Physical Chemistry
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 A → B, if the rate of formation of B is 0.5 mol/L/s, what is the rate of disappearance of A?
A.
0.5 mol/L/s
B.
1.0 mol/L/s
C.
0.25 mol/L/s
D.
2.0 mol/L/s
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Solution
The rate of disappearance of A is equal to the rate of formation of B, multiplied by the stoichiometric coefficients. Here, it is 1.0 mol/L/s.
Correct Answer: B — 1.0 mol/L/s
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Q. For a reaction A → B, if the rate of reaction doubles when the concentration of A is doubled, what is the order of the reaction with respect to A?
A.
Zero order
B.
First order
C.
Second order
D.
Third order
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Solution
If doubling the concentration of A doubles the rate, the reaction is first order with respect to A.
Correct Answer: B — First order
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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 equilibrium, if the concentration of products increases, what will happen to the equilibrium position?
A.
Shift to the left
B.
Shift to the right
C.
No change
D.
Depends on temperature
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Solution
According to Le Chatelier's principle, if the concentration of products increases, the equilibrium will shift to the left.
Correct Answer: A — Shift to the left
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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
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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 = (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 solution to obey Raoult's Law, the interactions between solute and solvent must be:
A.
Stronger than those in the pure components.
B.
Weaker than those in the pure components.
C.
Similar to those in the pure components.
D.
Non-existent.
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Solution
For a solution to obey Raoult's Law, the interactions between solute and solvent must be similar to those in the pure components.
Correct Answer: C — Similar to those in the pure components.
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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) 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 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:
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. For a zero-order reaction, how does the rate change with concentration?
A.
Increases linearly
B.
Decreases linearly
C.
Remains constant
D.
Increases exponentially
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Solution
In a zero-order reaction, the rate is constant and does not depend on the concentration of reactants.
Correct Answer: C — Remains constant
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Q. For a zero-order reaction, how does the rate change with respect to concentration?
A.
Increases linearly
B.
Decreases linearly
C.
Remains constant
D.
Increases exponentially
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Solution
In a zero-order reaction, the rate is constant and does not depend on the concentration of the reactants.
Correct Answer: C — Remains constant
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Q. For an electron in a 3d orbital, what are the possible values of l?
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Solution
For d orbitals, the azimuthal quantum number l = 2.
Correct Answer: C — 2
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Q. For an electron in a 3d orbital, what are the possible values of m_l?
A.
-2, -1, 0, 1, 2
B.
-1, 0, 1
C.
0, 1
D.
0, 1, 2
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Solution
For l=2 (d orbital), m_l can take values from -l to +l, which are -2, -1, 0, 1, 2.
Correct Answer: A — -2, -1, 0, 1, 2
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Q. For an electron in a 3p orbital, what are the possible values of m_l?
A.
-1, 0, +1
B.
0, +1, +2
C.
-2, -1, 0
D.
0, -1, -2
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Solution
For a p orbital, l=1, so m_l can take values -1, 0, +1.
Correct Answer: A — -1, 0, +1
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Q. For an electron in a 3p orbital, what are the possible values of the magnetic quantum number (m_l)?
A.
-1, 0, +1
B.
0, +1, +2
C.
-2, -1, 0
D.
1, 2, 3
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Solution
For a p orbital (l=1), m_l can take values -1, 0, +1.
Correct Answer: A — -1, 0, +1
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Q. For an electron in a 5d orbital, what are the possible values of m_l?
A.
-2, -1, 0, 1, 2
B.
-3, -2, -1, 0, 1, 2, 3
C.
0, 1, 2
D.
-1, 0, 1
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Solution
For l=2 (d orbital), m_l can take values from -2 to +2, which are -2, -1, 0, 1, 2.
Correct Answer: A — -2, -1, 0, 1, 2
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Q. For an ideal solution, if the mole fraction of the solvent is 0.75, what is the vapor pressure of the solution if the vapor pressure of the pure solvent is 100 mmHg?
A.
75 mmHg
B.
100 mmHg
C.
25 mmHg
D.
50 mmHg
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Solution
According to Raoult's Law, the vapor pressure of the solution is 0.75 * 100 mmHg = 75 mmHg.
Correct Answer: A — 75 mmHg
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