Q. A chemical reaction releases 250 J of heat. If the reaction occurs at constant pressure, what is the change in enthalpy?
A.-250 J
B.250 J
C.0 J
D.500 J
Solution
At constant pressure, ΔH = q = -250 J.
Correct Answer: A — -250 J
Q. A process absorbs 300 J of heat and does 100 J of work. What is the change in internal energy (ΔU)?
A.200 J
B.300 J
C.400 J
D.100 J
Solution
ΔU = q - W = 300 J - 100 J = 200 J.
Correct Answer: A — 200 J
Q. A reaction has an enthalpy change of 200 kJ for the formation of 1 mole of product. What is the enthalpy change for the formation of 0.5 moles of product?
A.100 kJ
B.200 kJ
C.50 kJ
D.400 kJ
Solution
ΔH for 0.5 moles = 200 kJ * 0.5 = 100 kJ.
Correct Answer: A — 100 kJ
Q. According to Graham's Law, the rate of effusion of a gas is inversely proportional to what?
A.Its molar mass
B.Its temperature
C.Its pressure
D.Its volume
Solution
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Correct Answer: A — Its molar mass
Q. According to the Kinetic Molecular Theory, which of the following is NOT a postulate?
A.Gas particles are in constant random motion.
B.Gas particles occupy a definite volume.
C.Collisions between gas particles are elastic.
D.The average kinetic energy is proportional to temperature.
Solution
Gas particles do not occupy a definite volume; they fill the container they are in.
Correct Answer: B — Gas particles occupy a definite volume.
Q. According to Werner's theory, what type of isomerism is primarily observed in coordination compounds?
A.Geometric isomerism
B.Optical isomerism
C.Structural isomerism
D.All of the above
Solution
Werner's theory accounts for various types of isomerism, including geometric, optical, and structural isomerism in coordination compounds.
Correct Answer: D — All of the above
Q. At constant temperature, what happens to the pressure of a gas if its volume is halved?
A.Pressure doubles
B.Pressure halves
C.Pressure remains the same
D.Pressure quadruples
Solution
According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. Halving the volume doubles the pressure.
Correct Answer: A — Pressure doubles
Q. At constant temperature, what happens to the pressure of a gas if its volume is doubled?
A.Pressure doubles
B.Pressure halves
C.Pressure remains the same
D.Pressure quadruples
Solution
According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. If the volume is doubled, the pressure is halved.
Correct Answer: B — Pressure halves
Q. At what temperature does the Kelvin scale start?
A.0°C
B.100°C
C.273.15°C
D.0 K
Solution
The Kelvin scale starts at absolute zero, which is 0 K or -273.15°C.
Correct Answer: D — 0 K
Q. Calculate the change in enthalpy (ΔH) when 2 moles of a substance absorb 500 J of heat at constant pressure.
A.250 J/mol
B.500 J/mol
C.1000 J/mol
D.125 J/mol
Solution
ΔH = q/n = 500 J / 2 mol = 250 J/mol.
Correct Answer: A — 250 J/mol
Q. Calculate the dipole moment of HCl given that the bond length is 1.27 Å and the charge separation is 0.33 e.
A.1.1 D
B.0.4 D
C.0.8 D
D.0.2 D
Solution
Dipole moment (μ) = charge (q) × distance (d). μ = 0.33 e × 1.27 × 10^-10 m = 4.2 × 10^-29 C·m = 1.1 D.
Correct Answer: A — 1.1 D
Q. Calculate the enthalpy change (ΔH) for the reaction: 2NO(g) + O2(g) → 2NO2(g) given the following bond enthalpies: N≡N = 941 kJ/mol, O=O = 498 kJ/mol, N=O = 201 kJ/mol.
A.-180 kJ
B.-200 kJ
C.-220 kJ
D.-240 kJ
Solution
ΔH = [2(201) + 498] - [2(941)] = -220 kJ.
Correct Answer: C — -220 kJ
Q. Calculate the formal charge on the nitrogen atom in NO3-.
A.0
B.-1
C.+1
D.+2
Solution
Formal charge = valence electrons - (non-bonding electrons + 1/2 bonding electrons). For nitrogen in NO3-, it is 5 - (0 + 1/2(6)) = 0.
Correct Answer: A — 0
Q. Calculate the ionization energy of hydrogen in eV if the energy level is -13.6 eV.
A.13.6 eV
B.1.24 eV
C.3.4 eV
D.0.85 eV
Solution
Ionization energy is the energy required to remove an electron from the ground state. For hydrogen, it is 13.6 eV.
Correct Answer: A — 13.6 eV
Q. Determine the bond angle in a tetrahedral molecule like CH4.
A.90°
B.109.5°
C.120°
D.180°
Solution
In a tetrahedral geometry, the bond angles are approximately 109.5° due to the arrangement of four electron pairs.
Correct Answer: B — 109.5°
Q. For a first-order reaction, if the half-life is 10 minutes, what will be the half-life if the concentration is doubled?
A.10 minutes
B.20 minutes
C.5 minutes
D.It cannot be determined
Solution
The half-life of a first-order reaction is independent of concentration; it remains 10 minutes.
Correct Answer: A — 10 minutes
Q. For a reaction A → B, if the rate constant k is 0.1 s^-1, what is the time required for the concentration of A to decrease to 25% of its initial value?
A.10 seconds
B.20 seconds
C.30 seconds
D.40 seconds
Solution
For a first-order reaction, t = (ln(2)/k) * ln([A]0/[A]t). To reach 25%, t = (ln(2)/0.1) * ln(4) = 20 seconds.
Correct Answer: B — 20 seconds
Q. For a reaction at equilibrium, if the concentration of products is increased, what will happen to the position of equilibrium according to Le Chatelier's principle?
A.Shift to the right
B.Shift to the left
C.No change
D.Depends on temperature
Solution
According to Le Chatelier's principle, increasing the concentration of products will shift the equilibrium position to the left to counteract the change.
Correct Answer: B — Shift to the left
Q. For a reaction at equilibrium, if the temperature is increased and ΔH is positive, what will happen to the position of equilibrium?
A.Shift to the right
B.Shift to the left
C.No change
D.Depends on concentration
Solution
According to Le Chatelier's principle, increasing temperature shifts the equilibrium to favor the endothermic direction (right).
Correct Answer: A — Shift to the right
Q. For a reaction that is first order in A and second order in B, what is the overall order of the reaction?
A.1
B.2
C.3
D.4
Solution
The overall order of a reaction is the sum of the orders with respect to each reactant. Here, it is 1 (for A) + 2 (for B) = 3.
Correct Answer: C — 3
Q. For a reaction with a negative ΔH and a positive ΔS, what can be said about the spontaneity at high temperatures?
A.The reaction is non-spontaneous.
B.The reaction is spontaneous.
C.The reaction is at equilibrium.
D.The spontaneity cannot be determined.
Solution
A negative ΔH and a positive ΔS indicate that the reaction is spontaneous at high temperatures, as ΔG will be negative.
Correct Answer: B — The reaction is spontaneous.
Q. For a reaction with a rate constant of 0.1 s^-1, what is the half-life for a first-order reaction?
A.0.693 s
B.6.93 s
C.10 s
D.0.1 s
Solution
The half-life for a first-order reaction is given by t1/2 = 0.693/k. Thus, t1/2 = 0.693/0.1 = 6.93 s.
Correct Answer: A — 0.693 s
Q. For a reaction with an activation energy of 50 kJ/mol, what is the effect of a 10 kJ/mol increase in activation energy on the rate constant at a constant temperature?
A.Rate constant increases
B.Rate constant decreases
C.Rate constant remains the same
D.Rate constant becomes zero
Solution
An increase in activation energy decreases the rate constant according to the Arrhenius equation, k = Ae^(-Ea/RT), where an increase in Ea results in a smaller k.
Correct Answer: B — Rate constant decreases
Q. For a reaction with an activation energy of 50 kJ/mol, what is the effect of increasing the temperature from 300 K to 350 K on the rate constant?
A.Rate constant decreases
B.Rate constant remains the same
C.Rate constant increases
D.Rate constant doubles
Solution
According to the Arrhenius equation, an increase in temperature generally increases the rate constant due to the exponential dependence on temperature.
Correct Answer: C — Rate constant increases
Q. For a reaction with an activation energy of 50 kJ/mol, what is the rate constant at 350 K if the rate constant at 300 K is 0.1 s^-1?
A.0.2 s^-1
B.0.3 s^-1
C.0.4 s^-1
D.0.5 s^-1
Solution
Using the Arrhenius equation, k2 = k1 * e^[-Ea/R(1/T2 - 1/T1)]. Substituting values gives k2 ≈ 0.4 s^-1.
Correct Answer: C — 0.4 s^-1
Q. For a reaction with ΔH = -120 kJ, how much heat is absorbed when 0.25 moles of reactants are converted?
A.30 kJ
B.60 kJ
C.120 kJ
D.15 kJ
Solution
Heat absorbed = ΔH * n = -120 kJ * 0.25 = -30 kJ.
Correct Answer: B — 60 kJ
Q. For a zero-order reaction, if the initial concentration is 0.5 M and the rate constant is 0.1 M/s, how long will it take for the concentration to drop to 0.2 M?
A.3 s
B.5 s
C.7 s
D.10 s
Solution
For a zero-order reaction, [A] = [A]0 - kt. Thus, 0.2 M = 0.5 M - (0.1 M/s)t, leading to t = (0.5 - 0.2) / 0.1 = 3 s.
Correct Answer: B — 5 s
Q. For an ideal gas, which equation relates the change in internal energy to heat and work?
A.ΔU = Q + W
B.ΔU = Q - W
C.ΔU = W - Q
D.ΔU = Q * W
Solution
The first law of thermodynamics states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).
Correct Answer: B — ΔU = Q - W
Q. For the equilibrium reaction 2NO2(g) ⇌ N2O4(g), what is the effect of increasing the temperature?
A.Shifts equilibrium to the right
B.Shifts equilibrium to the left
C.No effect on equilibrium
D.Increases the concentration of NO2
Solution
If the reaction is exothermic, increasing the temperature will shift the equilibrium to the left, favoring the formation of reactants (NO2).
Correct Answer: B — Shifts equilibrium to the left
Q. For the equilibrium reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), what happens if O2 is removed?
A.Equilibrium shifts to the right
B.Equilibrium shifts to the left
C.No change
D.Increases the temperature
Solution
Removing O2 decreases its concentration, causing the equilibrium to shift to the left to produce more O2, in accordance with Le Chatelier's Principle.
Correct Answer: B — Equilibrium shifts to the left
