UNE GENERAL CHEMISTRY II MIDTERM
ACTUAL EXAM QUESTIONS & ANSWERS
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beneath it to quickly clarify complex kinetics,
mathematical ICE tables, and acid-base buffer
systems. This highly organized, scannable guide is
specifically formatted to guarantee a top-tier score
and save you hours of study time.
1. Question 1 [Chemical Kinetics]: Which of the following
factors will increase the rate of a chemical reaction by
lowering the activation energy?
o A) Increasing the temperature
o B) Increasing the concentration of reactants
o C) Adding a catalyst
o D) Decreasing the surface area
,Answer: C) Adding a catalystRationale: A catalyst provides an
alternative reaction pathway with a lower activation energy,
thereby increasing the fraction of collisions that have
sufficient energy to react. Increasing temperature increases
kinetic energy but does not lower the activation energy itself.
2. Question 2 [Chemical Equilibrium]: If a reaction is zero-
order with respect to a reactant, what happens to the
rate of the reaction when the concentration of that
reactant is tripled?
o A) The rate triples
o B) The rate increases by a factor of nine
o C) The rate remains unchanged
o D) The rate decreases by half
Answer: C) The rate remains unchangedRationale: By definition,
the rate of a zero-order reaction is independent of the
concentration of the reactant (Rate = k[A]^0 = k). Therefore,
changing the concentration has zero effect on the overall
rate.
3. Question 3 [Acid-Base Chemistry]: The half-life of a first-
order reaction is 45 minutes. What is the rate constant
(k) for this reaction?
o A) 0.0154 min^-1
o B) 0.0222 min^-1
o C) 31.2 min^-1
, o D) 0.0111 min^-1
Answer: A) 0.0154 min^-1Rationale: For a first-order reaction,
t_1/2 = 0.693 / k. Rearranging for k gives k = 0. min =
0.0154 min^-1.
4. Question 4 [Chemical Kinetics]: The initial rate of a
reaction, A + B → C, is doubled when the concentration
of A is doubled, but quadrupled when the concentration
of B is doubled. What is the correct rate law?
o A) \(\text{Rate} = k[A][B]\)
o B) \(\text{Rate} = k[A]^2[B]\)
o C) \(\text{Rate} = k[A][B]^2\)
o D) \(\text{Rate} = k[A]^2[B]^2\)
Answer: C) \(\text{Rate} =\) k[A][B]^2Rationale: Doubling [A]
doubles the rate (2¹ = 2), meaning the reaction is first-order
with respect to A. Doubling [B] quadruples the rate (2² = 4),
meaning the reaction is second-order with respect to B.
Combining these gives \(\text{Rate} =\) k[A][B]^2.
5. Question 5 [Chemical Kinetics]: For a second-order
reaction, what are the standard units of the rate
constant (k), assuming time is in seconds and
concentration is in Molarity?
o A) M⁻¹s⁻¹
o B) M⋅s⁻¹
o C) s⁻¹
, o D) M⁻²s⁻¹
Answer: A) (\text{M}^{-1}\text{s}^{-1}Rationale: The unit of
reaction rate is always \text{M/s}. For a second-order
reaction, \text{Rate} = k[\text{Solute}]^2, so \text{M/s} =
k(\text{M}^2). Solving for the units of k gives \text{M}^{-
1}\text{s}^{-1} )(or \(\text{L}\cdot\text{mol}^{-1}\text{s}^{-1}\)).
6. Question 6 [Chemical Kinetics]: A reaction displays a
linear plot when \(\ln[A]\) is graphed against time. What
is the order of this reaction?
o A) Zero order
o B) First order
o C) Second order
o D) Third order
Answer: B) First orderRationale: The integrated rate law for a
first-order reaction is \(\ln[A]_t = -kt + \ln\)[A]_0. This matches
the equation of a straight line (y = mx + b), where a plot of
\(\ln \)[A] vs. time yields a straight line with a slope of -k.
7. Question 7 [Chemical Kinetics]: If the activation energy
(\(E_{a}\)) of a forward reaction is 75 kJ/mol and the
enthalpy change (Δ H) is -25 kJ/mol, what is the
activation energy for the reverse reaction?
o A) 50 kJ/mol
o B) 100 kJ/mol
ACTUAL EXAM QUESTIONS & ANSWERS
BUNDLE 100% VERIFIED PREMIUM GUIDE
Master your upcoming exam with this comprehensive
premium study bank featuring 100% verified actual
questions, answers, and analytical rationales tailored
perfectly to the UNE curriculum. Every multiple-choice
item delivers a step-by-step breakdown directly
beneath it to quickly clarify complex kinetics,
mathematical ICE tables, and acid-base buffer
systems. This highly organized, scannable guide is
specifically formatted to guarantee a top-tier score
and save you hours of study time.
1. Question 1 [Chemical Kinetics]: Which of the following
factors will increase the rate of a chemical reaction by
lowering the activation energy?
o A) Increasing the temperature
o B) Increasing the concentration of reactants
o C) Adding a catalyst
o D) Decreasing the surface area
,Answer: C) Adding a catalystRationale: A catalyst provides an
alternative reaction pathway with a lower activation energy,
thereby increasing the fraction of collisions that have
sufficient energy to react. Increasing temperature increases
kinetic energy but does not lower the activation energy itself.
2. Question 2 [Chemical Equilibrium]: If a reaction is zero-
order with respect to a reactant, what happens to the
rate of the reaction when the concentration of that
reactant is tripled?
o A) The rate triples
o B) The rate increases by a factor of nine
o C) The rate remains unchanged
o D) The rate decreases by half
Answer: C) The rate remains unchangedRationale: By definition,
the rate of a zero-order reaction is independent of the
concentration of the reactant (Rate = k[A]^0 = k). Therefore,
changing the concentration has zero effect on the overall
rate.
3. Question 3 [Acid-Base Chemistry]: The half-life of a first-
order reaction is 45 minutes. What is the rate constant
(k) for this reaction?
o A) 0.0154 min^-1
o B) 0.0222 min^-1
o C) 31.2 min^-1
, o D) 0.0111 min^-1
Answer: A) 0.0154 min^-1Rationale: For a first-order reaction,
t_1/2 = 0.693 / k. Rearranging for k gives k = 0. min =
0.0154 min^-1.
4. Question 4 [Chemical Kinetics]: The initial rate of a
reaction, A + B → C, is doubled when the concentration
of A is doubled, but quadrupled when the concentration
of B is doubled. What is the correct rate law?
o A) \(\text{Rate} = k[A][B]\)
o B) \(\text{Rate} = k[A]^2[B]\)
o C) \(\text{Rate} = k[A][B]^2\)
o D) \(\text{Rate} = k[A]^2[B]^2\)
Answer: C) \(\text{Rate} =\) k[A][B]^2Rationale: Doubling [A]
doubles the rate (2¹ = 2), meaning the reaction is first-order
with respect to A. Doubling [B] quadruples the rate (2² = 4),
meaning the reaction is second-order with respect to B.
Combining these gives \(\text{Rate} =\) k[A][B]^2.
5. Question 5 [Chemical Kinetics]: For a second-order
reaction, what are the standard units of the rate
constant (k), assuming time is in seconds and
concentration is in Molarity?
o A) M⁻¹s⁻¹
o B) M⋅s⁻¹
o C) s⁻¹
, o D) M⁻²s⁻¹
Answer: A) (\text{M}^{-1}\text{s}^{-1}Rationale: The unit of
reaction rate is always \text{M/s}. For a second-order
reaction, \text{Rate} = k[\text{Solute}]^2, so \text{M/s} =
k(\text{M}^2). Solving for the units of k gives \text{M}^{-
1}\text{s}^{-1} )(or \(\text{L}\cdot\text{mol}^{-1}\text{s}^{-1}\)).
6. Question 6 [Chemical Kinetics]: A reaction displays a
linear plot when \(\ln[A]\) is graphed against time. What
is the order of this reaction?
o A) Zero order
o B) First order
o C) Second order
o D) Third order
Answer: B) First orderRationale: The integrated rate law for a
first-order reaction is \(\ln[A]_t = -kt + \ln\)[A]_0. This matches
the equation of a straight line (y = mx + b), where a plot of
\(\ln \)[A] vs. time yields a straight line with a slope of -k.
7. Question 7 [Chemical Kinetics]: If the activation energy
(\(E_{a}\)) of a forward reaction is 75 kJ/mol and the
enthalpy change (Δ H) is -25 kJ/mol, what is the
activation energy for the reverse reaction?
o A) 50 kJ/mol
o B) 100 kJ/mol
