WorksheetsChapter 19: Thermodynamics and Spontaneity Concepts
Total questions: 61
Worksheet time: 1hrs 19mins
Match the terms in the following equation to what they represent.
S = k lnW
S
Entropy
k
Bolrzmann's Constanct (R/Avogadro's num)
ln
Natural Log
W
# of energy equivalent ways to arrange c
Which of the following is always true for a spontaneous reaction at constant temperature and pressure?
ΔH >0 & ΔS<0
ΔG >0
ΔG = 0
ΔS<0
ΔG<0
The equation ΔG = ΔH - TΔS helps predict the (a) of a chemical reaction.
In an endothermic reaction (ΔH > 0), the reaction can still be spontaneous if ΔS is (a) and the temperature is (b) .
Sort the following changes based on their entropy (ΔS) change.
ice melting
dissolving salt in water
boiling water
water freezing
compressing a gas
Match the thermodynamic condition with its description.
ΔG < 0
Spontaneous
ΔG > 0
Nonspontaneous
ΔG = 0
Equilibrium
Explain why a reaction with ΔH > 0 and ΔS > 0 becomes spontaneous only at high temperatures.
Entropy generally increases when (a)
Match the phase transition with the direction of entropy change.
Solid → Liquid
Liquid → Gas
Liquid → Solid
Gas → Liquid
An (a) process increases the entropy of the surroundings.
An (b) process decreases the entropy of the surroundings.
Which of the following is the correct formula for calculating entropy change (ΔS) at constant temperature?
ΔS = T/q
ΔS = q × T
ΔS = q/T
ΔS = T/q²
Reversible processes are idealized conditions where the system and surroundings are always at equilibrium.
True
False
Depends on the conditions
Explain why ΔS is positive when ice melts but negative when water condenses.
ΔSuniverse = ΔSsystem (a) ΔSsurroundings
The magnitude of ΔSsurroundings increases as the temperature of the surroundings increases.
True
False
Depends on the conditions
Classify each process as "Spontaneous at Low Temp" or "Nonspontaneous at High Temp":
Water freezing
Water vapor condensing
Water freezing in hot surroundings
Ice melting at room temp
Put the steps in order to calculate ΔSsurroundings from ΔHreaction (system).
Convert temperature to Kelvin
Plug in ΔH (in J)
Use the formula ΔSsurroundings = –ΔH / T
Solve for ΔSsurroundings
Explain why an exothermic reaction may still be nonspontaneous at high temperatures.
Match the term to its correct description.
ΔSsystem
within the reacting system
ΔSsurroundings
caused by heat exchanged with the surrou
ΔSuniverse
Total Δentropy that determines spontan
T−ΔH
used to calculate entropy in surrounding
TΔH
wrong equation
Match each ΔH and ΔS combination to the correct spontaneity condition:
If a reaction is exothermic (ΔH < 0) and the entropy decreases (ΔS < 0), the reaction is spontaneous only at ______ temperatures.
low
high
all
no
Place each condition into the category of "Spontaneous" or "Nonspontaneous" at 25°C.
ΔH = –100 kJ, ΔS = +150 J/K
ΔH = +50 kJ, ΔS = –200 J/K
ΔH = –80 kJ, ΔS = –300 J/K
ΔH = +75 kJ, ΔS = +250 J/K
Put the steps to calculate the temperature at which a reaction becomes spontaneous in order.
Set ΔG = 0
Use the formula 0 = ΔH – TΔS
Rearrange to solve for T: T = ΔH / ΔS
Plug in ΔH and ΔS values
The decomposition of CCl₄ has ΔH = +95.7 kJ and ΔS = +142.2 J/K. Will this reaction be spontaneous at 25°C?
Yes, because both ΔH and ΔS are positive
No, because ΔG is positive at 25°C
Yes, because temperature has no effect
No, because ΔS is too small
If Q < K for a reaction mixture, which of the following is true?
The system is at equilibrium
The reaction will proceed in the reverse direction
The reaction is spontaneous in the forward direction
ΔG = 0
What happens when Q > K?
The reaction is spontaneous in the forward direction
The reaction proceeds in the reverse direction
The system is at equilibrium
ΔG < 0
When Q (a) K, the system has a negative ΔG and the reaction proceeds spontaneously forward.
Match each Q/K relationship to the correct statement.
Q < K
Reaction proceeds forward (ΔG < 0)
Q > K
Reaction proceeds in reverse (ΔG > 0)
Q = K
System is at equilibrium (ΔG = 0)
You have a reaction where [products] = 0.1 M and [reactants] = 1.0 M. The value of K = 0.5. What can you conclude?
Q > K, reverse reaction is favored
Q < K, forward reaction is favored
Q = K, system is at equilibrium
Cannot determine without temperature
Which of the following expressions represents the reaction quotient (Q) for the reaction below?
The reaction quotient, Q, is calculated using the concentrations of (a) at any point in time.
Solids and pure liquids are included when calculating Q and K.
True
False
For the reaction, which is the correct Q expression?
Match each variable to its description.
Q
Ratio of concentrations at any time
K
Ratio of concentrations at equilibrium
[ ]
Molar concentration of a species
(s), (l)
Not included in Q or K
(g), (aq)
Do include in Q or K
Put these steps in order to determine whether a system is at equilibrium using Q and K.
Write the expression for Q.
Plug in current concentrations into the Q expression.
Compare Q to K.
Decide direction of the reaction (forward, reverse, or at equilibrium).
When is the value of K calculated?
Before a reaction begins
After a reaction is complete
When the system is at equilibrium
When ΔG = 0
Explain the difference between Q and K and how they help determine the direction a reaction will proceed.
Which of the following is the correct formula for calculating standard entropy change of a reaction (ΔS°rxn)?
ΔS°rxn = ΣS°(reactants) – ΣS°(products)
ΔS°rxn = ΣS°(products) – ΣS°(reactants)
ΔS°rxn = ΣΔH° / T
ΔS°rxn = q / T
At 25°C and 1 atm, the standard state for a gas is its (a) form at that pressure.
The standard molar entropy of a perfect crystal at 0 K is zero.
True
False
Arrange these in order of increasing standard molar entropy (S°):
SO₃(g)
Cl₂(g)
Kr(g)
Which of the following factors increases the entropy of a substance?
Lower molecular complexity
Decreasing temperature
Converting gas to solid
Dissolving a solid in water
Given the reaction:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
Using the S° values below, what is the sign of ΔS°rxn?
NH₃(g) = 192.8 J/mol·K
O₂(g) = 205.2 J/mol·K
NO(g) = 210.8 J/mol·K
H₂O(g) = 188.8 J/mol·K
Positive
Negative
Zero
Cannot be determined
Which of the following factors affect the standard molar entropy (S°) of a substance?
Molecular complexity
Phase of the substance (solid, liquid, gas)
Molar mass
Allotropes (atomic arrangement in the same state)
Number of protons in the nucleus
Which conditions describe a system in its standard state?
A gas at 1 atm
A liquid at 50°C
A solution at 1 M concentration
A pure solid at 25°C and 1 atm
A mixture of gases at 2 atm
Which of the following reactions would likely result in a positive ΔS°rxn?
2H₂O₂(l) → 2H₂O(l) + O₂(g)
CaCO₃(s) → CaO(s) + CO₂(g)
H₂(g) + Cl₂(g) → 2HCl(g)
2SO₂(g) + O₂(g) → 2SO₃(g)
NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq)
According to the third law of thermodynamics:
The entropy of a perfect crystal at 0 K is zero
Entropy can be measured on an absolute scale
The entropy of a substance increases with temperature
Entropy has no absolute reference point
All solids have zero entropy
Match each law of thermodynamics with its correct description:
The entropy of the universe tends to increase in any spontaneous process
Second Law
The entropy of a perfect crystal at absolute zero is exactly zero
Third Law
Energy cannot be created or destroyed, only transferred or transformed
First Law
What is ΔG at 298 K for a reaction where ΔH = –100.0 kJ and ΔS = –200.0 J/K?
Answer in kJ.
Calculate ΔG°rxn for the following reaction using standard free energies of formation:
Given:
ΔG°f (C₂H₄) = 68.1 kJ/mol
ΔG°f (H₂) = 0 kJ/mol
ΔG°f (C₂H₆) = –32.9 kJ/mol
Put these steps in the correct order to calculate ΔG°rxn from standard free energies of formation:
List the ΔGf° values of all reactants and products
Multiply each ΔGf° by its stoichiometric coefficient
Add up the products and reactants separately
Subtract: ΔG°rxn = ΣG°products – ΣG°reactants
A reaction has ΔH = +150 kJ and ΔS = +250 J/K. At what temperature (in K) does the reaction become spontaneous? ΔG becomes negative when (a) .
Match the method of calculating ΔG with the situation when it’s used.
You know ΔH and ΔS at a specific temperature
ΔG = ΔH – TΔS
You have standard free energy of formation values
ΔG = ΣG°products – ΣG°reactants
You’re calculating ΔG from equilibrium constant K
ΔG = –RT ln(K)
What is the correct equation that relates standard Gibbs free energy (ΔG°) to the equilibrium constant (K)?
ΔG° = ΔH – TΔS
ΔG° = –RT ln K
ΔG° = q/T
ΔG° = RT ln K
If K > 1, then ΔG° is (a) , and the reaction is (b) under standard conditions.
Match the relationship between K and ΔG° with the correct spontaneity condition.
ΔG° is negative, reaction is spontaneous
K > 1
ΔG° = 0, system is at equilibrium
K = 1
ΔG° is positive, reverse reaction is spontaneous
K < 1
At equilibrium, Q = K and ΔG = (a) .
Which of the following conditions would result in a spontaneous reaction under standard conditions?
ΔG° < 0
K > 1
Q > K
ΔG° > 0
ln K > 0
When ΔG°rxn is positive, K will be less than 1.
True
False
Put the steps in the correct order to calculate K from ΔG°rxn.
Use ΔG° = –RT ln K
Plug in the value of ΔG° in joules and T in kelvin
Rearrange the equation to solve for ln K
Use the inverse log (e^x) to solve for K
Which equation describes how K changes with temperature using two temperatures?
ΔG = –RT ln Q
ln(K₂/K₁) = –ΔH°/R (1/T₂ – 1/T₁)
ln K = ΔH°/RT
ΔG = q/T
