Thermodynamics
Q. At constant temperature and pressure, if ΔH is positive and ΔS is negative, what is the sign of ΔG?
A.
Always negative
B.
Always positive
C.
Depends on temperature
D.
Zero
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Solution
If ΔH is positive and ΔS is negative, ΔG will always be positive.
Correct Answer: B — Always positive
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Q. At constant temperature and pressure, if ΔH is positive and ΔS is negative, what can be said about ΔG?
A.
ΔG is positive
B.
ΔG is negative
C.
ΔG is zero
D.
ΔG can be either positive or negative
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Solution
If ΔH is positive and ΔS is negative, ΔG will be positive.
Correct Answer: A — ΔG is positive
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Q. At what temperature does a reaction become spontaneous if ΔH = 50 kJ and ΔS = 0.1 kJ/K?
A.
500 K
B.
250 K
C.
1000 K
D.
200 K
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Solution
Set ΔG = 0: 0 = ΔH - TΔS; T = ΔH/ΔS = 50 kJ / 0.1 kJ/K = 500 K.
Correct Answer: A — 500 K
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Q. At what temperature does the Gibbs Free Energy change from negative to positive?
A.
At absolute zero
B.
At the melting point
C.
At the boiling point
D.
At the transition temperature
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Solution
The Gibbs Free Energy changes from negative to positive at the transition temperature, where the system shifts from one phase to another.
Correct Answer: D — At the transition temperature
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Q. For a process with ΔH = 200 kJ and ΔS = 0.5 kJ/K, what is ΔG at 400 K?
A.
200 kJ
B.
180 kJ
C.
220 kJ
D.
160 kJ
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Solution
ΔG = ΔH - TΔS = 200 kJ - 400 K * 0.5 kJ/K = 200 kJ - 200 kJ = 0 kJ.
Correct Answer: B — 180 kJ
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Q. For a reaction at constant temperature and pressure, which of the following is true?
A.
ΔG = ΔH + TΔS
B.
ΔG = ΔH - TΔS
C.
ΔG = TΔS - ΔH
D.
ΔG = ΔS - ΔH
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Solution
The correct relationship at constant temperature and pressure is ΔG = ΔH - TΔS.
Correct Answer: B — ΔG = ΔH - TΔS
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Q. For a reaction at standard conditions, if ΔG° is negative, what can be said about the equilibrium constant (K)?
A.
K < 1
B.
K = 1
C.
K > 1
D.
K is undefined
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Solution
If ΔG° is negative, the equilibrium constant K is greater than 1.
Correct Answer: C — K > 1
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Q. For a reaction at standard conditions, if ΔG° is positive, what can be said about the reaction?
A.
The reaction is spontaneous in the forward direction.
B.
The reaction is spontaneous in the reverse direction.
C.
The reaction is at equilibrium.
D.
The reaction is impossible.
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Solution
A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction, thus spontaneous in the reverse.
Correct Answer: B — The reaction is spontaneous in the reverse direction.
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Q. For a reaction at standard conditions, if ΔG° is positive, what does it imply?
A.
The reaction is spontaneous in the forward direction.
B.
The reaction is at equilibrium.
C.
The reaction is non-spontaneous in the forward direction.
D.
The reaction will proceed rapidly.
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Solution
A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction under standard conditions.
Correct Answer: C — The reaction is non-spontaneous in the forward direction.
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Q. For a reaction at standard conditions, if ΔG° is positive, what does it indicate?
A.
The reaction is spontaneous in the forward direction.
B.
The reaction is non-spontaneous in the forward direction.
C.
The reaction is at equilibrium.
D.
The reaction is spontaneous in the reverse direction.
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Solution
A positive ΔG° indicates that the reaction is non-spontaneous in the forward direction.
Correct Answer: B — The reaction is non-spontaneous in the forward direction.
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Q. For a reaction with ΔH = 100 kJ and ΔS = 200 J/K, at what temperature will the reaction become spontaneous?
A.
500 K
B.
250 K
C.
200 K
D.
100 K
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Solution
To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS. Thus, T = ΔH/ΔS = (100,000 J)/(200 J/K) = 500 K.
Correct Answer: A — 500 K
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Q. For a reaction with ΔH = 100 kJ/mol and ΔS = 200 J/mol·K, at what temperature will the reaction become spontaneous?
A.
500 K
B.
250 K
C.
200 K
D.
100 K
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Solution
To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS, thus T = ΔH/ΔS = 100,000 J / 200 J/K = 500 K.
Correct Answer: A — 500 K
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Q. For a reaction with ΔH = 50 kJ/mol and ΔS = 100 J/mol·K, at what temperature will the reaction become spontaneous?
A.
500 K
B.
250 K
C.
1000 K
D.
200 K
Show solution
Solution
To find the temperature at which the reaction becomes spontaneous, set ΔG = 0: 0 = ΔH - TΔS, thus T = ΔH/ΔS = (50,000 J/mol) / (100 J/mol·K) = 500 K.
Correct Answer: A — 500 K
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Q. For a reversible process, the change in entropy is given by which of the following?
A.
ΔS = Q/T
B.
ΔS = W/T
C.
ΔS = Q + W
D.
ΔS = 0
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Solution
For a reversible process, the change in entropy is given by ΔS = Q/T, where Q is the heat exchanged and T is the temperature.
Correct Answer: A — ΔS = Q/T
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Q. For a reversible process, the change in entropy of the system is equal to the heat absorbed divided by the temperature. This is expressed as:
A.
ΔS = Q/T
B.
ΔS = T/Q
C.
ΔS = Q + T
D.
ΔS = Q - T
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Solution
For a reversible process, the change in entropy (ΔS) is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.
Correct Answer: A — ΔS = Q/T
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Q. For a reversible process, the change in entropy of the system is equal to the heat absorbed divided by the temperature. What is the formula?
A.
ΔS = Q/T
B.
ΔS = T/Q
C.
ΔS = Q*T
D.
ΔS = Q + T
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Solution
The change in entropy (ΔS) for a reversible process is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature.
Correct Answer: A — ΔS = Q/T
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Q. For a reversible process, the change in entropy of the universe is:
A.
Zero
B.
Positive
C.
Negative
D.
Undefined
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Solution
For a reversible process, the change in entropy of the universe is zero, as the system and surroundings are in equilibrium.
Correct Answer: A — Zero
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Q. For a reversible process, the efficiency of a Carnot engine is given by which formula?
A.
1 - (T2/T1)
B.
T1/T2
C.
T2/T1
D.
1 - (T1/T2)
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Solution
The efficiency of a Carnot engine is given by η = 1 - (T2/T1), where T1 is the temperature of the hot reservoir and T2 is the temperature of the cold reservoir.
Correct Answer: A — 1 - (T2/T1)
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Q. For a spontaneous process, the change in entropy of the universe must be:
A.
Zero
B.
Positive
C.
Negative
D.
Undefined
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Solution
For a spontaneous process, the total entropy change of the universe (system + surroundings) must be positive.
Correct Answer: B — Positive
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Q. For a spontaneous process, the change in Gibbs free energy (ΔG) is related to entropy (ΔS) by which of the following equations?
A.
ΔG = ΔH + TΔS
B.
ΔG = ΔH - TΔS
C.
ΔG = TΔS - ΔH
D.
ΔG = ΔS - ΔH
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Solution
The correct relationship is ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous process.
Correct Answer: B — ΔG = ΔH - TΔS
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Q. For a spontaneous process, the change in Gibbs free energy (ΔG) is related to entropy (ΔS) how?
A.
ΔG = ΔH - TΔS
B.
ΔG = TΔS - ΔH
C.
ΔG = ΔS - ΔH
D.
ΔG = ΔH + TΔS
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Solution
The relationship is given by ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous process.
Correct Answer: A — ΔG = ΔH - TΔS
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Q. For a spontaneous process, the change in Gibbs free energy (ΔG) is:
A.
Positive
B.
Negative
C.
Zero
D.
Undefined
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Solution
For a process to be spontaneous, the change in Gibbs free energy (ΔG) must be negative.
Correct Answer: B — Negative
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Q. If 100 J of heat is added to a system at a constant temperature of 300 K, what is the change in entropy?
A.
0.33 J/K
B.
0.25 J/K
C.
0.5 J/K
D.
0.75 J/K
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Solution
The change in entropy ΔS = Q/T = 100 J / 300 K = 0.33 J/K.
Correct Answer: A — 0.33 J/K
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Q. If a reaction has a ΔG of +5 kJ/mol, what can be inferred?
A.
The reaction is spontaneous
B.
The reaction is non-spontaneous
C.
The reaction is at equilibrium
D.
The reaction is exothermic
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Solution
A ΔG of +5 kJ/mol indicates that the reaction is non-spontaneous.
Correct Answer: B — The reaction is non-spontaneous
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Q. If a reaction has ΔH = 100 kJ and ΔS = -200 J/K, what is ΔG at 298 K?
A.
0 kJ
B.
100 kJ
C.
200 kJ
D.
300 kJ
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Solution
ΔG = ΔH - TΔS = 100 kJ - 298 K * (-0.2 kJ/K) = 100 kJ + 59.6 kJ = 159.6 kJ.
Correct Answer: D — 300 kJ
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Q. If the enthalpy of a system increases, the process is considered _____.
A.
exothermic
B.
endothermic
C.
isothermal
D.
adiabatic
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Solution
If the enthalpy of a system increases, the process is considered endothermic.
Correct Answer: B — endothermic
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Q. If the enthalpy of a system increases, what can be inferred about the system?
A.
It is losing heat
B.
It is gaining heat
C.
It is at equilibrium
D.
It is undergoing a phase change
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Solution
An increase in enthalpy indicates that the system is gaining heat.
Correct Answer: B — It is gaining heat
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Q. If the enthalpy of reaction is -100 kJ, what can be said about the reaction?
A.
It absorbs heat
B.
It releases heat
C.
It is at equilibrium
D.
It requires energy input
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Solution
A negative enthalpy change indicates that the reaction releases heat, making it exothermic.
Correct Answer: B — It releases heat
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Q. If the entropy of a system increases, what can be inferred about the spontaneity of the process?
A.
The process is non-spontaneous
B.
The process is spontaneous
C.
The process is at equilibrium
D.
None of the above
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Solution
An increase in entropy indicates that the process is spontaneous, as per the second law of thermodynamics.
Correct Answer: B — The process is spontaneous
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Q. If the entropy of a system increases, what is the effect on Gibbs Free Energy at constant temperature?
A.
ΔG increases
B.
ΔG decreases
C.
ΔG remains constant
D.
ΔG becomes zero
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Solution
If entropy increases, ΔG decreases, making the reaction more favorable.
Correct Answer: B — ΔG decreases
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