CHEM-131 Knewton Alta 9.1 Homework Energy & Thermochemistry
When 1.34 g Zn(s) react with an excess of HCl, 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction: Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) Select the correct answer below: −418 kJmol Zn −153 kJmol Zn −42.6 kJmol Zn −20.5 kJmol Zn
-153 kJmol Zn First, you need to find the moles of zinc that reacted in this equation by dividing the grams given by the molar mass from the periodic table. 1.34 grams Zn×1 mol Zn65.38g=0.0205 moles Zn Next, you need to find the number of kJ released per mole of Zn based on the fact that 3.14 kJ are released when exactly 0.0205 moles Zn react. ΔH=qrxnnZn=−3.14 kJ0.0205 moles Zn=−153 kJmol Zn Remember, the value is negative because the reaction is producing (not absorbing) heat, meaning it is exothermic.
What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)? Report your answer in scientific notation. Your answer should have two significant figures. Use −890.8kJmol for the molar heat of combustion of methane.
-2.9 X 10^4 kJ First, calculate the number of moles of CH4 using a formula mass of 16.043gmol from the periodic table. 530g×1.0mol16.043g=33.036mol Next, calculate the total change in enthalpy. ΔH=33.036mol×−890.8kJmol=−29,428kJ The answer should have two significant figures, so round to −2.9×104kJ.
Given that CH4 (g)+Cl2 (g)→CH3Cl (g)+HCl (g) has an enthalpy change of −99.6 kJ, and CH3Cl (g)+Cl2 (g)→CH2Cl2 (g)+HCl (g) has an enthalpy change of −105.8 kJ, what is the change in enthalpy for CH4 (g)+2Cl2 (g)→CH2Cl2 (g)+2HCl (g)? Your answer should have three significant figures.
-205 kJ Since this represents two consecutive processes to yield a particular product, we can see that nothing needs to be done to either equation. They can simply be added together as they are, and things will cancel to yield the desired product. −99.6 kJ-105.8 kJ=−205 kJ
When 2.16g of H2 reacts with excess O2 by the following equation, 258kJ of heat are released. What is the change of enthalpy associated with the reaction of 1.00mol of hydrogen gas? 2H2+O2⟶2H2O
-241 kJmol H2 Since heat is being produced, the forward reaction is exothermic. Therefore, change in enthalpy associated with the reaction of 2.16g of H2 is −258kJ . Now, convert the mass of H2 to moles using the molar mass. 2.16gH2×1.00 molH22.016gH2=1.0714molH2 Now use the amount ofH2inmolto determine the enthalpy change accociated with the reaction of1.00molofH2. ΔH=qrxnnH2=−258kJ1.0714molH2=−240.8kJ/molH2 The answer should have three significant figures, so round to −241kJmol H2.
Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2 kJ What is the heat of formation of SO2 in kilojoules? S (s)+O2 (g)→SO2 (g) Your answer should have four significant figures.
-296.1 kJ Reverse the second equation to get SO2 as a product, combine the equations, then divide the combined equation by two, as follows. Reverse the second equation and change the sign of its enthalpy. 2S (s)+3O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ 2SO3 (g)←2SO2 (g)+O2 (g) has an enthalpy change of 198.2 kJ Combine the equations and the enthalpies. 2S (s)+2O2 (g)→2SO2 (g) −790.4 kJ+198.2 kJ=−592.2 kJ Divide the combined equation and the enthalpy by two to obtain the heat of formation equation. S (s)+O2 (g)→SO2 (g)= −296.1 kJ Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.
Calculate ΔH∘298 in kilojoules for the process: Sb(s)+52Cl2(g)⟶SbCl5(g) Given the information below: Sb(s)+32Cl2(g)⟶SbCl3(g)ΔH∘298=-314 kJSbCl3(g)+Cl2(g)⟶SbCl5(g)ΔH∘298=-80.0 kJ
-394 kJ Note that the chemical reactions corresponding to the enthalpy values given already add up to the target overall reaction. Therefore, we can simply sum the given values of ΔH to obtain the enthalpy of the overall reaction. ΔHtotal=−314 kJ+(−80.0 kJ)=−394 kJ
If the heat of formation for CO2 is −393.5kJmol, what is the change in enthalpy when 5.00 g of CO2 are formed? Your answer should have three significant figures. (Round your answer to first decimal place).
-44.7 kJ First, calculate the number of moles of CO2 using a formula mass of 44.009gmol from the periodic table: 5.00g×1.00mol44.009g=0.11361mol Next, calculate the total change in enthalpy: ΔH=0.11361mol×(−393.5kJmol)=−44.706kJ The answer should have three significant figures, so round to −44.7kJ.
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at 25.0∘C. After the reaction, the final temperature of the water is 33.2∘C . Calculate the heat of combustion for 1.00mol of octane. Use 4.184Jg∘C for the specific heat capacity of water. Your answer should have three significant figures.
-5490 kj First find the heat of combustion of 1g of octane (C8H18) using the given information, and then use the molar mass to determine the heat of combustion for 1mol of octane. First recognize that the heat released by the combustion reaction is entirely absorbed by the water and the calorimeter. 0qrxn=qrxn+qwater+qcalorimeter=−(qwater+qcalorimeter) qrxn=−[(1200.g)(4.184Jg∘C)(33.2∘C−25.0∘C)+(837J∘C)(33.2∘C−25.0∘C)]qrxn=−48033.96J Therefore, the heat of combustion for 1.00g of octane is about −48.034kJg. Now, multiply this value by the molar mass to determine the heat of combustion for 1mol of octane. qrxnqrxn=114.232gmol×−48.034kJg×1mol≈−5487.02kJ Rounding the answer to 3 significant figures, we find that the molar heat of combustion for 1mol of octane is approximately −5490kJ.
Given the following equations, determine the standard enthalpy of formation (ΔH∘f) for one mole of ICl3(g). I2(g)+3Cl2(g)⟶2ICl3(g)ΔH∘298=−214 kJI2(s)⟶I2(g)ΔH∘298=38 kJ Note that iodine (I2) is a solid at standard conditions. Your answer should have two significant figures.
-88kJ ΔH∘f is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. We are trying to find the standard enthalpy of formation of ICl3(g), which is equal to ΔH∘ for the reaction below. Recall that iodine is a solid at standard conditions. 12I2(s)+32Cl2(g)⟶ICl3(g)ΔH∘f=? Looking at the reactions, we see that the reaction for which we want to find ΔH is the sum of the two reactions with known values of ΔH, so we can sum their ΔH values. Step 1: I2(s)⟶I2(g)ΔH∘=38 kJStep 2: I2(g)+3Cl2(g)⟶2ICl3(g)ΔH∘=−214 kJSum: I2(s)+3Cl2(g)⟶2ICl3(g)ΔH∘=−176 kJ Divide both sides of the summed equation and the enthalpy by two to obtain the target equation and the corresponding enthalpy. 12I2(s)+32Cl2(g)⟶ICl3(g)ΔH∘f=−88 kJ Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.
Consider the reaction described by the following chemical equation. 2HN3(l)+2NO(g)⟶H2O2(l)+4N2(g) What is the enthalpy change associated with the production of 1mol of H2O2 if a reaction that produces 2.50g of H2O2 releases 65.9kJ of heat? Assume the reaction occurs under conditions of constant pressure. Report the answer to three significant figures. Include a negative sign if appropriate.
-897 kJmol This reaction occurs under conditions of constant pressure, so the enthalpy change associated with the production of 2.50g of H2O2 is equal to the heat of reaction. ΔHrxn=qrxn=−65.9kJ Note that the value is negative because heat is released in this reaction (the reaction is exothermic). Use the molar mass to convert the mass of H2O2 to moles of H2O2. 2.50gH2O2(1molH2O234.014gH2O2)=0.073499molH2O2 To calculate the enthalpy change per mole of H2O2, divide by the number of moles of H2O2. ΔH=−65.9kJ0.073499mol=−896.6kJmol Rounding the answer to three significant figures, the enthalpy change associated with the production of 1mol of H2O2 is ΔH=−897kJmol Molar mass values are per 1 mole exactly, so "1 mol" in a numerator or denominator involving molar mass does not constrain the number of significant figures.
Calculate the change in enthalpy associated with the combustion of 322 g of ethanol. Report your answer in scientific notation. Your answer should have three significant figures. C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH∘c=−1366.8kJ/mol
-9.55*10^3 kJ First, calculate the number of moles of C2H5OH using a formula mass of 46.069gmol from the periodic table: 322g×1.00mol46.069g=6.9895mol Next, calculate the total change in enthalpy: ΔH=6.9895mol×(−1,366.8kJmol)=−9,553.3kJ The answer should have three significant figures, so round to −9.55×103kJ.
Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber brakes of an airliner heat up to 312∘C. As the brakes start to cool down, the heat is absorbed by the 123 kg rubber tires. What is the specific heat of the tires if the final temperature of both the brakes and the tires at thermal equilibrium is 172∘C? Enter the answer with three significant figures. Use 1.400Jg∘C for the specific heat of carbon fiber.
.921 J/g degrees C Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation). Therefore, heat given off by the brakes = −heat taken in by tires, or: −qbrakes=qtires The equation used to calculate the quantity of heat energy exchanged in this process is: −qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires First we must convert the mass of the tires and the brakes from kg to g. massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g masstires=123 kg×1,000. g1 kg=1.23×105 g Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C. −(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C) ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C The answer should have three significant figures, so round to 0.921Jg∘C.
A piece of unknown metal weighs 217 g. When the metal piece absorbs 1.43 kJ of heat, its temperature increases from 24.5∘C to 39.1∘C. Determine the specific heat of this metal, and predict its identity using the chart above. Select the correct answer below: 0.00045 J/g∘C; the metal is not any of those listed above 0.131 J/g∘C, the metal is likely to be lead 0.712 J/g∘C, the metal is likely to be silicon 0.451 J/g∘C, the metal is likely to be iron
0.451 J/g∘C, the metal is likely to be iron First, convert kJ to J by multiplying by 1000: 1.43 kJ×1000 JkJ=1430 J Second, plug the given values into the equation and solve for specific heat (c). qc=mcΔT=qmΔT=1430 J(217 g)(39.1∘C−24.5∘C)=0.451Jg ∘C Therefore, the specific heat capacity is about 0.451 J/g ∘C, and the metal is likely iron (Fe).
If two lattice structures have the same interionic distance, but the first lattice contains ions of 1+ and 2− charge, while the second contains ions of 2+ and 1− charge, what will be the ratio of their lattice energies? 1 to 2 2 to 1 1 to 1 4 to 1
1 to 1 Each lattice has charges of magnitudes 1 and 2, and it will not matter what their sign is, so they will have identical lattice energies.
If two lattice structures have the same interionic distance, but the first lattice contains ions of 1+ and 1− charge, while the second contains ions of 2+ and 2− charge, what will be the ratio of their lattice energies? Select the correct answer below: 1 to 2 2 to 1 1 to 4 4 to 1
1 to 4 Since each charge doubles, the lattice energy must quadruple.
A 10.0 g gold ring with a specific heat 0.129 Jg∘C at 24.00∘C is placed in a calorimeter with 118 g of water at 1.00∘C. What will be the final temperature of the system? Round your answer to two decimal place. Use 4.184Jg∘C for the specific heat of water.
1.06 degrees C Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the metal must be gained by the water. Therefore, heat given off by the metal = −heat taken in by water, or: −qring=qwater The equation used to calculate the quantity of heat energy exchanged in this process is: −qring=−cring mring ΔTring=cwater mwater ΔTwater=qwater Heat stops flowing when the two samples are at the same temperature, so the final temperature of the water will be the same as the final temperature of the ring. Substitute in the known values for the equation above and rearrange to solve for Tf. −(0.129Jg∘C)(10.0 g)(Tf−24.00∘C)=(4.184Jg∘C)(118. g)(Tf−1.00∘C) Simplify by multiplying specific heat and mass. −(1.29J∘C)(Tf−24.0∘C)=(493.7J∘C)(Tf−1.00∘C) Simplify by multiplying heat capacity by temperature. −(1.29TfJ∘C)+30.96J=(493.7TfJ∘C)−493.7J Combine like terms. −494.99Tf J∘C=−524.66 J Tf=1.0599∘C The answer should have three significant figures so round to 1.06∘C.
Calculate the heat capacity of 18.5 g iron given that 2,982 J is needed to raise the temperature 194 ∘C. Select the correct answer below: 6.3J∘C 12.4J∘C 10.3J∘C 15.4J∘C
15.4J∘C C=qΔT2,982 J194 ∘C=15.4J∘C
The SI unit for energy is the Joule. Which of the following is/are equal to 1 joule? Select all that apply: 1N×m 1kg×ms 1kg×m2s2 1Nm
1st and 3rd options are correct: 1N×m 1kg×m2s2 The joule is equal to the amount of work that is done when a 1N force is applied over a distance of 1m. Therefore, the joule has units of N×m or kgm2s2, which are all SI units themselves. The calorie with a lower case "c" has been the traditional unit of energy defined as the amount of energy needed to raise 1 gram of water 1∘C. In an effort to standardize the definition of a calorie and create an SI unit for energy, the Joule was created and defined as the energy used when 1 Newton of force moves and object 1 meter. Then a calorie was defined as being equal to 4.184 J. The Calorie with a capital "C" is the largest unit of energy because it is what we recognize as the food calorie which is 1 Calorie=1,000 calories=1 kcal . Therefore, the Calorie > calorie > joule, and the Newton is a unit of force that is used in the definition of the joule. The Newton is not a unit of energy.
David is cooking a brisket and needs his casserole dish to be exactly 150∘C. It takes 32,000 J to heat the casserole dish to 150∘C from 0∘C. What is the heat capacity of the dish in joules per degrees Celsius? Round your answer to the nearest whole number.
213 joules per degree Celsius To find the heat capacity of the dish, we can divide the given energy that the dish absorbed by its resultant temperature change according to the following equation: Ccasserole dish=qΔT First calculate ΔT using the following formula: ΔT=Tfinal−Tinitial ΔT=150∘C−0∘C=150∘C Next, substitute in the known values and solve for Ccasserole dish. C=32,000 J150∘C≈213.33 J∘C Rounding to the nearest whole number, the value is 213J∘C.
If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what will be the final temperature of the object? Report your answer with the correct number of significant figures.
27.2 degree C The final temperature can be determined by rearranging the specific heat capacity equation for ΔT and substituting the known values of mass, specific heat, and the heat released (which is represented by a negative number). qcm=ΔT −450 J(2.89 Jg∘C)(20.0 g)=ΔT -7.79∘C=ΔT The negative sign of ΔT makes sense, as we know that energy is released in this exothermic process: (Tinitial+ΔT=Tfinal) 35.0∘C+(-7.79∘C)=27.2∘C
In the heat of formation equation for CH3CH2OH, what would be the coefficient for hydrogen gas?
3 Since there are six hydrogen atoms in one molecule of ethanol, we will need three H2 molecules on the left as shown in the balanced reaction below. 2C(s)+3H2(g)+12O2(g)⟶CH3CH2OH(l)
The enthalpy of formation for ammonia (NH3) is −46.19 kJ/mol. What does this mean? Select the correct answer below: 46.19 kJ of energy are released for every mole of ammonia that is formed from its elements in their most stable states under standard conditions. 46.19 kJ of energy are absorbed for every mole of ammonia that is formed from its elements in their most stable states under standard conditions. 46.19 kJ of energy are needed to create ammonia from its elements in their most stable states under standard conditions. 46.19 kJ of energy are converted to chemical energy when ammonia is formed from its elements in their most stable states under standard conditions.
46.19 kJ of energy are released for every mole of ammonia that is formed from its elements in their most stable states under standard conditions. Since the enthalpy of formation for ammonia is a negative (−) sign we know it is an exothermic process. This means that 46.19 kJ of energy are released for every mole of ammonia that is formed.
How much heat, in joules, must be added to a 500 g iron skillet to increase its temperature from 25∘C to 250∘C? The specific heat of iron is 0.451Jg∘C. Select the correct answer below: −6.2×104 J −6.2×105 J 5.1×104 J 5.1×105 J
5.1×104 J q=(specific heat)×(mass of a substance)×(temperature change)q=(0.451Jg ∘C)×(500 g)×(250∘C−25∘C)q=(0.451Jg ∘C)×(500 g)×(225∘C) q=5.1×104 J
5.0 g of C8H18 are burned in a bomb calorimeter with 750 g water, and there is 10.0∘C increase in temperature when 32.0 kJ of heat is released. What is the heat capacity of the calorimeter?
62 J/degree C qrxn=−[qwater+qbomb]qrxn=−(cwatermΔT+CcalorimeterΔT)−32,000 J=−[(4.184 Jg∘C)(750 g)(10.0∘C)+(Ccalorimeter)(10.0∘C)]−32,000 J=−31,380 J−(Ccalorimeter)(10.0∘C)−620 J=−(Ccalorimeter)(10.0∘C)Ccalorimeter=62 J∘C
If the heat of combustion of hydrogen gas (H2) is −285.8kJmol, how many grams of H2 must combust in order to release 1.2×103kJ of heat? Your answer should have two significant figures.
8.5g First, calculate the number of moles of hydrogen gas that must combust. Note that a release of heat corresponds to a negative value of ΔH. −1.2×103kJ×1.0mol−285.8kJ=4.199mol Next, calculate the number of grams of hydrogen, using a formula mass of 2.016gmol for H2(g). 4.199mol×2.016gmol=8.465g The answer should have two significant figures, so round to 8.5g.
Which of the following is true regarding the enthalpy change (ΔH) of a reaction? Select all that apply: Increasing the coefficients of the reaction increases the enthalpy change, ΔH. A positive enthalpy change (ΔH) indicates an exothermic reaction because heat is being absorbed. A change in the physical state of the reactants and products will affect the enthalpy change, ΔH. A negative enthalpy change (ΔH) indicates an endothermic reaction because heat is being released.
A. and C. are correct: Increasing the coefficients of the reaction increases the enthalpy change, ΔH. A change in the physical state of the reactants and products will affect the enthalpy change, ΔH. If the coefficients of the reaction increase, then the enthalpy change increases as well. If you double the coefficients, the enthalpy change doubles. If you cut the coefficients in half, the enthalpy change gets cut in half. Changing the physical state of the reactants also affects enthalpy change. A reaction that produces a gas will have a different enthalpy change than one that produces a liquid. Finally, enthalpy changes are negative if heat is released (exothermic) and positive if heat is absorbed (endothermic).
The _______ breaks down the formation of an ionic solid into a series of individual steps. Born-Haber cycle Bohr cycle Hess cycle none of the above
Born-Haber cycle The Born-Haber cycle breaks down the formation of an ionic solid into individual steps. Those steps are the standard enthalpy of formation of the compound, the ionization energy of the metal, the electron affinity of the nonmetal, the enthalpy of sublimation of the metal, the bond dissociation energy of the nonmetal, and the lattice energy of the compound.
Which is the largest unit of energy? Select the correct answer below: joule calorie Calorie Newton
Calorie The Calorie, spelled with a capital letter C, or food calorie, is 1,000 calories, each of which is larger than a joule. The Newton is a unit of force.
4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What is the specific heat of the metal? Use 4.184Jg∘C for the specific heat of water. Water has a density of 1 g mL. Select the correct answer below: 2.2 J/g∘C 3.5 J/g∘C 4.184 J/g∘C 850 J/g∘C
First, you need to find the q value for the water using the equation q=cmT Where: c is the specific heat of water (4.184) m is the amount of the substance present in grams. For water, mL of water is converted to g of water because the density of water is 1 g mL. Therefore, 35 mL=35 g ΔT is the change in temperature, 34.5−28.7=5.8∘C q=(4.184Jg∘C)(35 g)(5.8∘C) 850 J is the amount of heat absorbed. This is also equal to the amount of heat released. qabsorbed=850 J=qreleased Now we can calculate the specific heat of the metal. Use the same equation but with the given values for the metal: q=cmΔT Where: q is the heat absorbed/released (850J) c is the unknown specific heat m is the amount of metal, 4.82 g ΔT is the change in temperature, 115.0−34.5=80.5∘C 850 J=(c)(4.82 g)(80.5∘C) Rearrange to solve for c. c=850 J(4.82 g)(80.5∘C) c=2.19Jg∘C The answer should have two significant figures. Round your answer to 2.2Jg∘C.
Which of the following is true? Select the correct answer below: Heat and energy are lost during chemical reactions. A warmer object will have lower kinetic energy. A colder object will have higher kinetic energy. Heat flow will continue between objects until the substances in question are at the same temperature.
Heat flow will continue between objects until the substances in question are at the same temperature. Heat (q) is the transfer of thermal energy between two bodies at different temperatures. Heat flow increases the thermal energy of one body and decreases the thermal energy of the other. Heat flow will continue between objects until the substances in question are at the same temperature. Thermal energy is kinetic energy associated with the random motion of atoms and molecules. A colder object will have lower kinetic energy and a warmer object will have higher kinetic energy. Matter undergoing chemical reactions and physical changes can release or absorb heat.
Magnesium oxide (MgO) and sodium chloride (NaCl) have the same crystal structure. Which of these has the higher lattice energy? Select the correct answer below: MgO NaCl They have the same lattice energy.
MgO MgO would have the higher lattice energy because the Z values of both the cation and the anion in MgO are greater, and the interionic distance of MgO is smaller than that of NaCl.
Which of the following compounds would require dividing the bond dissociation energy in half when calculating the lattice enthalpy? Select the correct answer below: NaCl CaBr2 ScF3 Ba3P2
NaCl Since exactly 1 mol of solid Na combines with exactly 12 mol of gaseous Cl2 to form precisely 1 mol of NaCl(s), the bond dissociation energy will need to be divided in half.
What is energy? Select the correct answer below: The motion an object possesses The capacity to supply heat or do work The downward flow of temperature in an object all of the above
The capacity to supply heat or do work Energy is the capacity to supply heat or do work. The two main types of energy are potential energy and kinetic energy.
A 1g sample of copper and a 50g sample of copper are both heated. Which item has the higher specific heat (c)? Select the correct answer below: The 1g sample The 50g sample They have the same specific heat capacities There is not enough information to know
They have the same specific heat capacities Although the 50g sample is more massive than the 1g sample, they are both are made of the same material, thus they both have the same specific heat. Specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). Hence, specific heat is independent of the mass of the substance.
When a substance undergoes a chemical reaction in a closed system, which of the following will NOT change for the system? (select all that apply) Select all that apply: temperature energy density mass
energy mass Due to the Law of Conservation of Energy (energy can be neither created or destroyed) and the Law of Conservation of Mass (there is no detectable change in the total amount of matter during a chemical change), neither energy nor mass has detectable change when undergoing a chemical reaction in a closed system. The total quantity of matter and energy in the universe is fixed.
Which of the following are types of energy? (select all that apply) Select all that apply: metabolism heat chemical potential kinetic thermochemistry
heat chemical potential kinetic Heat, chemical, potential and kinetic are all types of energy. Potential energy is the energy an object has by virtue of its position in a field and it converts to kinetic energy which is the energy of motion. Thermochemistry is the study of heat exchange during chemical reactions and metabolism is the process of converting chemical energy to usable energy in a living organism.
A hot shower is associated with _____________ thermal energy, and the heat will flow from a __________ object to a ___________ object. Select the correct answer below: low; hot, cold low; cold, hot high; hot, cold high; cold, hot
high; hot, cold Thermal energy is the kinetic energy associated with the random motion of atoms and molecules. In a hot shower, the particles are moving more quickly than they would in a cold shower. Additionally, heat is the exchange of thermal energy, and heat will always flow from a hot object to a cold object.
Which of the following quantities can be calculated using the Born-Haber cycle? ionization energy dissociation energy electron affinity heat of vaporization
ionization energy dissociation energy electron affinity Ionization energy, dissociation energy and electron affinity are all involved in the Born-Haber cycle. Each of these energy terms can be calculated using the general equation associated with the cycle, reproduced below, so long as the values of all other terms are known: ΔHlattice=νΔHf−νΔHsub−νD−∑ν(IE)−∑ν(EA)
Which of the following is an example of an endothermic process? Select the correct answer below: melting ice cubes burning sugar a candle burning condensation of rain from water vapor
melting ice cubes Melting ice cubes are an example of an endothermic process, because heat energy is taken from the surroundings when ice cubes melt. Burning sugar and a candle burning are both chemical reactions which give off heat energy, so they are both exothermic. When water vapor condenses, more energetic gas molecules become less energetic liquid molecules; the heat energy from the more energetic molecules in the gas phase is given off to the environment, making it an exothermic process.
Which symbol represents the amount of heat exchange in a reaction? Select the correct answer below: C m q c
q The amount of heat exchanged in a chemical reaction is represented by the letter q. The equations below are used to calculate this amount. q=CΔT and q=cmΔT Where q is the heat exchange C is heat capacity c is specific heat capacity ΔT is the change in temperature
Heat is abbreviated as: Select the correct answer below: h q H T
q The symbol for heat is a lowercase q. A lowercase h represents height, and an uppercase H represents enthalpy. An uppercase T represents temperature.
An ionic compound is stable due to: Select the correct answer below: the size and charges of its ions the electrostatic attraction between its positive and negative ions the overall mass of the compound all of the above
the electrostatic attraction between its positive and negative ions The strong electrostatic attraction between the cations (+) and anions (−) in an ionic compound due to their opposite charges allows for an extremely stable bond to form.
Which of the following thermodynamic quantities are included in the Born-Haber cycle? Select the correct answer below: the heat of vaporization of the metal the heat of fusion of the metal the heat of sublimation of the metal the heat of deposition of the metal
the heat of sublimation of the metal Metals are generally found as solids in their elemental form (with the exception of Hg), so we need to know the enthalpy of sublimation of the metal in order to convert the metal into the necessary phase that completes the cycle.
What is work, in a scientific context? Select the correct answer below: the process of causing matter to move against an opposing force the absolute motion an object possesses the capacity to supply heat the flow of temperature in an object
the process of causing matter to move against an opposing force Work is the process of causing matter to move against an opposing force. This means that if a force does work on an object, the point to which the force is applied is moved in the direction of the force.
Determine the standard enthalpy of reaction (ΔH∘rxn) given the enthalpies of formation below. 2Ag2S(s)+2H2O(l)⟶4Ag(s)+2H2S(g)+O2(g) ΔH∘fAg2S(s) = -32.6 kJmol ΔH∘fH2O(l)= -285.8 kJmol ΔH∘fAg(s)= 0kJmol ΔH∘fH2S(g)= −20.6kJmol ΔH∘fO2(g)= 0kJmol Select the correct answer below: ΔH∘reaction=148.9kJmol ΔH∘reaction=297.8kJmol ΔH∘reaction=595.6kJmol ΔH∘reaction=−297.8kJmol
ΔH∘reaction=595.6kJmol To determine the standard enthalpy of reaction, ΔH∘rxn, use the following equation. ΔH∘reaction=∑ n⋅ΔH∘f,products−∑ n⋅ΔH∘f,reactants Substitute the given enthalpies of formation into the equation. ΔH∘reactionΔH∘reactionΔH∘reaction=(nAgΔH∘fAg(s)+nH2SΔH∘fH2S(g)+nO2ΔH∘fO2(g))−(nAg2SΔH∘fAg2S(s)+nH2OΔH∘fH2O(l))=(0kJ/mol+2(−20.6kJ/mol)+0kJ/mol)−(2(−32.6kJ/mol)+2(−285.8kJ/mol))=595.6kJ/mol Therefore the enthalpy of the reaction is approximately 595.6kJmol.
If 4.40 g of glucose are burned in a bomb calorimeter, the temperature of the 800 g of water in the calorimeter increases from 15.0∘C to 18.7∘C. If the heat capacity of the calorimeter is 550J∘C, what is the value of q for the combustion of the glucose sample? The specific heat capacity of water is 4.184Jg∘C. Select the correct answer below: −14.4 kJ 14.4 kJ −16.3 kJ 16.3 kJ
−14.4 kJ To determine the qrxn, use the following equation: qrxn=−(qwater+qbomb)qrxn=−(cwatermΔT+ccalorimeterΔT) Plug in the values given in the question.qrxn=−[(4.184Jg∘C)(800 g)(3.7∘C)+(550 J∘C)(3.7∘C)]qrxn=−(12,384.64 J+2,035 J)qrxn=−14,420 J=−14.4 kJ
CO(g)+H2(g)+O2(g)⟶CO2(g)+H2O(g) What is ΔH for the reaction above, given the following reaction enthalpies 2C(s)+O2(g) → 2CO(g) = ΔH1=−222 kJ C(s)+O2(g) → CO2(g) =ΔH2=−394 kJ 2H2(g)+O2(g) → 2H2O(g) =ΔH3=−484 kJ
−525 kJ To find the unknown enthalpy change, find a combination of three given reactions that can add to the reaction of interest. Start by identifying the contributions with the fewest repeated components. Reaction (1) has the only occurrence of CO(g), with a coefficient of 2 as a product. CO(g) is a reactant with a coefficient of 1 in the net reactions, so reaction (1) has to be multiplied by 1/2 (one half or .5) and reversed, while ΔH1 is multiplied by −1/2 (negative one half) Reaction (2) has the only occurrence of CO2(g), which appears as a product with a coefficient of 1 in reaction (2) and in the net reaction, so it contributes as is. The contribution from reaction (3) can be determined by looking at the coefficients of either H2(g) or H2O(l) and either way (3) and its enthalpy change are multiplied by 1/2 to match the coefficients in the net reaction. Combining all these changes, you can see that the three modified reactions add to give the desired reaction, and the enthalpy change is calculated by adding reaction enthalpies of the modified reactions. CO(g)→C(s)+1/2O2(g) = −1/2ΔH1 = 111kJ C(s)+O2(g)→CO2(g) = ΔH2=−394 kJ H2(g)+1/2O2(g)→H2O(g)= 1/2ΔH3 = =−242 kJ CO(g)+H2(g)+O2(g)→CO2(g)+H2O(g) = ΔH = −525 kJ
How much heat is produced by the combustion of 125 g of acetylene (C2H2)? C2H2(g)+52O2(g)⟶2CO2(g)+H2O(l)ΔH∘=−1301.1 kJ/mol Select the correct answer below: 1.62×105 kJ −4.81×103kJ 3.2×103kJ −6.25×103 kJ
−6.25×10^3 kJ First you need to use the periodic table to find the molar mass of acetylene. Using the table we find that the molar mass is 26.036 g/mol. Second, convert the given mass of acetylene to moles of acetylene using its molar mass. 125 g acetylene26.036 g/mol=4.801 moles acetylene Third, find the energy (in kJ) released from that number of moles of acetylene by multiplying by the enthalpy of combustion given. 4.801 moles acetylene×−1301.1 kJ/mol=−6.25×103 kJ of heat produced
A 1.00 g sample of hydrazine (N2H4) is burned in a bomb calorimeter containing 1200 g of water that experiences a temperature change of 3.54∘C. If the heat capacity of the calorimeter is 840 J C, what is the heat of combustion for 1 mole of the sample? Select the correct answer below: −913 kJ mol −665 kJ mol −392 kJ mol −20.7 kJ mol
−665 kJ mol First write an equation to relate the heat of reaction to the heat transferred to the water and the calorimeter. qrxn=−(qwater+qbomb) Now, we can determine the amount of heat transferred to the water and the calorimeter as follows. qrxnqrxnqrxnqrxn=−(cwatermΔT+ccalorimeterΔT)=−[(4.184Jg∘C)(1,200g)(3.54∘C)+(840 J∘C)(3.54∘C)]=−20,747J=−20.747kJ The calculation above is for the heat associated with the combustion of 1 g of N2H4. To determine the heat associated with the combustion 1 mole of N2H4, multiply this value by the molar mass of N2H4. (32.046gmol)(−20.747kJg)=−664.86kJmol Therefore, after rounding the value to three significant figures, we find that the heat associated with the combustion of 1mol of N2H4 is about −665kJmol.
