General Problems
7.71. A constant-pressure calorimeter similar to the one shown in Figure 7.7 can be used to study the reaction of magnesium with hydronium ion:
Mg(s) + H3O+(aq) ¨ Mg2+(aq) + H2(g)
100.0 mL of 0.5 M hydrochloric acid, HCl(aq), was put in the calorimeter; its temperature was 19.32 ¡C. Then, 0.1372 g of Mg metal was added to the acid and the temperature observed until it reached a maximum of 25.69 ¡C.
(a) What is DHreaction, in kJ per mole of Mg reacted?
(b) Calculate DEreaction, in kJ per mole of Mg reacted.
(c) How could you measure DEreaction directly? Are data like those in this problem precise enough to measure any difference between DEreaction and DHreaction? Explain why you answer as you do.
Answer to 7.71:
(a) The balanced stoichiometric equation is:
Mg(s) + 2H3O+(aq) ¨ Mg2+(aq) + H2(g) + 2H2O(l)
Assume that the magnesium reacts completely, so that (0.1372 g)/(24.31 gmol- 1) = 5.644 x 10- 3 mol of Mg reacts. The original acidic solution contained 5 x 10- 2 mol H3O+ (= (0.5 M)(0.1 L)), which is about five times as much as needed to react with all the Mg. Assume that the specific heat of the solution is 4.184 JK- 1g- 1, that 100.0 mL is 100.0 g, and that the heat capacity of the calorimeter is zero. The energy change in the calorimeter is:
qP = (100.0 g)(25.69 - 19.32 K)(4.184 JK- 1g- 1) = 2.67 kJ
DHreaction = (2.67 kJ)/(5.644 x 10- 3 mol) = 472 kJmol- 1
(b) Since a gas is formed, we need to correct for the pressure-volume work done. One mole of gas is formed for every mole of Mg reacted, so Dn = 1 and we have:
DE = DH - DnRT = 472 kJmol- 1 - (1 mol)(8.314 JK- 1mol- 1)(298 K)
DE = 472 kJmol- 1 - 2.47 kJ (per mol Mg) È 470 kJmol- 1
(c) You could measure DE directly by carrying out the reaction in a sealed container (like a bomb calorimeter) with some arrangement to drop the Mg in the acid when the container had been sealed. In theory, the precision of the data above is good enough to measure the half percent difference between DE and DH. However, in order actually to get data that are accurate enough to observe this difference, you would have to take account of the actual mass of the calorimetric liquid, its true specific heat, and (probably most important) the heat capacity of the calorimeter.
7.72.
Several elegant experiments have been done
to
determine the amount of work that molecular motors from living cells can do. The motor in the illustration[jab1]
consists of seven proteins, three a, three b, and
a g,
that come together as shown. The motor is too small (about 10 x 10 x 8
nm) to be observed, so a long actin filament is attached to the motor shaft,
the g
protein, and movement of the filament is observed under a microscope. The
rotation of the shaft is in 120¡ steps. Each step appears to
require the coupled hydrolysis of one ATP molecule, reaction (7.37).
Calculations based on the observed motion indicate that the motor is able to
produce a force of at least 100 x 10- 12 N.
In one complete revolution, the top of the motor shaft moves about 1 nm.
(a) How much work does the motor do in one complete revolution? (1 J = 1 Nm).
(b) How much energy (enthalpy) is provided by ATP during one complete revolution?
(c) What is the approximate efficiency of this motor? Explain how you arrive at your conclusion.
Answer to 7.72:
(a) Recall that work is equal to force times the distance through which the force acts. In one revolution the shaft moves about 1 nm = 1 x 10- 9 m so:
work in a revolution = (100 x 10- 12 N)(1 x 10- 9 m) = 1 x 10- 19 Nm (or J)
(b) Our estimate in this chapter was that the hydrolysis of ATP produces about 21 kJmol- 1. Hydrolysis of one molecule of ATP would produce about 3.5 x 10- 20 J. Since hydrolysis of three molecules of ATP is required for each revolution of the motor, the energy provided by the ATP hydrolysis is about 3(3.5 x 10- 20 J) = 1 x 10- 19 J.
(c) The enthalpy supplied by the ATP is essentially the same as the work done by the motor. The motor is 100% efficient in converting its energy source into work. Even if our estimates are off somewhat, this molecular motor appears to be an extremely efficient device.
7.73. Calculate the DH¡reaction for the hydrolysis of a molecule of glycylglycine to two molecules of glycine using the DH¡combustion data for glycylglycine and glycine. Assume that the nitrogen atoms in both molecules are oxidized to NO2. Explain your procedure.
DH¡combustion for glycine, H2NCH2C(O)OH = - 981 kJmol-1
DH¡combustion for
glycylglycine,H2NCH2C(O)NHCH2C(O)OH=- 1,996 kJmol-1
Answer to 7.73:
The balanced equation for the combustion of glycylglycine:
C4H8O3N2 + 13/2 O2
ˆ 2NO2 +
4CO2 + 4H2O DH reaction=
-1966 kJ
The balanced equation for the combustion of glycine needs to be doubled and used in the reverse direction:
2NO2 +
4CO2 + 5H2O ˆ 2C2H5O2N +
13/2 O2
DH reaction= (2mol)(981 kJmol-1) = 1962 kJmol-1
Summing these two equations gives the overall reaction which is
the balanced equation for the hydrolysis of glycylglycine. Summing the corresponding heats of
combustion gives the DH hydrolysis
for 1 mole of glycylglycine:
C4H8O3N2 + 4H2O ˆ 2C2H5O2N DH hydrolysis= -4 kJmol-1
7.74.
The diagram shows a glass demonstration
version of a fire syringe[jab2]
(also called a fire piston). A tiny amount of a flammable material, cloth or
paper, is placed at the closed end of the syringe. The plunger is inserted and
then thrust quickly into the piston, compressing the air inside to about 1/20th of its initial volume.
The temperature of the air rises almost instantly to about 1000 K and ignites
the flammable material, which burns with a bright flash.
(a) The internal energy of a gas depends on its temperature. When the temperature of the gas changes, its change in internal energy depends on its constant-volume molar heat capacity (CV), the number of moles of gas, and the temperature change: DE = n x CV x DT. CV for air is 21 kJmol- 1. What is DE for the gas compressed in a fire syringe that is 15 cm long and has a cross-sectional area of 0.20 cm2? One mole of gas at 300 K occupies about 25 L.
(b) The compression in a fire syringe is so rapid that the gas has no time to transfer thermal energy to its surroundings during the compression. How much work is done on the gas during the compression? What pressure (in atm) on the plunger is required to obtain this much work? Explain clearly how you obtain your answer. (1 Latm = 101.3 J)
(c) To get a better ÒfeelÓ for the effort required to operate a fire syringe, calculate the force in pounds required to produce the pressure you calculated in part (b). Use the fact that a pressure of 1 atm = 15 poundin- 2. Show clearly how you do the necessary unit conversions. Is the force required a reasonable amount for a human to produce?
(d) Make a molecular level drawing showing
the motion of the molecules of air in the syringe before, during, and
immediately after the compression. Clearly show how they obtain the increase in
energy you calculated in part (a).
(e) Why does the amount of flammable material in the syringe have to be small? If the temperature of the gas reaches 1000 K, couldnÕt it set fire to a larger quantity? Hint: Consider the heat capacity (or specific heat) of a piece of solid cloth or paper.
Answer to 7.74:
(a) To get DE, we need the number of moles of gas in the syringe and the temperature change upon compression. The temperature change is from room temperature, about 300 K, to about 1000 K: DT = 700 K. The initial volume of gas in the cylinder is:
volume of gas = (15 cm)(0.20 cm2) = 3.0 cm3 (mL)
This volume of gas is 1.2 x 10- 4 mol ( = (3.0 mL)/(25000 mLmol- 1), so the internal energy change is:
DE = n x CV x DT = (1.2 x 10- 4 mol)(21 Jmol- 1K- 1)(700 K) = 1.8 J
(b) If no thermal energy is exchanged with the surroundings during the compression, then q is zero for the compression and the work done on the gas is equal to its change in internal energy: w = DE = 1.8 J.This work is the result of a decrease in volume of the gas to 1/20th of its initial volume (part of the description at the beginning of the problem). The change in volume, DV, is a decrease of 19/20th of the initial volume = - 19/20(3.0 mL) = - 2.85 mL = - 2.85 x 10- 3 L. The work on the gas is:
w = 1.8 J = 1.8 x 10- 2 Latm = - PDV = - P (- 2.85 x 10- 3 L)
P = (1.8 x 10- 2 Latm)/(2.85 x 10- 3 L) = 6.2 atm
(c)You have to multiply pressure in atmospheres by (15 poundin- 2/1 atm) to get pressure in poundin- 2. The result is about 93 poundin- 2. A square inch, in2, is (2.54 cm)2 = 6.5 cm2. To convert the area of the piston to square inches, we have to multiply the area in cm2 by (1 in2/6.5 cm2). The result is 3.0 x 10- 2 in2. Therefore, it requires about 3 pounds of force (= (93 poundin- 2)(3.0 x 10- 2 in2)) to produce a pressure of 93 poundin- 2 on the tiny piston. Three pounds of force is easy for most teenagers or adults to exert, so the fire piston should be easy to use.
(d) Before the compression, the
molecules are moving randomly and relatively slowly (short arrows). During the
compression, the molecules have a large component of motion in the direction of
the compression superimposed on their random motion. When the compression
stops, the large component of motion gets randomized and the molecules are
again moving randomly but at much greater speed (longer arrows) corresponding
to a higher temperature.
(e) The maximum temperature is not the relevant variable here. The amount of energy available in the compressed gas is what you have to consider. The calculation in part (a) shows that the internal energy change of the gas is a little less than 2 J. Handbooks show that the specific heat of paper or cotton (both made of cellulose, a glucose polymer) is in the ballpark of 1 Jg- 1K- 1. The kindling temperature of paper is 451 ¡F (which is the source of the title, ÒFahrenheit 451,Ó of a Ray Bradbury story about book burning) or about 233 ¡C. Thus, you have to raise the temperature of paper about 200 degrees from room temperature to get it to burn. One gram of paper would require about 200 J of energy to increase its temperature this much. Since you only have about 2 J of energy in the compressed air of the fire piston, you have enough energy to raise the temperature of about 10 mg of paper or cotton to its kindling point. That is why you use a wisp of cotton in the demonstration. To start a larger fire, you use the glowing tinder to ignite some further small pieces of material and finally build up to a real fire. The match was a great invention and quickly made fire pistons obsolete.
7.75. Given these atomization reactions and their corresponding enthalpy changes, calculate DH¡ for the combustion of 1 mol of PH3(g) to yield H2O(g) and P2O5(g). Show your reasoning clearly.
PH3(g) ¨ P(g) + 3H(g) DH¡ = 965 kJmol-1
O2(g) ¨ 2O(g) DH¡ = 490 kJmol-1
H2O(g) ¨ 2H(g) + O(g) DH¡ = 930 kJmol-1
P2O5(g) ¨ 2P(g)
+ 5O(g)
DH¡ = 3382 kJmol-1
Answer
to 7.75:
Begin this problem by figuring out the balanced equation:
2PH3(g) + 4O2(g) --> P2O5(g) +
3H2O(g)
The atomization reactions need to be combined in such a way that their sum is equal to the balanced equation. As needed, some reactions will need to be used more than once. For example, the atomization of PH3(g) is used twice so 2(PH3(g) ˆ P(g) + 3H(g)) becomes 2PH3(g) ˆ 2P(g) + 6H(g) and DH for the reaction is also doubled as given below. Other reactions are needed in the reverse direction, which changes the sign of DH from (+) to (-).
Reactions
from given data:
2PH3(g) --> 2P(g) +
6H(g) DH =
2mol(965 kJmol-1) = 1,930 kJ
4O2(g) --> 8O(g) DH =
4mol(490 kJmol-1) = 1,960 kJ
2P(g) + 5O(g) --> P2O5(g) DH =
-3,382 kJ
6H(g) + 3O(g) ˆ 3H2O(g) DH =
3mol(-930 kJmol-1) = -2,790
kJ
Summing
these four equations does, in fact, generate the balanced equation:
overall equation: 2PH3(g) +4O2(g) ˆ P2O5(g) + 3H2O(g)
Summing the corresponding DH values gives an overall DH which equals
-2,282 kJ.
Note: This DH value is for 2 moles of PH3. Therefore, for the combustion of 1 mole of PH3, DH = -1,141 kJ.
