GENERAL CHEMISTRY II COMPREHENSIVE
FINAL ASSESSMENT EXPERT CRAFTED
QUESTIONS WITH DETAILED CORRECT
VERIFIED ANSWERS AND EXPLANATORY
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General Chemistry II Comprehensive Exam
1. A laboratory technician monitors the decomposition of
dinitrogen pentoxide: \(2\text{N}_2\text{O}_5(g) \rightarrow
4\text{NO}_2(g) + \text{O}_2(g)\). At a specific time interval,
the rate of disappearance of \(\text{N}_2\text{O}_5\) is
determined to be \(2.4 \times 10^{-4}\text{ M}\cdot\text{s}^{-
1}\). What is the corresponding rate of appearance of
\(\text{NO}_2(g)\) during this identical time frame? [1]
A) \(1.2 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
B) \(2.4 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
C) \(4.8 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
D) \(9.6 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
Correct Answer: C
Rationale: According to reaction stoichiometry, the relative rate of a
reaction is given by \(\text{Rate} = -
\frac{1}{2}\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} =
+\frac{1}{4}\frac{\Delta[\text{NO}_2]}{\Delta t}\). Therefore, the rate
of appearance of \(\text{NO}_{2}\) is exactly twice the rate of
disappearance of \(\text{N}_2\text{O}_5\). Multiplying \(2.4 \times
10^{-4}\text{ M}\cdot\text{s}^{-1}\) by 2 yields \(4.8 \times 10^{-
4}\text{ M}\cdot\text{s}^{-1}\).
2. An experimental kinetic study of the generic reaction
\(\text{A} + 2\text{B} \rightarrow \text{C}\) shows that
doubling the concentration of reactant \(\text{A}\) while
,holding \(\text{B}\) constant quadruples the initial rate.
Doubling the concentration of reactant \(\text{B}\) while
holding \(\text{A}\) constant has no effect on the initial rate.
What is the overall order of this reaction?
A) First-order
B) Second-order
C) Third-order
D) Zero-order
Correct Answer: B
Rationale: Because quadrupling the rate (\(2^2 = 4\)) occurs when
reactant \(\text{A}\) is doubled, the reaction is second-order with
respect to \(\text{A}\). Because changing the concentration of
\(\text{B}\) causes zero change in rate (\(2^0 = 1\)), the reaction is
zero-order with respect to \(\text{B}\). The overall reaction order is the
sum of the individual orders: \(2 + 0 = 2\). [1, 2]
3. A first-order chemical reaction has a specific rate constant
(\(k\)) of \(3.5 \times 10^{-3}\text{ s}^{-1}\) at \(25^{\circ
}\text{C}\). If the initial concentration of the reactant is
\(0.80\text{ M}\), what concentration of this reactant remains
unreacted after exactly \(5.0\text{ minutes}\) have elapsed?
A) \(0.72\text{ M}\)
B) \(0.54\text{ M}\)
C) \(0.28\text{ M}\)
D) \(0.11\text{ M}\)
Correct Answer: C
Rationale: For a first-order reaction, the integrated rate law is
\(\ln[\text{A}]_t = -kt + \ln[\text{A}]_0\). First, convert time to
seconds: \(5.0\text{ min} \times 60\text{ s/min} = 300\text{ s}\).
Substitute the values: \(\ln[\text{A}]_t = -(3.5 \times 10^{-3}\text{
s}^{-1})(300\text{ s}) + \ln(0.80) = -1.05 + (-0.223) = -1.273\). Taking
the exponential of both sides yields \([\text{A}]_t = e^{-1.273} =
0.28\text{ M}\).
4. The radioactive decay of an isotope follows precise first-
order kinetics. If a sample decays to \(12.5\%\) of its original
,mass over a time span of exactly \(24.0\text{ days}\), what is
the half-life (\(t_{1/2}\)) of this particular isotope? [1]
A) \(8.0\text{ days}\)
B) \(6.0\text{ days}\)
C) \(12.0\text{ days}\)
D) \(3.0\text{ days}\)
Correct Answer: A
Rationale: A reduction to \(12.5\%\) remaining signifies that the
sample has undergone exactly three half-lives (\(100\% \rightarrow
50\% \rightarrow 25\% \rightarrow 12.5\%\)). To find the duration of
a single half-life, divide the total elapsed time by the number of half-
lives: \(24.0\text{ days} / 3 = 8.0\text{ days}\). [1]
5. According to Collision Theory and the Arrhenius equation,
which statement best explains why increasing the temperature
of a reaction mixture exponentially accelerates the rate of a
chemical reaction?
A) The activation energy (\(E_{a}\)) threshold shifts to a lower energy
state.
B) The reaction mechanism changes to prioritize a lower-order rate law.
C) A significantly larger fraction of reactant molecules possess
kinetic energy that meets or exceeds \(E_{a}\).
D) The frequency factor (\(A\)) drops due to fewer steric collisions.
Correct Answer: C
Rationale: Temperature is a direct measure of average kinetic energy.
According to the Maxwell-Boltzmann distribution, an increase in
temperature flattens and extends the curve, drastically increasing the
fraction of molecules with sufficient energy to overcome the activation
energy barrier (\(E_{a}\)). The value of \(E_{a}\) itself remains
unchanged unless a catalyst is introduced. [1]
6. A proposed mechanism for the reaction \(2\text{NO}_2(g) +
\text{F}_2(g) \rightarrow 2\text{NO}_2\text{F}(g)\) consists
of the following elementary steps:
Step 1 (Slow): \(\text{NO}_2 + \text{F}_2 \rightarrow
\text{NO}_2\text{F} + \text{F}\)
, Step 2 (Fast): \(\text{NO}_2 + \text{F} \rightarrow
\text{NO}_2\text{F}\)
Which chemical species acts as a reaction intermediate, and
what is the rate law derived from this mechanism?
A) \(\text{NO}_2\text{F}\) is an intermediate; \(\text{Rate} =
k[\text{NO}_2]^2[\text{F}_2]\)
B) \(\text{F}\) is an intermediate; \(\text{Rate} =
k[\text{NO}_2][\text{F}_2]\)
C) \(\text{F}_{2}\) is an intermediate; \(\text{Rate} =
k[\text{NO}_2][\text{F}]\)
D) \(\text{F}\) is a catalyst; \(\text{Rate} = k[\text{NO}_2]^2\) [1]
Correct Answer: B
Rationale: An intermediate is produced in an early step and consumed
in a later step (\(\text{F}\) atom). The rate law of a multi-step reaction
is governed strictly by its rate-determining (slow) step. Because Step 1
is the slow step, its molecularity directly dictates the rate law, yielding
\(\text{Rate} = k[\text{NO}_2][\text{F}_2]\). [1, 2, 3, 4]
7. How does the addition of a heterogeneous catalyst modify a
reaction profile to accelerate the speed at which a chemical
system reaches dynamic equilibrium?
A) It provides an alternative reaction pathway possessing a
lower activation energy for both the forward and reverse
reactions.
B) It systematically alters the enthalpy change (\(\Delta H\)) to make
the reaction highly exothermic.
C) It increases the equilibrium constant (\(K_{c}\)), forcing the
production of more products.
D) It selectively increases the kinetic energy of the forward step without
affecting the reverse step. [1]
Correct Answer: A
Rationale: A catalyst speeds up a reaction by offering a completely
different pathway with a lower activation energy (\(E_{a}\)). Because
it lowers the barrier for both the forward and reverse directions
equally, it speeds up both rates uniformly, allowing the system to
achieve equilibrium faster without changing the actual position of
equilibrium (\(K_{c}\) stays constant). [1, 2, 3]
FINAL ASSESSMENT EXPERT CRAFTED
QUESTIONS WITH DETAILED CORRECT
VERIFIED ANSWERS AND EXPLANATORY
RATIONALES GRADE A+ ACHIEVEMENT
RESOURCE | 100% SUCCESS MATERIAL |
INSTANT DOWNLOAD
General Chemistry II Comprehensive Exam
1. A laboratory technician monitors the decomposition of
dinitrogen pentoxide: \(2\text{N}_2\text{O}_5(g) \rightarrow
4\text{NO}_2(g) + \text{O}_2(g)\). At a specific time interval,
the rate of disappearance of \(\text{N}_2\text{O}_5\) is
determined to be \(2.4 \times 10^{-4}\text{ M}\cdot\text{s}^{-
1}\). What is the corresponding rate of appearance of
\(\text{NO}_2(g)\) during this identical time frame? [1]
A) \(1.2 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
B) \(2.4 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
C) \(4.8 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
D) \(9.6 \times 10^{-4}\text{ M}\cdot\text{s}^{-1}\)
Correct Answer: C
Rationale: According to reaction stoichiometry, the relative rate of a
reaction is given by \(\text{Rate} = -
\frac{1}{2}\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} =
+\frac{1}{4}\frac{\Delta[\text{NO}_2]}{\Delta t}\). Therefore, the rate
of appearance of \(\text{NO}_{2}\) is exactly twice the rate of
disappearance of \(\text{N}_2\text{O}_5\). Multiplying \(2.4 \times
10^{-4}\text{ M}\cdot\text{s}^{-1}\) by 2 yields \(4.8 \times 10^{-
4}\text{ M}\cdot\text{s}^{-1}\).
2. An experimental kinetic study of the generic reaction
\(\text{A} + 2\text{B} \rightarrow \text{C}\) shows that
doubling the concentration of reactant \(\text{A}\) while
,holding \(\text{B}\) constant quadruples the initial rate.
Doubling the concentration of reactant \(\text{B}\) while
holding \(\text{A}\) constant has no effect on the initial rate.
What is the overall order of this reaction?
A) First-order
B) Second-order
C) Third-order
D) Zero-order
Correct Answer: B
Rationale: Because quadrupling the rate (\(2^2 = 4\)) occurs when
reactant \(\text{A}\) is doubled, the reaction is second-order with
respect to \(\text{A}\). Because changing the concentration of
\(\text{B}\) causes zero change in rate (\(2^0 = 1\)), the reaction is
zero-order with respect to \(\text{B}\). The overall reaction order is the
sum of the individual orders: \(2 + 0 = 2\). [1, 2]
3. A first-order chemical reaction has a specific rate constant
(\(k\)) of \(3.5 \times 10^{-3}\text{ s}^{-1}\) at \(25^{\circ
}\text{C}\). If the initial concentration of the reactant is
\(0.80\text{ M}\), what concentration of this reactant remains
unreacted after exactly \(5.0\text{ minutes}\) have elapsed?
A) \(0.72\text{ M}\)
B) \(0.54\text{ M}\)
C) \(0.28\text{ M}\)
D) \(0.11\text{ M}\)
Correct Answer: C
Rationale: For a first-order reaction, the integrated rate law is
\(\ln[\text{A}]_t = -kt + \ln[\text{A}]_0\). First, convert time to
seconds: \(5.0\text{ min} \times 60\text{ s/min} = 300\text{ s}\).
Substitute the values: \(\ln[\text{A}]_t = -(3.5 \times 10^{-3}\text{
s}^{-1})(300\text{ s}) + \ln(0.80) = -1.05 + (-0.223) = -1.273\). Taking
the exponential of both sides yields \([\text{A}]_t = e^{-1.273} =
0.28\text{ M}\).
4. The radioactive decay of an isotope follows precise first-
order kinetics. If a sample decays to \(12.5\%\) of its original
,mass over a time span of exactly \(24.0\text{ days}\), what is
the half-life (\(t_{1/2}\)) of this particular isotope? [1]
A) \(8.0\text{ days}\)
B) \(6.0\text{ days}\)
C) \(12.0\text{ days}\)
D) \(3.0\text{ days}\)
Correct Answer: A
Rationale: A reduction to \(12.5\%\) remaining signifies that the
sample has undergone exactly three half-lives (\(100\% \rightarrow
50\% \rightarrow 25\% \rightarrow 12.5\%\)). To find the duration of
a single half-life, divide the total elapsed time by the number of half-
lives: \(24.0\text{ days} / 3 = 8.0\text{ days}\). [1]
5. According to Collision Theory and the Arrhenius equation,
which statement best explains why increasing the temperature
of a reaction mixture exponentially accelerates the rate of a
chemical reaction?
A) The activation energy (\(E_{a}\)) threshold shifts to a lower energy
state.
B) The reaction mechanism changes to prioritize a lower-order rate law.
C) A significantly larger fraction of reactant molecules possess
kinetic energy that meets or exceeds \(E_{a}\).
D) The frequency factor (\(A\)) drops due to fewer steric collisions.
Correct Answer: C
Rationale: Temperature is a direct measure of average kinetic energy.
According to the Maxwell-Boltzmann distribution, an increase in
temperature flattens and extends the curve, drastically increasing the
fraction of molecules with sufficient energy to overcome the activation
energy barrier (\(E_{a}\)). The value of \(E_{a}\) itself remains
unchanged unless a catalyst is introduced. [1]
6. A proposed mechanism for the reaction \(2\text{NO}_2(g) +
\text{F}_2(g) \rightarrow 2\text{NO}_2\text{F}(g)\) consists
of the following elementary steps:
Step 1 (Slow): \(\text{NO}_2 + \text{F}_2 \rightarrow
\text{NO}_2\text{F} + \text{F}\)
, Step 2 (Fast): \(\text{NO}_2 + \text{F} \rightarrow
\text{NO}_2\text{F}\)
Which chemical species acts as a reaction intermediate, and
what is the rate law derived from this mechanism?
A) \(\text{NO}_2\text{F}\) is an intermediate; \(\text{Rate} =
k[\text{NO}_2]^2[\text{F}_2]\)
B) \(\text{F}\) is an intermediate; \(\text{Rate} =
k[\text{NO}_2][\text{F}_2]\)
C) \(\text{F}_{2}\) is an intermediate; \(\text{Rate} =
k[\text{NO}_2][\text{F}]\)
D) \(\text{F}\) is a catalyst; \(\text{Rate} = k[\text{NO}_2]^2\) [1]
Correct Answer: B
Rationale: An intermediate is produced in an early step and consumed
in a later step (\(\text{F}\) atom). The rate law of a multi-step reaction
is governed strictly by its rate-determining (slow) step. Because Step 1
is the slow step, its molecularity directly dictates the rate law, yielding
\(\text{Rate} = k[\text{NO}_2][\text{F}_2]\). [1, 2, 3, 4]
7. How does the addition of a heterogeneous catalyst modify a
reaction profile to accelerate the speed at which a chemical
system reaches dynamic equilibrium?
A) It provides an alternative reaction pathway possessing a
lower activation energy for both the forward and reverse
reactions.
B) It systematically alters the enthalpy change (\(\Delta H\)) to make
the reaction highly exothermic.
C) It increases the equilibrium constant (\(K_{c}\)), forcing the
production of more products.
D) It selectively increases the kinetic energy of the forward step without
affecting the reverse step. [1]
Correct Answer: A
Rationale: A catalyst speeds up a reaction by offering a completely
different pathway with a lower activation energy (\(E_{a}\)). Because
it lowers the barrier for both the forward and reverse directions
equally, it speeds up both rates uniformly, allowing the system to
achieve equilibrium faster without changing the actual position of
equilibrium (\(K_{c}\) stays constant). [1, 2, 3]
