Predict the sign of ΔS of the system for both of the following.
I. 2 C(graphite) + O2(g) → 2 CO(g)
II. C4H10(g) → C4H10(l)
ΔS should be negative for I and negative for II.
ΔS should be negative for I and positive for II.
ΔS should be positive for I and negative for II.
ΔS should be positive for I and positive for II.
Sodium reacts violently with water according to the equation: 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
The resulting solution has a higher temperature than the water prior to the addition of sodium. What are the signs of and for this reaction?
ΔH° is negative and ΔS° is negative.
ΔH° is negative and ΔS° is positive.
ΔH° is positive and ΔS° is negative.
ΔH° is positive and ΔS° is positive.
The brown color associated with photochemical smog is due to NO2(g), which is involved in an equilibrium with N2O4(g) in the atmosphere.
2 NO2(g) ?N2O4(g)
Predict the signs of the enthalpy and entropy change for the forward reaction.
The enthalpy change is negative and the entropy change is negative.
The enthalpy change is negative and the entropy change is positive.
The enthalpy change is positive and the entropy change is negative.
The enthalpy change is positive and the entropy change is positive.
What is W in Boltzmann's formula, S = k ln W?
a fraction indicating the probability of obtaining a result
a random number
the number of ways of obtaining the state
the worktimesAvogadro's number
Which electron on an atom of copper would have the highest value of W in the Boltzmann formula?
3s
3d
4s
4p
An electron in an oxygen p orbital on which of the following would have the highest entropy?
CH3CH2OH
CH3CH2O–
CH3CO2OH
CH3CO2–
What is k in Boltzmann's formula, S = k ln W?
the degeneracy of the state
the equilibrium constant for the process
theuniversal gas constant divided by Avogadro's number
the universalgas constanttimes Avogadro's number
The entropy change associated with the expansion of one mole of an ideal gas from an initial volume of Vi to a final volume of Vf at constant temperature is given by the equation, ΔS = R ln (Vf/Vi). What is the entropy change associated with the expansion of three moles of an ideal gas from an initial volume of Vi to a final volume of Vf at constant temperature?
ΔS = R ln (Vf/Vi)
ΔS = 3 mol × R ln (Vf/Vi)
ΔS = R ln (Vf × 23/Vi)
ΔS = R ln (Vf × 3!/Vi)
What is the entropy change associated with the expansion of one mole of an ideal gas from an initial volume of V to a final volume of V of 2.50V at constant temperature?
ΔS = 2.50 R ln (Vf/Vi)
ΔS = -2.50 R ln (Vf/Vi)
ΔS = R ln 2.50
ΔS = -R ln 2.50
Predict the sign of ΔS for each of the following processes, which occur at constant temperature.
The volume of 2.0 moles of O2(g) increases from 44 L to 52 L.
The pressure of 2.0 moles of O2(g) increases from 1.0 atm to 1.2 atm.
I: ΔS= negative; II: ΔS= negative
I: ΔS= negative; II: ΔS= positive
I: ΔS= positive; II: ΔS= negative
I: ΔS= positive; II: ΔS= positive
Assume a heteronuclear diatomic molecule, AB, forms a one-dimensional crystal by lining up along the x-axis. Also assume that each molecule can only have one of six possible orientations, corresponding to atom A facing in either the positive or negative direction along the x-, y-, or z-axis. If the molecules are arranged randomly in the six directions, the molar entropy at absolute zero should be
R ln 6.
R ln 66.
R ln 6!
0.
The Boltzmann formula is S = k ln W. A perfect crystal has a molar entropy of 0 at absolute zero because
W = 0.
W = 1.
W = NA.
k = 1.
What is the sign of ?S for each of the following processes?
The separation of gaseous molecules of UF6, into 238UF6 and 235UF6 at constant
temperature and pressure.
The dissolving of I2(s) in CCl4(l).
ΔS is negative for I and negative for II.
ΔS is negative for I and positive for II.
ΔS is positive for I and negative for II.
ΔS is positive for I and positive for II.
Which has the lowest entropy?
CH3OH(s, –25°C)
CH3OH(s, –15°C)
CH3OH(l, 15°C)
CH3OH(l, 25°C)
Which has the highest entropy in each set?
I. H2O(s), H2O(l), H2O(g) at 0.1°C, 4.58 atm
II. H2O(l) at 0°C, H2O(l) at 25°C, H2O(l) at 100°C (all at 1.0 atm pressure)
H2O(l) in set I and H2O(l) at 0°C in set II
H2O(s) in set I and H2O(l) at 100°C in set II
H2O(g) in set I and H2O(l) at 0°C in set II
H2O(g) in set I and H2O(l) at 100°C in set II
Which provides the greatest increase in entropy? A) H2O (s, 0°C) → H2O (l, 0°C)
B) H2O (l, 0°C) → H2O (l, 25°C)
C) H2O (g, 0.1°C) → H2O (s, 0.1°C)
D) H2O (l, 100°C) → H2O (g, 100°C)
According to the third law of thermodynamics,
energy is conserved in any transformation of matter.
the entropy increases for any spontaneous process.
the entropy of a perfectly ordered, crystalline substance is zero at 0 Kelvin.
the entropy of the universe increases for any spontaneous process.
Which of the following statements must be true for the entropy of a pure solid to be zero?
The temperature must be 0 K.
The solid must be crystalline, not amorphous.
The solid must be perfectly ordered.
The solid must be an element.
I
I and II
I, II, and III
I, II, III, and IV
Under which of the following conditions would one mole of Ne have the highest entropy, S?
27°C and 25 L
137°C and 25 L
27°C and 35 L
137°C and 35 L
Which has the highest standard molar entropy at 25°C?
Al(s)
Al(l)
Al(g)
All three should have a standard molar entropy of zero.
Which has the highest standard molar entropy at 25°C?
F2(g)
Cl2(g)
Br2(g)
I2(g)
Which of the following gas molecules has the greatest standard molar entropy at 25°C?
C2H2
CH2CH2
CH3CH3
All have the same entropy.
Which substance has the highest standard molar entropy at 25°C ?
C(graphite)
C2H4(g)
CH3OH(l)
MgCO3(s)
Which one of the following has the lowest standard molar entropy, S°, at 25°C?"
C8H18(s)
C8H18(l)
C) C12H26(s)
D) C12H26(l)
Calculate ΔS° for the following reaction. N2(g) + 2 O2(g) → 2 NO2(g)
A) -156.5 J/K
B) -121.5 J/K
C) 15.5 J/K
D) 636.5 J/K
ΔS° = – 198.7 J/K for the reaction shown below. Calculate S° for NH3(g). N2(g) + 3 H2(g) → 2 NH3(g)
61.7 J/K?mol
123.4 J/K?mol
192.3 J/K?mol
384.6 J/K?mol
Calculate ΔS° for the formation of one mole of solid sodium bromide from the elements at 25°C.
A) -116.7 J/K
B) -81.2 J/K
C) -40.5 J/K
D) 86.8 J/K
The standard molar entropy for Br2(g) is 245.46 J/(mol ? K) at 25°C. Given that ΔS° =
104.58 J/K for the dissociation of one mole of Br2(g) into Br(g) at 25°C, find the standard molar entropy for Br(g) at 25°C.
A) 70.44 J/(mol ? K)
B) 140.08 J/(mol ? K)
C) 175.02 J/(mol ? K)
D) 350.04 J/(mol ? K)
Which of the three laws of thermodynamics provides a criterion for spontaneity?
the first law of thermodynamics
the second law of thermodynamics
the third law of thermodynamics
both the second and third laws of thermodynamics
Which of the following is a criterion for spontaneity that holds for any process?
ΔG < 0
ΔG > 0
ΔStotal < 0
ΔStotal > 0
According to the second law of thermodynamics, all reactions proceed spontaneously in the direction that increases the entropy of the
surroundings.
system.
system – surroundings
system + surroundings
For a process to be at equilibrium, it is necessary that
ΔSsys = ΔSsurr.
ΔSsys = - ΔSsurr.
ΔSsys = 0.
ΔSsys = 0 and ΔSsurr = 0.
For a spontaneous process
energy and entropy are conserved.
energy is conserved and the entropy of the system and surroundings increases.
the energy of the system and the surroundings decreases and the entropy of the system and surroundings increases.
both the energy and the entropy of the system and surroundings decrease.
For the process
CaCO3(calcite) → CaCO3(aragonite) ΔH° = -0.21 kJ, ΔS° = -4.2 J/K Assuming that the surroundings can be considered a large heat reservoir at 25°C, calculate ΔSsurr and ΔStotal for the process at 25°C and 1 atm pressure. Is the process spontaneous at 25°C and 1 atm pressure?
ΔSsurr = 4.2 J/K, Δtotal = 0, not spontaneous
ΔSsurr = 0.7 J/K, ΔStotal = -3.5 J/K, not spontaneous
ΔSsurr = -0.7 J/K, ΔStotal = -4.9 J/K, not spontaneous
During perspiration,
the entropy of the water evaporated decreases and the entropy of the body decreases.
the entropy of the water evaporated decreases and the entropy of the body increases.
the entropy of the water evaporated increases and the entropy of the body decreases.
the entropy of the water evaporated increases and the entropy of the body increases.
A hot penny is dropped into cold water inside a polystyrene foam cup. Assuming negligible heat loss to the atmosphere and the cup,
the decrease in entropy of the penny is equal to the increase in entropy of the water.
| ΔSpenny | = | ΔSwater |
the decrease in entropy of the penny is less than the increase in entropy of the water.
| ΔSpenny | < | ΔSwater |
the decrease in entropy of the penny is more than the increase in entropy of the water.
| ΔSpenny | > | ΔSwater |
the entropy of both the penny and the water increases.
At constant pressure and temperature, which statement is true?
All reactions for which ?H < 0 are spontaneous.
All reactions for which ?S < 0 are spontaneous.
All reactions for which ?G < 0 are spontaneous.
All reactions for which K < 1 are spontaneous.
Why is the sign of ΔG rather than the sign of ΔStotal generally used to determine the spontaneity of a chemical reaction?
ΔG can be used for processes that occur under any conditions.
ΔG involves thermodynamic functions of the system only.
Free energy is easier to understand than entropy.
Entropy is based on probability and is therefore less reliable.
Other than only PV work, what reaction conditions must be satisfied for the sign of ΔG to be used as a criterion for spontaneity?
constant volume and pressure
constant temperature and pressure
constant temperature and volume
constant volume only
For the reaction 3 C2H2(g) → C6H6(l) at 25°C, the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K. Calculate the standard free energy change at 25°C.
948 kJ
-503 kJ
-618 kJ
D) -1061 kJ
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