chemistry chapter 6
A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed. HCl(aq) + NH3(aq) ® NH4Cl(aq) The temperature increase is 2.34°C. Calculate DH per mole of HCl and NH3 reacted.
A) -154 kJ/mol 100 mL x 0.200 M = 20 mmol = 0.020 mol total volume 200 mL; assume 200 g and 4.184 J/gK 2.34 K x 480 J/K = 1123 J 200 g x 4.184 x 2.34 K = 1958 J 1123 + 1958 = 3081 J 3081 J/ 0.020 mol = 154000 J/mol
Octane (C8H18) undergoes combustion according to the following thermochemical equation: 2C8H18(l) + 25O2(g) ® 16CO2(g) + 18H2O(l) DH°rxn = -11,020 kJ/mol. Given that DH°f[CO2(g)] = -393.5 kJ/mol and DH°f[H2O(l)] = -285.8 kJ/mol, calculate the standard enthalpy of formation of octane
A) -210 kJ/mol
The enthalpy change when a strong acid is neutralized by strong base is -56.1 kJ/mol. If 135 mL of 0.450 M HI at 23.15°C is mixed with 145 mL of 0.500 M NaOH, also at 23.15°C, what will the maximum temperature reached by the resulting solution? (Assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 J/g·°C, and that the density of the final solution is that of water.)
B) 26.06°C
How much heat is required to raise the temperature of 1.5 x 103 g of water from 45°F to 130.°F? The specific heat of water is 4.184 J/g·°C.
B) 3.0 x 102 kJ (1500 x 4.184)(54-7.2) x 1/1000= 3.0
A beaker contains 115 g of ethanol at 18.2°C. If the ethanol absorbs 1125 J of heat without losing heat to the surroundings, what will be the final temperature of the ethanol? The specific heat of ethanol is 2.46 J/g×°C.
D) 22.2°C 1125j = (115g x 2.46j/g) T-18.2= T = 22.2 °C You know the mass, the energy, the specific heat of ethanol (and whatever units), the only thing you don't know is the temperature change. Substitute your numbers and solve for the temperature change. Add the temperature change to the starting temperature, the result is the final temperature.
For which of these reactions will the difference between DH° and DE° be the greatest?
D) 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(l)
If 10.6 moles of water at 35°C absorbs 12.30 kJ, what is the final temperature of the water? The specific heat of water is 4.184 J/g·°C
D) 50.°C Using Q=m*c*(finaltemp-initialtemp), you just plug in answers (don't forget to convert 12.3 kJ to Joules). You'll also have to convert 10.6 mol H2O to grams. 1 mol H2O is approx. 17g, so 10.6mol would be 180.2g. Solve the equation 12300=180.2*4.184*(finaltemp-35). Final answer is 51.31 C.
Calculate the amount of work done, in joules, when 2.5 mole of H2O vaporizes at 1.0 atm and 25°C. Assume the volume of liquid H2O is negligible compared to that of vapor. (1 L·atm = 101.3 J) work =-pV
D) 6.19 kJ v=nrt/p h20=(2.5 mol)(0.0821L atm/molK)(298K)/1.0 atm = 61.2L w=-pV -(1.0atm)(61.2L) = -61.2 atm convert to J: -61.2 L atm(101.3J/L atm) = 6200J or 6.2 x 103J = -619
Given H2(g) + (1/2)O2(g) ® H2O(l), DH° = -286 kJ/mol, determine the standard enthalpy change for the reaction 2H2O(l) ® 2H2(g) + O2(g).
D) DH° = +572 kJ/mol
6.4) Which of the following processes is endothermic? A) O2(g) + 2H2(g) = 2H2O(g) B) H2O(g) = H2O(l) C) 3O2(g) + 2CH3OH(g) = 2CO2(g) + 2H2O(g) D) H2O(s) = H2O(l)
D) H2O(s) = H2O(l)
Which of the following processes always results in an increase in the energy of a system? A) The system loses heat and does work on the surroundings. B) The system gains heat and does work on the surroundings. C) The system loses heat and has work done on it by the surroundings. D) The system gains heat and has work done on it by the surroundings. E) None of these is always true
D) The system gains heat and has work done on it by the surroundings
6.2 Heat is A) a measure of temperature. B) a measure of the change in temperature. C) a measure of thermal energy. D) a measure of thermal energy transferred between two bodies at different temperature
D) a measure of thermal energy transferred between two bodies at different temperature
Potential energy is A) the energy stored within the structural units of chemical substances. B) the energy associated with the random motion of atoms and molecules. C) solar energy, i.e. energy that comes from the sun. D) energy available by virtue of an object's position.
D) energy available by virtue of an object's position
At 25°C, the standard enthalpy of formation of anhydrous sodium carbonate is -1130.9 kJ/mol, whereas the standard enthalpy of formation of sodium carbonate monohydrate is -1430.1 kJ/mol. Determine DH° at 25°C for the reaction Na2CO3(s) + H2O(l) ® Na2CO3·H2O(s). (Given: DH°f[H2O(l)] = -285.8 kJ/mol)
E) -13.4 kJ/mol
A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of work done on the system.
E) 1.81 × 104 J w=-pv -(10.0)(20.0-2.0)= -180 -180 x 101.3 J= 1.81 x 10^4 J
Given 2Al(s) + (3/2)O2(g) ® Al2O3(s), DH°f = -1,670 kJ/mol for Al2O3 (s). Determine DH° for the reaction 2Al2O3(s) ® 4Al(s) + 3O2(g).
E) 3,340 kJ/mol
Which of the following has a DH°f = 0 kJ/mol? A) CO2(g) B) O3(g) C) Cl-(aq) D) NH3(aq) E) I2(s)
E) I2(s)
Which of the following has a DH°f = 0 kJ/mol? A) NO(g) B) CS2(l) C) Fe2+(aq) D) H2O(l) E) N2(g)
E) N2(g)
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C. Zn(s) + 2H+(aq) = Zn2+(aq) + H2(g)
E) w = -19.3 kJ We first find the number of moles of hydrogen gas formed in the reaction 50gzn 1/65.38Zn x 1/1= 0.765 moles The next step is to find the volume occupied by the hydrogen gas under the given conditions. This is the change in volume. v=nrt/p 0.765 x 0.0821 x 303.15k/1atm= 19.03 w=pv -(1atm)(19.03 =-19.03
A gas is allowed to expand, at constant temperature, from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w, and DE?
A
Calculate the standard enthalpy change for the reaction 2A + 2A2 + 4AB + B ® 5A2B Given: 2A + B ® A2B DH° = - 25.0 kJ/mol 2A2B ® 2AB + A2 DH° = 35.0 kJ/mol
A) - 95.0 kJ/mol
During volcanic eruptions, hydrogen sulfide gas is given off and oxidized by air according to the following chemical equation: 2H2S(g) + 3O2(g) ® 2SO2(g) + 2H2O(g) Calculate the standard enthalpy change for the above reaction given: 3S(s) + 2H2O(g) ® 2H2S(g) + SO2(g) DH° = 146.9 kJ/mol S(s) + O2(g) ® SO2(g) DH° = -296.4 kJ/mol
A) -1036.1 kJ/mol
The heat of solution of LiCl is -37.1 kJ/mol, and the lattice energy of LiCl(s) is 828 kJ/mol. Calculate the total heat of hydration of 1.00 mol of gas phase Li+ ions and Cl- ions
A) -865 kJ
Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: heat of combustion of C2H4(g) = -1411 kJ/mol; DH°f[CO2(g)] = -393.5 kJ/mol; DH°f[H2O(l)] = -285.8 kJ/mol
A) 52 kJ/mol
An exothermic reaction causes the surroundings to
A) increase in temperature
Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graph) + O2 ® CO2(g) DH° = -393.5 kJ/mol H2(g) + (1/2)O2 ® H2O(l) DH° = -285.8 kJ/mol CH3OH(l) + (3/2)O2(g) ® CO2(g) + 2H2O(l) DH° = -726.4 kJ/mol
B) -238.7 kJ/mol
Ethanol undergoes combustion in oxygen to produce carbon dioxide gas and liquid water. The standard heat of combustion of ethanol, C2H5OH(l), is -1366.8 kJ/mol. Given that DH°f[CO2(g)] = -393.5 kJ/mol and DH°f[H2O(l)] = -285.8 kJ/mol, what is the standard enthalpy of formation of ethanol?
B) -277.6 kJ/mol
The total heat of hydration of 1.00 mol of gas phase Li+ ions and Cl - ions is -865 kJ. The lattice energy of LiCl(s) is 828 kJ/mol. Calculate the heat of solution of LiCl
B) -37 kJ/mol
Calculate the standard enthalpy change for the reaction 2C8H18(l) + 17O2(g) ® 16CO(g) + 18H2O(l). Given: 2C8H18(l) + 25O2(g) ®16CO2(g) + 18H2O(l) DH° = -11,020 kJ/mol 2CO(g) + O2(g) ® 2CO2(g) DH° = -566.0 kJ/mol
B) -6,492 kJ/mol
Calculate the standard enthalpy change for the reaction 2C8H18(l) + 21O2(g) ® 8CO(g) + 8CO2(g) + 18H2O(l). Given: 2C8H18(l) + 25O2(g) ® 16CO2(g) + 18H2O(l) DH° = -11,020 kJ/mol 2CO(g) + O2(g) ® 2CO2(g) DH° = -566.0 kJ/mol
B) -8,756 kJ/mol
Ethanol (C2H5OH) burns according to the equation C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l), DH°rxn = -1367 kJ/mol. How much heat is released when 35.0 g of ethanol is burned?
B) 1,040 kJ 35 x 1/46mol x 1367/1
The enthalpy change when a strong acid is neutralized by strong base is -56.1 kJ/mol. If 12.0 mL of 6.00 M HBr at 21.30°C is mixed with 300. mL of 0.250 M NaOH, also at 21.30°C, what will the maximum temperature reached by the resulting solution? (Assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 J/g·°C, and that the density of the final solution is that of water.)
B) 24.40°C
Calcium oxide and water react in an exothermic reaction: CaO(s) + H2O(l) = Ca(OH)2(s) DH°rxn = -64.8 kJ/mol How much heat would be liberated when 7.15 g CaO(s) is dropped into a beaker containing 152g H2O?
B) 8.26 kJ
Concerning the reaction C(graphite) + O2(g) =CO2(g) DH° = -393 kJ/mol how many grams of C(graphite) must be burned to release 275 kJ of heat?
B) 8.40 g From equation, one can infer that 1 mol carbon on combustion releases 393 KJ of energy >12 g carbon on combustion releases 393 KJ >to get 275 KJ of energy, one would need 12 x 275 / 393 = 8.39694656 g
Thermal energy is A) the energy stored within the structural units of chemical substances. B) the energy associated with the random motion of atoms and molecules. C) solar energy, i.e. energy that comes from the sun. D) energy available by virtue of an object's position
B) the energy associated with the random motion of atoms and molecules.
The heat of solution of ammonium chloride is 15.2 kJ/mol. If a 6.134 g sample of NH4Cl is added to 65.0 mL of water in a calorimeter at 24.5°C, what is the minimum temperature reached by the solution? (The specific heat of water = 4.18 J/g·°C; the heat capacity of the calorimeter = 365. J/°C.)
C) 21.9°C
The bond enthalpy of the Br-Cl bond is equal to DH° for the reaction BrCl(g) ® Br(g) + Cl(g). Use the following data to find the bond enthalpy of the Br-Cl bond. Br2(l) ® Br2(g) DH° = 30.91 kJ/mol Br2(g) ® 2Br(g) DH° = 192.9 kJ/mol Cl2(g) ® 2Cl(g) DH° = 243.4 kJ/mol Br2(l) + Cl2(g) ® 2BrCl(g) DH° = 29.2 kJ/mol
C) 219.0 kJ/mol
Radiant energy is A) the energy stored within the structural units of chemical substances. B) the energy associated with the random motion of atoms and molecules. C) solar energy, i.e. energy that comes from the sun. D) energy available by virtue of an object's position
C) solar energy, i.e. energy that comes from the sun.
Given the specific heat for aluminum is 0.900 J/g·°C, how much heat is released when a 3.8 g sample of Al cools from 450.0°C to 25°C.
D) 1.5 kJ MST (3.8g x 0.900) (450-25)=1.5
Methanol (CH3OH) burns according to the equation 2CH3OH(l) + 3O2(g) =2CO2(g) + 4H2O(l), DH°rxn = -1454 kJ/mol. How much heat, in kilojoules, is given off when 75.0 g of methanol is burned?
D) 1.70 × 103 kJ 75x 1/32 x 1454/2= 1.70 x 103 (see prob 6.26 in book)
The combustion of pentane produces heat according to the equation C5H12(l) + 8O2(g) =5CO2(g) + 6H2O(l) DH°rxn= -3,510 kJ/mol How many grams of CO2 are produced per 2.50 × 103 kJ of heat released?
D) 157 g 44co2 x 5/1 x 2500 /3510 =157
The heat of solution of ammonium nitrate is 26.2 kJ/mol. If a 5.368 g sample of NH4NO3 is added to 40.0 mL of water in a calorimeter at 23.5°C, what is the minimum temperature reached by the solution? (The specific heat of water = 4.18 J/g·°C; the heat capacity of the calorimeter = 650. J/°C.)
D) 21.4°C
Section: 6.5 Aluminum metal has a specific heat of 0.900 J/g·°C. Calculate the amount of heat required to raise the temperature of 10.5 moles of Al from 30.5 °C to 225°C.
D) 49.6 kJ MST You have 10.5 moles of Al. If you multiply by the atomic mass of Al, you get the grams of Al. The amount of energy needed is the grams of Al * heat capacity * temperature change. 10.5 moles x 27al=283.5g (283.5 x 0.900)(30.5-225)= 4926 4926 x 1/1000= 49.6
The reaction that represents the standard enthalpy of formation for benzene (C6H6) is: A) 6 C(diamond) + 3 H2(g) = C6H6(l) B) 6 C(graphite) + 6 H(g) = C6H6(l) C) C6H6(l) + 15/2 O2(g) = 6 CO2(g) + 3 H2O(g) D) 6 C(graphite) + 3 H2(g) = C6H6(l) E) C6H6(l) =6 C(graphite) + 3 H2(g)
D) 6 C(graphite) + 3 H2(g) ® C6H6(l)
Which of the following processes is endothermic, given the following: S(s) + O2(g) = SO2(g) DH = -299 kJ/mol S(s) + 3/2 O2(g) = SO3(g) DH = -395 kJ/mol S(s) + O2(g) ® SO2(g) DH = -299 kJ/mol S(s) + 3/2 O2(g) ® SO3(g) DH = -395 kJ/mol A) 2 S(s) + 2 O2(g) = 2 SO2(g) B) ½ S(s) + ½ O2(g) =½ SO2(g) C) 2 S(s) + 5/2 O2(g) = SO2(g) + SO3(g) D) SO3(g) = S(s) + 3/2 O2(g) E) 2 S(s) + 3 O2(g) = 2 SO3(g)
D) SO3(g) = S(s) + 3/2 O2(g)
The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) = 8CO2(g) + 10H2O(l) DH°rxn= -5,314 kJ/mol How many grams of butane must be burned to release 1.00 × 104 kJ of heat
E) 219 g 58mol butane x 2/10 x 10000/-5314
An average home in Colorado requires 20. GJ of heat per month. How many grams of natural gas (methane) must be burned to supply this energy? CH4(g) + 2O2= CO2(g) + 2H2O(l) DH°rxn= -890.4 kJ/mol
E) 3.6 × 105 g 20 giga joules = 20 e6 kilojoules 20e6 kJ @ 16 grams CH4 / 890.4kJ = 3.59 e5 grams of CH4 Your answer: B.3.6 x 105 g
The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) =8CO2(g) + 10H2O(l) DH°rxn= -5,314 kJ/mol How many grams of CO2 are produced per 1.00 × 104 kJ of heat released?
E) 662 g 44co2 x 8/10 x 10000 /-5314= 662
The specific heat of gold is 0.129 J/g·°C. What is the molar heat capacity of gold?
C) 25.4 J/mol·°C molar mass of Au = 197 0.129 J/gC = 0.129 J/(1/197 mol)C = 25.413 J/molC j.investi · 7 years
An endothermic reaction causes the surroundings to
D) decrease in temperature
When 0.560 g of Na(s) reacts with excess F2(g) to form NaF(s), 13.8 kJ of heat is evolved at standard-state conditions. What is the standard enthalpy of formation (DH°f) of NaF(s)?
A) -570 kJ/mol Enthalpy of formation is the change in enthalpy for the formation of one mole of the compound; therefore you just have to change the data you have proportionally to what it would be for one mole of the compound. Start with the balanced equation first, just to ensure the coefficient on the substance you're given is 1, remembering that the coefficient on your product must be 1. Na + (1/2)F2 --> NaF So no problem, since the coefficient of Na is 1. Convert the mass of Na you're given to moles: (0.560 g Na)(1 mol Na/22.99 g Na) = 0.0244 mol Na. So from the data, 13.8 kJ heat is evolved by the reaction of 0.0244 mol Na. To find the heat evolved by 1 mol Na, just divide the amount of heat by the number of moles: 13.8 kJ/0.0244 mol which gives 566.5 kJ/mol. Finally, pay attention to whether heat is evolved or required to determine the sign of ΔH. Heat is evolved here, meaning the reaction is exothermic, so ΔH = -566 kJ/mol, which is closest to answer E
When 18.5 g of HgO(s) is decomposed to form Hg(l) and O2(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is the standard enthalpy of formation (DH°f) of HgO(s)?
A) -90.7 kJ/mol mole= 18.5g/216.58HgO=0.0854 0.0854mole/7.75kj= 90.7
A piece of copper with a mass of 218 g has a heat capacity of 83.9 J/°C. What is the specific heat of copper?
A) 0.385 J/g·°C Specific heat capacity = heat / g C Specific heat = 83.9 J / (218 g x 1 C) = 0.385 J/gC
question 50. How much heat (kJ) is evolved when 4.50 g of Fe2O3 is reacted with excess carbon monoxide using the equation below? Fe2O3(s) + 3CO(g) = 2 Fe(s) + 3 CO2(g) DH°rxn = - 24.8 kJ/mol,
A) 0.699 kJ 4.50 g Fe2O3) / (159.6887 g Fe2O3/mol) x (-24.8 kJ/mol) = -0.699 kJ
When 0.7521 g of benzoic acid was burned in a calorimeter containing 1,000. g of water, a temperature rise of 3.60°C was observed. What is the heat capacity of the bomb calorimeter, excluding the water? The heat of combustion of benzoic acid is -26.42 kJ/g.
A) 1.34 kJ/°C Q= mSt or St or mS (look at the units of your S (specific heat capacity) to find out if you are multiplying it by mass or T as well!) Qrxn is the benzoic acid. In these kinds of questions, it is usually some sort of organic compound. The specific heat capacity of benzoic acid is in kJ/g. So we need to use Q=mS. Qrxn=0.7521(26.42)=19.870482 KJ (because S was in kJ) Now we deal with Qeverything else. Q calorimeter is always in the q everything else. Look at the world problem and try to determine if there is anything else. We are given mass of water, so it is probably incorporated in as well (Qwater). So Qrxn= Qcalorimeter + Qwater The specific heat capacity of water is 4.184 J/g-c. We need to multiply by mass and T! Qwater=1000(4.184)(3.6)=15062.4 J (because S is in J) Q reaction is in kJ, Q water in J. Fix this. Multiply Qrxn by 1000 so it =19870.42 J. 19870.482 = 15062.4 + Qcalorimeter Qcalorimeter = 4808.082 We want S of Q calorimeter! Well, what do you want your units of S to be? The answer is in KJ/C. So I guess we could use that. 4808.082=St (no need for mass, look at our units-KJ/C) T=3.6 So S=1335 J/C Divide this by 1000. S=1.34 KJ/C
Determine the heat given off to the surroundings when 9.0 g of aluminum reacts according to the equation 2Al + Fe2O3 = Al2O3 + 2Fe, DH°rxn= -849 kJ/mol
A) 1.4 × 102 kJ you obtain -849 kJ with 2Al = 2*27=54g of aluminium with 9g there will be -849 kJ*9/54= -849/6=1.4 *10^2 kJ
Styrene, C8H8, is one of the substances used in the production of synthetic rubber. When styrene burns in oxygen to form carbon dioxide and liquid water under standard-state conditions at 25°C, 42.62 kJ are released per gram of styrene. Find the standard enthalpy of formation of styrene at 25°C. (Given: DH°f[CO2(g)] = -393.5 kJ/mol, DH°f[H2O(l)] = -285.8 kJ/mol, DH°f[H2O(g)] = -241.8 kJ/mol)
A) 147.8 kJ/mol
The reaction that represents the standard enthalpy of formation for acetone (CH3COCH3), a common ingredient in nail polish remover is: A) 3 C(graphite) + 3 H2(g) + ½ O2(g) = CH3COCH3(l) B) 6 C(diamond) + 6 H2(g) + O2(g) = 2 CH3COCH3(l) C) 3 C(diamond) + 3 H2(g) + ½ O2(g) = CH3COCH3(l) D) CH3COCH3(l) =3 C(graphite) + 3 H2(g) + ½ O2(g) E) CH3COCH3(l) + 4 O2(g) = 3 CO2(g) + 3 H2O(g)
A) 3 C(graphite) + 3 H2(g) + ½ O2(g) = CH3COCH3(l)
Which of the following processes is exothermic? A) CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(l) B) CO2(g) + 2 H2O(l) = CH4(g) + 2 O2(g) C) CO2(s) =CO2(g) D) H2O(l) =H2O(g) E) 6 H2O(g) + 4 CO2(g) = 2 C2H6(g) + 7 O2(g)
A) CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(l)
6.3 According to the first law of thermodynamics A) Energy is neither lost nor gained in any energy transformations. B) Perpetual motion is possible. C) Energy is conserved in quality but not in quantity. D) Energy is being created as time passes. We have more energy in the universe now than when time began
A) Energy is neither lost nor gained in any energy transformations
Chemical energy is A) the energy stored within the structural units of chemical substances. B) the energy associated with the random motion of atoms and molecules. C) solar energy, i.e. energy that comes from the sun. D) energy available by virtue of an object's position
A) the energy stored within the structural units of chemical substances.
A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. The total heat capacity of the calorimeter plus water was 5,760 J/°C. If the temperature rise of the calorimeter with water was 0.570°C, calculate the enthalpy of combustion of magnesium.
B) -602 kJ/mol Heat produced = Heat capacity of calorimeter x delta T = -(5,760 J / C)(0.570 C) = -3280 J = -3.28 kJ (-3.28 kJ / 0.1326 g Mg) x (24.3 g Mg / 1 mole Mg) = -601 kJ / mole Mg
The heat of solution of KCl is 17.2 kJ/mol and the lattice energy of KCl(s) is 701.2 kJ/mol. Calculate the total heat of hydration of 1.00 mol of gas phase K+ ions and Cl- ions.
B) -684 kJ
Calculate the amount of heat necessary to raise the temperature of 135.0 g of water from 50.4°F to 85.0°F. The specific heat of water = 4.184 J/g·°C. convert to celsius
B) 10.9 kJ (135 x 4.184) (10.22-31.11) x 1/1000
Calculate the heat required when 2.50 mol of A reacts with excess B and A2B according to the reaction: 2A + B + A2B ® 2AB + A2 Given: 2A + B ® A2B DH° = - 25.0 kJ/mol 2A2B ® 2AB + A2 DH° = 35.0 kJ/mol
B) 12.5 kJ
Naphthalene combustion can be used to calibrate the heat capacity of a bomb calorimeter. The heat of combustion of naphthalene is -40.1 kJ/g. When 0.8210 g of naphthalene was burned in a calorimeter containing 1,000. g of water, a temperature rise of 4.21°C was observed. What is the heat capacity of the bomb calorimeter excluding the water? specific heat of h2o=4.184
B) 3.64 kJ/°C Burning 0.821g naphthalene yields 0.821*40.1 = 32.92 kJ This heated water+calorimeter through 4.21°C The specific heat of water is 4.184 kJ/kg•°C. To heat 1,000g H2O through 4.21°C requires 4.21*4.184 =17.61 kJ Heating the calorimeter therefore required 32.92-17.61 = 15.31 kJ Heat capacity of the calorimeter is 15.31/4.21 = 3.635 kJ/°C
A 22.0 g block of copper at 45°C absorbs 2.50 kJ of heat. Given the specific heat of Cu is 0.385 J/g·°C what will be the final temperature of the Cu?
B) 340.°C
10.1 g CaO is dropped into a styrofoam coffee cup containing 157 g H2O at 18.0°C. If the following reaction occurs, then what temperature will the water reach, assuming that the cup is a perfect insulator and that the cup absorbs only a negligible amount of heat? (the specific heat of water = 4.18 J/g·°C)
B) 35.8°C normally, I would recommend first calculating moles of each to find the limiting reagent.... moles CaO = mass / molecular weight = 10.1g / 56 g/mole = 0.180 moles. moles H2O =157/18 = 8.72 so CaO is limiting reagent.. next calculate how much heat is liberated by dissolve the CaO like this.... .180 moles CaO x (64.8 kJ/mole) = 11.7 kJ that is the same amount of heat added to the water to heat it up. so find dT like this... since Δ H = m x Cp x dT dT = Δ H / (m x Cp) = 11.7 kJ x (1000 J/kJ) x (1 / 157 g) x (1g°C / 4.18 J) = 17.8°C since your water stated at 18.0 degrees and rose 17.8 degrees, the final temp would be 35.8°C b is the answer
A 135 g sample of H2O at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling process. What is the final temperature of the water? The specific heat of water is 4.184 J/g·°C.
B) 58°C
Given the thermochemical equation 2SO2(g) + O2(g) ® 2SO3(g), DH°rxn= -198 kJ/mol, how much heat is evolved when 600. g of SO2 is burned?
B) 928 kJ See pg 244 600so2 x 1/64.07gso2 x -198/2 =928
Three separate 3.5g blocks of Al, Cu, and Fe at 25 °C each absorb 0.505 kJ of heat. Which block reaches the highest temperature? The specific heats of Al, Cu, and Fe are 0.900 J/g·°C, 0.385J/g·°C, and 0.444 J/g·°C, respectively.
B) Cu
Suppose a 50.0 g block of silver (specific heat = 0.2350 J/g·°C) at 100°C is placed in contact with a 50.0 g block of iron (specific heat = 0.4494 J/g·°C) at 0°C, and the two blocks are insulated from the rest of the universe. The final temperature of the two blocks
B) will be lower than 50°C.
Calculate the standard enthalpy change for the reaction 4A + 2B ® 2AB + A2 Given: 2A + B ® A2B DH° = - 25.0 kJ/mol 2A2B ® 2AB + A2 DH° = 35.0 kJ/mol
C) - 15.0 kJ/mol
Pentaborane B5H9(s) burns vigorously in O2 to give B2O3(s) and H2O(l). Calculate DHrxn for the combustion of 5.00 mol of B5H9. DH°f[B2O3(s)] = -1,273.5 kJ/mol DH°f[B5H9(s)] = 73.2 kJ/mol DH°f[H2O(l)] = -285.8 kJ/mol
C) - 22,700 kJ
40. Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) ® 8CO2(g) + 10H2O(l) + 2N2(g) DH°rxn = -3857 kJ/mol Given that DH°f[CO2(g)] = -393.5 kJ/mol and DH°f[H2O(l)] = -285.8 kJ/mol, calculate the enthalpy of formation of glycine.
C) -537.2 kJ/mol
The reaction that represents the standard enthalpy of formation for sucrose (C12H22O11) is: A) C12H22O11(s) + 12 O2 = 12 CO2(g) + 11 H2O(g) B) 12 C(diamond) + 11 H2(g) + 11/2 O2(g) = C12H22O11(s) C) 12 C(graphite) + 11 H2(g) + 11/2 O2(g) = C12H22O11(s) D) 24 C(diamond) + 22 H2(g) + 11 O2(g) = 2 C12H22O11(s) E) C12H22O11(s) = 12 C(graphite) + 11 H2(g) + 11/2 O2(g)
C) 12 C(graphite) + 11 H2(g) + 11/2 O2(g) = C12H22O11(s)
Acetylene (C2H2) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. Given DH°f[CO2(g)] = -393.5 kJ/mol, DH°f[H2O(g)] = -241.8 kJ/mol, and DH°f [C2H2(g)] = 226.6 kJ/mol, how much energy is released (kJ) when 10.5 moles of acetylene is burned?
C) 13,200 kJ
Solid sodium peroxide (Na2O2) reacts with liquid water yielding aqueous sodium hydroxide and oxygen gas. How much heat is released when 250.0 L of oxygen gas is produced from the reaction of sodium peroxide and water if the reaction is carried out in an open container at 1.000 atm pressure and 25°C? (Given: DH°f[Na2O2(s)] = -510.9 kJ/mol; DH°f[NaOH(aq)] = -469.2 kJ/mol; DH°f[H2O(l)] = -285.8 kJ/mol)
C) 2900 kJ
At 25°C, the following heats of reaction are known: 2ClF(g) + O2(g) ® Cl2O(g) + F2O(g) DH°rxn = 167.4 kJ/mol 2ClF3(g) + 2O2(g) ® Cl2O(g) + 3F2O(g) DH°rxn = 341.4 kJ/mol 2F2(g) + O2(g) ® 2F2O(g) DH°rxn = -43.4 kJ/mol At the same temperature, use the above data to calculate the heat released (kJ) when 3.40 moles of ClF(g) reacts with excess F2: ClF(g) + F2(g) ® ClF3(g)
C) 370. kJ
Butane (C4H10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. Given DH°f [C4H10(g)] = -124.7 kJ/mol, DH°f[CO2(g)] = -393.5 kJ/mol, DH°f[H2O(g)] = -241.8 kJ/mol, how much energy is released (kJ) when 8.30 g of butane is burned?
C) 379 kJ
Given that CaO(s) + H2O(l) = Ca(OH)2(s), DH°rxn = -64.8 kJ/mol, how many grams of CaO must react in order to liberate 525 kJ of heat?
C) 455 g 1 /64.8 x 525 kj= 8.10 x = 8.10 moles 8.10 mol x 56.08 gCaO/mol = 454 g
Find the heat absorbed from the surroundings when 15 g of O2 reacts according to the equation O + O2 = O3, DH°rxn= -103 kJ/mol.
C) 48 kJ The molar mass of O2 is 32g so 15/32 so u have .47mols of O2. Enthrapy change is 103 per mole so 103x.47=48kJ
For which of these reactions will the difference between DH° and DE° be the smallest
C) H2(g) + Cl2(g) = 2HCl(g)
Section: 6.6 Which of the following processes is exothermic, given the following: N2(g) + 2 O2(g) =N2O4(l) DH° = 9.67 kJ/mol N2(g) + 2 O2(g) = 2 NO2(g) DH° = 67.70 kJ/mol A) 2 N2(g) + 4 O2(g) = 2 N2O4(l) B) ½ N2(g) + O2(g) = ½ N2O4(l) C) N2O4(l) = N2(g) + 2 O2(g) D) 2 N2(g) + 4 O2(g) = 2 NO2(g) + N2O4(l) E) 2 N2(g) + 4 O2(g) = 4 NO2(g)
C) N2O4(l) = N2(g) + 2 O2(g)
