Q. What is the enthalpy change for the reaction 2Na + Cl2 → 2NaCl?
-
A.
-411 kJ
-
B.
-240 kJ
-
C.
0 kJ
-
D.
411 kJ
Solution
The enthalpy change for the formation of NaCl from its elements is -411 kJ.
Correct Answer: A — -411 kJ
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Q. What is the enthalpy change for the reaction at constant pressure?
-
A.
ΔH = ΔU + PΔV
-
B.
ΔH = ΔU - PΔV
-
C.
ΔH = ΔU + VΔP
-
D.
ΔH = ΔU - VΔP
Solution
At constant pressure, the enthalpy change is given by ΔH = ΔU + PΔV.
Correct Answer: A — ΔH = ΔU + PΔV
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Q. What is the enthalpy change for the reaction N2(g) + 3H2(g) → 2NH3(g) at standard conditions?
-
A.
-92.4 kJ
-
B.
-45.9 kJ
-
C.
0 kJ
-
D.
0.5 kJ
Solution
The standard enthalpy change for the formation of ammonia is -92.4 kJ.
Correct Answer: A — -92.4 kJ
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Q. What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(g)?
-
A.
It is positive.
-
B.
It is negative.
-
C.
It is zero.
-
D.
It is dependent on temperature.
Solution
The formation of water from hydrogen and oxygen is an exothermic reaction, thus the enthalpy change is negative.
Correct Answer: B — It is negative.
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Q. What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(l)?
-
A.
-571.6 kJ
-
B.
-285.8 kJ
-
C.
0 kJ
-
D.
285.8 kJ
Solution
The enthalpy change for the formation of 2 moles of water is -571.6 kJ.
Correct Answer: A — -571.6 kJ
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Q. What is the enthalpy change for the reaction: C(s) + O2(g) -> CO2(g)?
-
A.
-393.5 kJ/mol
-
B.
-241.8 kJ/mol
-
C.
0 kJ/mol
-
D.
285.8 kJ/mol
Solution
The enthalpy change for the formation of CO2 from carbon and oxygen is -393.5 kJ/mol.
Correct Answer: A — -393.5 kJ/mol
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Q. What is the enthalpy change for the reaction: C(s) + O2(g) → CO2(g)?
-
A.
-393.5 kJ/mol
-
B.
-241.8 kJ/mol
-
C.
0 kJ/mol
-
D.
285.8 kJ/mol
Solution
The enthalpy change for the formation of CO2 from its elements is -393.5 kJ/mol.
Correct Answer: A — -393.5 kJ/mol
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Q. What is the enthalpy change for the reaction: CaCO3(s) → CaO(s) + CO2(g)?
-
A.
It is an endothermic reaction.
-
B.
It is an exothermic reaction.
-
C.
It has no enthalpy change.
-
D.
It is spontaneous at all temperatures.
Solution
The decomposition of calcium carbonate is an endothermic reaction, requiring heat input.
Correct Answer: A — It is an endothermic reaction.
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Q. What is the enthalpy change for the reaction: H2(g) + 1/2 O2(g) → H2O(l)?
-
A.
-285.8 kJ/mol
-
B.
0 kJ/mol
-
C.
-241.8 kJ/mol
-
D.
-572 kJ/mol
Solution
The standard enthalpy change for the formation of water from its elements is -241.8 kJ/mol.
Correct Answer: C — -241.8 kJ/mol
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Q. What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g)?
-
A.
It is always positive.
-
B.
It is always negative.
-
C.
It can be either positive or negative depending on conditions.
-
D.
It is zero.
Solution
The formation of ammonia from nitrogen and hydrogen is an exothermic reaction, thus the enthalpy change is negative.
Correct Answer: B — It is always negative.
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Q. What is the enthalpy change when 1 mole of NaCl is dissolved in water?
-
A.
-3.87 kJ
-
B.
0 kJ
-
C.
+3.87 kJ
-
D.
-7.0 kJ
Solution
The enthalpy change when 1 mole of NaCl is dissolved in water is approximately -3.87 kJ.
Correct Answer: A — -3.87 kJ
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Q. What is the enthalpy change when 1 mole of water vapor condenses to liquid water?
-
A.
It is positive.
-
B.
It is negative.
-
C.
It is zero.
-
D.
It is dependent on pressure.
Solution
The condensation of water vapor to liquid water releases heat, making the enthalpy change negative.
Correct Answer: B — It is negative.
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Q. What is the entropy change for a reversible isothermal expansion of an ideal gas?
-
A.
nR ln(Vf/Vi)
-
B.
0
-
C.
nR(Tf - Ti)
-
D.
nC ln(Vf/Vi)
Solution
The entropy change for a reversible isothermal expansion of an ideal gas is ΔS = nR ln(Vf/Vi).
Correct Answer: A — nR ln(Vf/Vi)
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Q. What is the entropy change for a reversible isothermal process?
-
A.
Zero
-
B.
nR ln(Vf/Vi)
-
C.
nR(Tf - Ti)
-
D.
nR ln(Tf/Ti)
Solution
The entropy change for a reversible isothermal process is ΔS = nR ln(Vf/Vi).
Correct Answer: B — nR ln(Vf/Vi)
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Q. What is the entropy change for a reversible process?
-
A.
Always positive
-
B.
Always negative
-
C.
Can be zero
-
D.
Depends on the path taken
Solution
For a reversible process, the entropy change can be zero if the process is isothermal and reversible.
Correct Answer: C — Can be zero
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Q. What is the entropy change for a system that undergoes a phase transition at constant temperature?
-
A.
ΔS = 0
-
B.
ΔS = Q/T
-
C.
ΔS = T/Q
-
D.
ΔS = Q + T
Solution
During a phase transition at constant temperature, the change in entropy is given by ΔS = Q/T, where Q is the heat absorbed or released.
Correct Answer: B — ΔS = Q/T
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Q. What is the entropy change for an ideal gas during an isothermal expansion?
-
A.
Zero
-
B.
nR ln(Vf/Vi)
-
C.
nC_v ln(Tf/Ti)
-
D.
nC_p ln(Tf/Ti)
Solution
The entropy change for an ideal gas during an isothermal expansion is ΔS = nR ln(Vf/Vi).
Correct Answer: B — nR ln(Vf/Vi)
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Q. What is the entropy change for the isothermal expansion of an ideal gas from volume V1 to V2 at temperature T?
-
A.
R ln(V2/V1)
-
B.
R (V2 - V1)/T
-
C.
0
-
D.
R (V1/V2)
Solution
The entropy change for an isothermal expansion is given by ΔS = nR ln(V2/V1). For 1 mole, ΔS = R ln(V2/V1).
Correct Answer: A — R ln(V2/V1)
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Q. What is the entropy change for the mixing of two ideal gases at constant temperature?
-
A.
0
-
B.
R ln(2)
-
C.
R ln(V1/V2)
-
D.
R ln(V1*V2)
Solution
The entropy change for the mixing of two ideal gases at constant temperature is ΔS = nR ln(2) for equal moles of each gas.
Correct Answer: B — R ln(2)
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Q. What is the entropy change when 1 mole of an ideal gas is heated at constant volume?
-
A.
0
-
B.
R ln(T2/T1)
-
C.
R (T2 - T1)
-
D.
R (T1/T2)
Solution
The change in entropy when heating an ideal gas at constant volume is given by ΔS = nC_v ln(T2/T1). For 1 mole, it simplifies to ΔS = R ln(T2/T1).
Correct Answer: B — R ln(T2/T1)
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Q. What is the entropy change when 1 mole of an ideal gas is heated at constant volume from temperature T1 to T2?
-
A.
R ln(T2/T1)
-
B.
R (T2 - T1)
-
C.
0
-
D.
R (T1/T2)
Solution
The change in entropy at constant volume is given by ΔS = nC_v ln(T2/T1). For 1 mole, ΔS = R ln(T2/T1).
Correct Answer: A — R ln(T2/T1)
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Q. What is the entropy change when 1 mole of ice at 0°C is converted to water at 0°C?
-
A.
0 J/K
-
B.
R ln(2)
-
C.
R
-
D.
Positive value
Solution
The phase change from ice to water at 0°C involves an increase in disorder, thus resulting in a positive change in entropy.
Correct Answer: D — Positive value
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Q. What is the entropy change when 2 moles of an ideal gas are compressed isothermally from volume V2 to V1?
-
A.
-R ln(V1/V2)
-
B.
R ln(V1/V2)
-
C.
0
-
D.
R (V2 - V1)
Solution
The change in entropy for an isothermal compression is ΔS = nR ln(V1/V2). For 2 moles, ΔS = 2R ln(V1/V2), which is negative since V1 < V2.
Correct Answer: A — -R ln(V1/V2)
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Q. What is the equation for the displacement of a damped harmonic oscillator?
-
A.
x(t) = A e^(-bt) cos(ωt)
-
B.
x(t) = A e^(bt) cos(ωt)
-
C.
x(t) = A cos(ωt)
-
D.
x(t) = A e^(-bt) sin(ωt)
Solution
The displacement of a damped harmonic oscillator is given by x(t) = A e^(-bt) cos(ωt), where b is the damping coefficient.
Correct Answer: A — x(t) = A e^(-bt) cos(ωt)
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Q. What is the equation of a circle with center at (-1, 2) and radius 4?
-
A.
(x + 1)² + (y - 2)² = 16
-
B.
(x - 1)² + (y + 2)² = 16
-
C.
(x + 1)² + (y + 2)² = 16
-
D.
(x - 1)² + (y - 2)² = 16
Solution
Using the standard form, the equation is (x + 1)² + (y - 2)² = 4² = 16.
Correct Answer: A — (x + 1)² + (y - 2)² = 16
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Q. What is the equation of a circle with center at (2, -3) and radius 5?
-
A.
(x - 2)² + (y + 3)² = 25
-
B.
(x + 2)² + (y - 3)² = 25
-
C.
(x - 2)² + (y - 3)² = 25
-
D.
(x + 2)² + (y + 3)² = 25
Solution
The standard form of a circle's equation is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.
Correct Answer: A — (x - 2)² + (y + 3)² = 25
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Q. What is the equation of a circle with center at (h, k) and radius r?
-
A.
(x - h)^2 + (y - k)^2 = r^2
-
B.
(x + h)^2 + (y + k)^2 = r^2
-
C.
(x - h)^2 - (y - k)^2 = r^2
-
D.
(x + h)^2 - (y + k)^2 = r^2
Solution
The equation of a circle with center at (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2.
Correct Answer: A — (x - h)^2 + (y - k)^2 = r^2
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Q. What is the equation of an ellipse with foci at (±c, 0) and vertices at (±a, 0)?
-
A.
x^2/a^2 + y^2/b^2 = 1
-
B.
y^2/a^2 + x^2/b^2 = 1
-
C.
x^2/b^2 + y^2/a^2 = 1
-
D.
y^2/b^2 + x^2/a^2 = 1
Solution
The standard form of the equation of an ellipse with horizontal major axis is x^2/a^2 + y^2/b^2 = 1.
Correct Answer: A — x^2/a^2 + y^2/b^2 = 1
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Q. What is the equation of motion for a damped harmonic oscillator?
-
A.
m d²x/dt² + b dx/dt + kx = 0
-
B.
m d²x/dt² + kx = 0
-
C.
m d²x/dt² + b dx/dt = 0
-
D.
m d²x/dt² + b dx/dt + kx = F(t)
Solution
The equation of motion for a damped harmonic oscillator is m d²x/dt² + b dx/dt + kx = 0.
Correct Answer: A — m d²x/dt² + b dx/dt + kx = 0
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Q. What is the equation of motion for a simple harmonic oscillator with amplitude A and angular frequency ω?
-
A.
x(t) = A cos(ωt)
-
B.
x(t) = A sin(ωt)
-
C.
x(t) = A e^(ωt)
-
D.
x(t) = A ωt
Solution
The equation of motion for SHM is x(t) = A cos(ωt) or x(t) = A sin(ωt).
Correct Answer: A — x(t) = A cos(ωt)
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