Which statement defines a spontaneous process in thermodynamics?
It proceeds without a continuous input of outside energy
✓
It happens quickly once it starts
It releases heat to the surroundings
It needs a catalyst before it can begin
Correct Answer
It proceeds without a continuous input of outside energy
A spontaneous process, once conditions allow it, continues on its own without energy being supplied continuously from outside the system. Happening quickly once it starts is about rate, which thermodynamics does not predict. Releasing heat to the surroundings is not required, since some spontaneous processes, such as ice melting above 0 °C, absorb heat. Needing a catalyst before it can begin is a kinetic matter that does not decide spontaneity.
Category: Spontaneity
Question 2
The conversion of diamond to graphite at room temperature has a negative ΔG°, yet diamonds remain unchanged for very long times. What explains this?
The conversion is nonspontaneous at room temperature
The conversion is spontaneous but extremely slow
✓
Diamond has a higher entropy than graphite
The two forms are at equilibrium at room temperature
Correct Answer
The conversion is spontaneous but extremely slow
A negative ΔG° means the change is thermodynamically favored, but spontaneity says nothing about rate, and this conversion has an enormous activation barrier. The conversion being nonspontaneous at room temperature contradicts its negative ΔG°. Diamond having a higher entropy than graphite is false, since the more rigid diamond lattice has the lower entropy. The two forms being at equilibrium would require ΔG to be zero.
Category: Spontaneity
Question 3
According to the Boltzmann equation S = k ln W, how does the entropy of a system change when its number of microstates doubles?
It doubles
It increases by 2k
It increases by k ln 2
✓
It increases by a factor of ln 2
Correct Answer
It increases by k ln 2
ΔS = k ln(Wf/Wi) = k ln(2W/W) = k ln 2, a small fixed increase. Saying it doubles confuses doubling W with doubling S; because S depends on the logarithm of W, S does not scale in proportion to W. Increasing by 2k would require ln(Wf/Wi) to equal 2. Increasing by a factor of ln 2 treats the change as a multiplication of S rather than an added amount.
Category: Entropy and Microstates
Question 4
For a single pure substance at the same temperature, how do the standard molar entropies of its phases compare?
Gas < liquid < solid
Liquid < solid < gas
Solid < gas < liquid
Solid < liquid < gas
✓
Correct Answer
Solid < liquid < gas
Particles in a solid are locked in place and have the fewest microstates, particles in a liquid can move past one another, and gas particles move freely through a large volume, so entropy rises from solid to liquid to gas. Gas < liquid < solid reverses the true order. Liquid < solid < gas wrongly puts the rigid solid above the liquid. Solid < gas < liquid wrongly ranks the liquid above the far more dispersed gas.
Category: Predicting Entropy Changes
Question 5
Which process has a negative entropy change for the system?
N2(g) + 3 H2(g) → 2 NH3(g)
✓
H2O(l) → H2O(g)
CaCO3(s) → CaO(s) + CO2(g)
Ar(g) at 1 atm → Ar(g) at 0.5 atm
Correct Answer
N2(g) + 3 H2(g) → 2 NH3(g)
Ammonia synthesis turns 4 moles of gas into 2 moles of gas, reducing the number of microstates, so ΔS is negative. H2O(l) → H2O(g) is vaporization, which greatly increases entropy. CaCO3(s) → CaO(s) + CO2(g) produces a gas from a solid, so entropy rises. Ar(g) at 1 atm → Ar(g) at 0.5 atm is an expansion into a larger volume, which increases entropy.
Category: Predicting Entropy Changes
Question 6
All four of these gases are at 298 K and 1 bar. Which one has the greatest standard molar entropy?
He(g)
C3H8(g)
✓
CH4(g)
C2H6(g)
Correct Answer
C3H8(g)
In the same phase and at the same temperature, larger and more complex molecules have more ways to vibrate and rotate, and so more microstates; propane is the largest and most complex here. He(g) is a single atom with only translational motion and has the lowest entropy of the group. CH4(g) is the smallest of the three hydrocarbons. C2H6(g) is more complex than methane but simpler than propane.
Category: Predicting Entropy Changes
Question 7
Ice melts reversibly at 273 K with ΔH(fusion) = 6.01 kJ/mol. What is the entropy change for melting 1 mol of ice, to three significant figures?
0.0220 J/(mol·K)
-22.0 J/(mol·K)
22.0 J/(mol·K)
✓
1.64 x 10^6 J/(mol·K)
Correct Answer
22.0 J/(mol·K)
For a reversible process at constant temperature, ΔS = qrev/T = 6010 J/mol / 273 K = 22.0 J/(mol·K), positive because the liquid has more microstates than the solid. The value 0.0220 J/(mol·K) keeps the heat in kilojoules but labels the result in joules. The value -22.0 J/(mol·K) has the wrong sign for a process that absorbs heat. The value 1.64 x 10^6 J/(mol·K) multiplies by the temperature instead of dividing.
Category: Entropy and Microstates
Question 8
Using S°(H2, g) = 130.7, S°(O2, g) = 205.2 and S°(H2O, l) = 70.0 J/(mol·K), what is ΔS° for 2 H2(g) + O2(g) → 2 H2O(l)?
-265.9 J/K
+326.6 J/K
-121.4 J/K
-326.6 J/K
✓
Correct Answer
-326.6 J/K
ΔS° = 2(70.0) - [2(130.7) + 205.2] = 140.0 - 466.6 = -326.6 J/K, negative because 3 moles of gas become a liquid. The value -265.9 J/K ignores the coefficients of 2. The value +326.6 J/K subtracts products from reactants instead of reactants from products. The value -121.4 J/K leaves out the entropy of O2.
Category: Third Law and Standard Entropy
Question 9
For a process taking place at equilibrium, what is the value of the entropy change of the universe?
Zero
✓
A positive value
A negative value
The same as ΔS of the system
Correct Answer
Zero
When a process is at equilibrium, it has no tendency to go in either direction, so ΔSuniv = 0. A positive value describes a spontaneous process. A negative value describes a nonspontaneous process, whose reverse is spontaneous. The same as ΔS of the system would hold only if the surroundings had no entropy change, which is not a condition for equilibrium.
Category: Second Law
Question 10
A reaction with ΔH = -92.2 kJ occurs at a constant 298 K and constant pressure. What is the entropy change of the surroundings, to three significant figures?
-309 J/K
+309 J/K
✓
+0.309 J/K
+2.75 x 10^7 J/K
Correct Answer
+309 J/K
The heat released by the system flows into the surroundings, so ΔSsurr = -ΔHsys/T = +92,200 J / 298 K = +309 J/K. The value -309 J/K has the wrong sign; an exothermic reaction increases the entropy of its surroundings. The value +0.309 J/K uses kilojoules but labels the answer in joules. The value +2.75 x 10^7 J/K multiplies by the temperature instead of dividing.
Category: Second Law
Question 11
For a process at constant temperature, ΔSsys = -120 J/K and ΔSsurr = +150 J/K. What is ΔSuniv, and what does it show?
-270 J/K; nonspontaneous
+270 J/K; spontaneous
+30 J/K; spontaneous
✓
+30 J/K; at equilibrium
Correct Answer
+30 J/K; spontaneous
ΔSuniv = ΔSsys + ΔSsurr = -120 + 150 = +30 J/K, and a positive entropy change of the universe means the process is spontaneous even though the system's entropy falls. The value -270 J/K subtracts the surroundings term and flips its sign. The value +270 J/K adds the magnitudes while ignoring the negative system term. Calling +30 J/K at equilibrium is wrong because equilibrium requires ΔSuniv to be exactly zero.
Category: Second Law
Question 12
According to the third law of thermodynamics, which of these has an entropy of 0 J/K?
Any element in its standard state at 298 K
A pure liquid at its freezing point
An ideal gas at 0 °C
A perfect crystal of a pure substance at 0 K
✓
Correct Answer
A perfect crystal of a pure substance at 0 K
A perfect, pure crystal at absolute zero has only one possible microstate, so S = k ln 1 = 0. Any element in its standard state at 298 K has a positive standard entropy, since it is far above absolute zero. A pure liquid at its freezing point still has many microstates. An ideal gas at 0 °C is at 273 K and has a large positive entropy.
Category: Third Law and Standard Entropy
Question 13
Reaction A has ΔG° = +45 kJ and reaction B has ΔG° = -70 kJ. When the two are added to give one coupled overall reaction, what is the overall ΔG°?
-25 kJ, spontaneous
✓
+115 kJ, nonspontaneous
-115 kJ, spontaneous
+25 kJ, nonspontaneous
Correct Answer
-25 kJ, spontaneous
Free energy is a state function, so the ΔG° values of added reactions simply add: +45 + (-70) = -25 kJ, and the negative total makes the coupled reaction spontaneous. This is how a favorable reaction can drive an unfavorable one. The value +115 kJ subtracts the second value instead of adding it. The value -115 kJ makes both values negative. The value +25 kJ has the sign reversed.
Category: Free Energy Calculations
Question 14
A process has ΔH = +40.0 kJ and ΔS = +100.0 J/K. What is ΔG at 300 K, to three significant figures?
-29,960 kJ
+10.0 kJ
✓
+70.0 kJ
-10.0 kJ
Correct Answer
+10.0 kJ
Converting ΔS to 0.100 kJ/K, ΔG = ΔH - TΔS = 40.0 kJ - (300 K)(0.100 kJ/K) = +10.0 kJ, so the process is nonspontaneous at 300 K. The value -29,960 kJ mixes kilojoules for ΔH with joules for ΔS. The value +70.0 kJ adds TΔS instead of subtracting it. The value -10.0 kJ subtracts ΔH from TΔS, reversing the sign.
Category: Free Energy Calculations
Question 15
A reaction has ΔH = +150 kJ and ΔS = +200 J/K, and these values change little with temperature. Under what condition is the reaction spontaneous?
Below 750 K
Above 0.75 K
Above 750 K
✓
At every temperature
Correct Answer
Above 750 K
With both ΔH and ΔS positive, ΔG = ΔH - TΔS becomes negative only once TΔS exceeds ΔH, which happens above T = 150,000 J / 200 J/K = 750 K. Below 750 K the positive ΔH dominates and ΔG is positive. Above 0.75 K comes from dividing 150 J by 200 J/K, forgetting to convert kilojoules to joules. At every temperature applies only to reactions with negative ΔH and positive ΔS.
Category: Temperature and Spontaneity
Question 16
A reaction has both ΔH < 0 and ΔS < 0. At which temperatures is it spontaneous?
Only at high temperatures
At all temperatures
At no temperature
Only at low temperatures
✓
Correct Answer
Only at low temperatures
With ΔH negative and ΔS negative, ΔG = ΔH - TΔS is negative when T is small enough that the unfavorable -TΔS term stays smaller than the favorable ΔH. Only at high temperatures fits reactions with positive ΔH and positive ΔS. At all temperatures fits negative ΔH with positive ΔS. At no temperature fits positive ΔH with negative ΔS.
Category: Temperature and Spontaneity
Question 17
Using ΔG°f values of -50.5 kJ/mol for CH4(g), -394.4 kJ/mol for CO2(g) and -237.1 kJ/mol for H2O(l), what is ΔG° for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)?
-818.1 kJ
✓
-581.0 kJ
-919.1 kJ
+818.1 kJ
Correct Answer
-818.1 kJ
ΔG° = [(-394.4) + 2(-237.1)] - [(-50.5) + 2(0)] = -868.6 + 50.5 = -818.1 kJ, with O2 contributing zero as an element in its standard state. The value -581.0 kJ ignores the coefficient of 2 on water. The value -919.1 kJ adds the methane value instead of subtracting it. The value +818.1 kJ subtracts products from reactants.
Category: Free Energy Calculations
Question 18
A reaction has a negative standard free energy change, ΔG° < 0. What does this indicate about its equilibrium constant K?
K is less than 1
K is greater than 1
✓
K is exactly 1
K is a negative number
Correct Answer
K is greater than 1
From ΔG° = -RT ln K, a negative ΔG° requires ln K to be positive, so K is greater than 1 and products are favored at equilibrium. K is less than 1 corresponds to a positive ΔG°. K is exactly 1 corresponds to ΔG° = 0. K is a negative number is impossible, since equilibrium constants are always positive.
Category: Free Energy and Equilibrium
Question 19
A reaction has K = 10 at 298 K. Using R = 8.314 J/(mol·K), what is ΔG° to three significant figures?
+5.70 kJ/mol
-2.48 kJ/mol
-5.70 kJ/mol
✓
-0.0191 kJ/mol
Correct Answer
-5.70 kJ/mol
ΔG° = -RT ln K = -(8.314 J/(mol·K))(298 K)(ln 10) = -(8.314)(298)(2.303) J/mol = -5.70 kJ/mol, negative because K is greater than 1. The value +5.70 kJ/mol drops the negative sign. The value -2.48 kJ/mol uses log 10 = 1 instead of the natural logarithm. The value -0.0191 kJ/mol leaves out the temperature.
Category: Free Energy and Equilibrium
Question 20
At a moment when a reaction mixture has a reaction quotient Q smaller than K, what is true according to ΔG = ΔG° + RT ln Q?
ΔG is positive and the reverse reaction proceeds
ΔG is zero and no net change occurs
ΔG equals ΔG° whatever the value of Q
ΔG is negative and the forward reaction proceeds
✓
Correct Answer
ΔG is negative and the forward reaction proceeds
Since ΔG° = -RT ln K, the equation can be written ΔG = RT ln(Q/K); when Q is less than K, the logarithm is negative, so ΔG is negative and the reaction moves forward toward equilibrium. ΔG being positive with the reverse reaction proceeding describes Q greater than K. ΔG being zero describes Q equal to K. ΔG equaling ΔG° whatever the value of Q is true only when Q happens to equal 1.
Category: Free Energy and Equilibrium