D. Water
B. Oxygen Explanation: When determining the enthalpy of a reaction, the elements have to be in their natural states and have to be in a synthesis reaction. Remember that for an enthalpy of formation, you must make 1 mole of the compound.
A balloon is expanding from 5L to 10L against an outside pressure of 1 bar while it absorbs 500J of heat. How does its temperature change? A. Decreases B. Stays constant C. Increases D. Need more info to answer this
B. Stays constant 1 atm=1.01325 bar P∆V=nR∆T (1 atm(1.01325 bar/1atm))(10L-5L)=nR∆T ∆T stays constant
The heat capacity of an ideal gas is greatest when which of the following is true? A. The gas is monatomic and pressure is constant B. The gas is diatomic and the pressure is constant C. The gas is diatomic and volume is constant D. The gas is monatomic and volume is constant
B. The gas is diatomic and the pressure is constant Explanation: Diatomic can store more energy (more ways) than monatomic. Molar Heat Capacities at a set temperature are usually higher for Cp than Cv for certain gases.
An 80.0 g sample of metal, initially at 96.0 °C, is placed into 150. g of water initially at 26.0 °C in a calorimeter. The final temperature of the water is 28.1 °C. What is the identity of the metal? (The specific heat of water is 4.18 J/g･°C.) A) Fe B) Sn C) Ag D) Cu
C) Ag The heat released by the metal and the heat absorbed by the water are the same.The specific heat capacity can be calculated using the mass, heat, and the change in temperature. qmetal=qwater mmetalcmetal∆Tmetal=mwatercwater∆Tmetal cmetal=mwatercwater∆Twater/mmetal∆Tmetal =(150g H2O)(4.18 Jg-1C-1)(28.1-26.0)/(80.0g metal)(96.0-28.1) =0.242Jg-1C-1 Then, it can be compared to the table of different metals' specific heat capacity. This is closest to the specific heat capacity for silver.
What property of water makes it a good medium for heat exchange in calorimetry? A) High liquid density B) High dielectric constant C) High specific heat capacity D) High refractive index E) High cryoscopic constant
C) High specific heat capacity Explanation: Because water has a high specific heat capacity, it can absorb a lot of heat energy without boiling or even changing its temperature as much as other materials would. This makes water a good heat exchange medium for calorimetry.
Which one of the equations below is an endothermic reaction? A) SrO (s) + CO₂ (g) → SrCO₃ (s)∆H° = -234 kJ/mol B) H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol C) H₂ (g) + C (s) + N₂ (g) → 2 HCN (g) ∆H° = 270.3 kJ/mol D) 2 K (s) + 2 H₂O (l) → 2 KOH (aq) + H₂ (g) ∆H° = -393.1 kJ/mol
C) H₂ (g) + C (s) + N₂ (g) → 2 HCN (g) ∆H° = 270.3 kJ/mol Explanation: An endothermic reaction has a positive ∆H. The positive sign indicates that the energy is being added to the chemical system. The reaction enthalpy is per mole of reaction.
Which of the following is NOT a statement of the First Law of Thermodynamics? A) Energy can neither be created nor destroyed. B) The total energy of the universe is constant. C) The total entropy of an isolated system cannot decrease. D) Energy can only be converted from one form to another.
C) The total entropy of an isolated system cannot decrease. Explanation: The First Law of Thermodynamics states that energy cannot be created nor destroyed. The statement that entropy must never decrease for an isolated system is based on the Second Law of Thermodynamics.
Based on the law of conservation of energy, the ___ of potential and kinetic energy is constant. A) product B) difference C) sum D) quotient E) logarithm
C) sum Explanation: The sum of kinetic and potential energy is constant. For example, a ball on a track has only potential energy before it starts rolling, and an equivalent amount of kinetic translational and rotational energy at the bottom of the track.
Given the equation S(s)+O2(g)=SO2(g), ∆H=-296kJ, which of the following statement(s) is (are) true? I. The reaction is exothermic II. When 0.500 mol sulfur is reacted, 149L of energy is released III. When 32.0g of sulfur is burned, 2.96 x 10^-5 J of energy is released A. None is true B. I and II are ture C. All are true D. Only II is true E. I and III are true
C. All are true
You have a 50.0-g sample of water at 20C. To it, you add a 50g sample of iron at 80C. The final temperature of the water should be A. Between 50C and 80C B. Above 80C C. Between 20C and 50C D. 50C
C. Between 20C and 50C -Qm=QH2O -(mCAT)m=mCATH2O -(50g(0.449 J/gC)(Tf-80C))=50g(4.184J/gC)(Tf-20C) -(22.45J/C(Tf-80C))=209.2J/C(Tf-20C) -22.45Tf+1796=209.2Tf-4184 *Plug into numeric solver Tf=25.8
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10C, the water is A. frozen solid B. Boiling C. Still a liquid D. Completely vaporized
C. Still a liquid Explanation: Q=mCAT AT=Q/mc AT=(5000J)/(15.5)(4.184) AT=77.09C 77.09+10=87.09C 5.0kJ (5000J)=15.5g(4.184J/gC)(10C) Energy required to vaporize water: 2260 J Energy required to freeze water: 334 J Boiling point of water at sea level is 100Cp
Which one is not a formulation of Hess's Law? A. Change in enthalpy from initial to final state is independent of the pathway B. From reactants to products, change in enthalpy is the same whether the reaction is in one step or in a series of steps C. The heat flow equals ∆E at constant volume but ∆H at constant pressure D. In a closed thermochemical cycle, the net enthalpy change is zero
C. The heat flow equals ∆E at constant volume but ∆H at constant pressure
For a particular process, q=-10kJ and w=25 kJ. Which of the following statements is true? A. ∆E=-35kJ B. System flows from the surroundings to the heat C. The surroundings do work on the system D. ∆E=-15kJ E. The system does work on the surroundings
C. The surroundings do work on the system Explanation: Since W>0; The surroundings do work on the system. If W<0, the system does work on the surroundings If Q>0, heat goes into the system If Q<0,, heat is released from the system
Which of the following is a state function? A) Pressure B) Volume C) Temperature D) All of the above
D) All of the above Explanation: Pressure, volume, and temperature are all examples of state functions. State functions are properties that have values that are pathway independent.
Which of the following has an heat of formation equaling zero? A) H B) N C) O D) F₂ E) None of the above
D) F₂ Explanation: Elements that are in their natural state have an enthalpy of formation that is equal to zero. F₂ is the only element on the list that is in its natural state. The others are diatomic molecules in their natural state.
For which one of the following is the enthalpy of the reaction the same as the enthalpy of formation? A) 2 H (g) + Se (s) → H₂Se (g) B) KOH (s) → K⁺(aq) + OH⁻ (aq) C) 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) D) Ni (s) + ½ O₂ (g) → NiO (s)
D) Ni (s) + ½ O₂ (g) → NiO (s) Explanation: Determine which enthalpy reaction has the elements in their natural states and is a synthesis reaction. Then, remember that for an enthalpy of formation, you must make 1 mole of the compound.
A plot of pressure-volume isotherms for an ideal gas at two different temperatures is shown below. In which of these processes would q > 0? A) X → Y B) Y → Z C) Z → X D) X → Y and Y → Z E) Y → Z and Z → X (See image on Chem 101)
D) X → Y and Y → Z Explanation: X → Y and Y → Z In the process of Y → Z, heat flows into the system so q > 0. In the process of X → Y, the gas is expanding, it is doing work on the system so q > 0. In the process Z → X, heat is flowing out of the system so q < 0.
You have a 50.0-g sample of water at 20C. To it, you add a 50g sample of water at 80C. The final temperature of the water should be A. Between 50C and 80C B. Above 80C C. Between 20C and 50C D. 50C
D. 50C Temperature of the same substance. (M1T1+M2T2)/(M1+M2) (50g(20C)+50(g)(80g))/(50g+50g)=50C
Which of the following is true? A. The surroundings do work on the system when an ideal gas expands against a constant external pressure. B. The internal energy of a system increases when more work is done by the system than heat was flowing into the system. C. The internal energy of a system decreases when work is done on the system and heat is flowing into the system D. The system does work on the surroundings when an ideal gas expands against a constant external pressure.
D. The system does work on the surroundings when an ideal gas expands against a constant external pressure. W=-P∆V
For a reaction with net mole increase of gases, the pressure-volume work is A. Positive B. Negative C. Zero D. Zero in a bomb calorimeter
D. Zero in a bomb calorimeter
In any Hess's law calculation, the enthalpy of formation for an element in its standard state is defined as: A) -100 kJ/mol B) 10 kJ/mol C) -10 kJ/mol D) 100 kJ/mol E) 0 kJ/mol
E) 0 kJ/mol Explanation: All elements in their standard states are defined as having an enthalpy of formation of 0 kJ/mol. It takes no energy for an element to be in its natural state.
An unknown solid is added to water in a calorimeter. The temperature of the water increases. What does this say about the reaction? A) The reaction is exothermic. B) The reaction is endothermic. C) The reaction has absorbed heat from the solution. D) The reaction has released heat to the solution. E) Both A and D.
E) Both A and D. Explanation: If the temperature of the water increased, this indicates that an exothermic reaction occurred and that the heat was released to the solution.
A 40.0 g ball traveling at a speed of 2.30 m/s has a kinetic energy of
KE=1/2mv^2 m=kg KE=1/2(0.04)(2.3m/s)^2 KE=0.106J
