ACS General Chemistry 2 (Gen Chem 2) Final Exam 2026–2028 |
Practice Test Questions & Answers | Complete Study Guide
Prepare for the ACS General Chemistry 2 (Gen Chem 2) Final Exam with this comprehensive
study guide featuring practice questions, verified answers, and detailed rationales. This resource
covers essential topics including chemical kinetics, chemical equilibrium, acids and bases,
buffers, solubility, thermodynamics, electrochemistry, nuclear chemistry, and coordination
compounds. Designed to reinforce fundamental chemistry concepts and improve exam readiness,
the material reflects the key content areas commonly assessed on the ACS Gen Chem 2 Final
Examination. Ideal for college students seeking a reliable resource to strengthen their
understanding, build confidence, and achieve success on the ACS Chemistry final exam.
Question 1
For a first-order reaction, a graph of which variables will yield a straight line with a
negative slope equal to \(-k\)?
A) \([A]\) versus time
B) \(1/[A]\) versus time
C) \(\ln[A]\) versus time
D) \(\ln(1/[A])\) versus time
Rationale: The integrated rate law for a first-order reaction is \(\ln[A]_t = -kt + \ln[A]_0\).
This matches the linear equation format \(y = mx + b\), where graphing \(\ln[A]\) on the
y-axis against time on the x-axis yields a straight line with a slope (\(m\)) of \(-k\).
Question 2
The decomposition of dinitrogen pentoxide is a first-order process with a rate constant
of \(4.80 \times 10^{-4}\text{ s}^{-1}\) at a specific temperature. What is the half-life of
this reaction?
A) \(2.08 \times 10^3\text{ s}\)
B) \(9.60 \times 10^{-4}\text{ s}\)
C) \(1.44 \times 10^3\text{ s}\)
D) \(6.93 \times 10^2\text{ s}\)
Rationale: For any first-order reaction, the half-life is independent of initial concentration
and calculated using the formula \(t_{1/2} = \frac{\ln(2)}{k}\). Plugging in the given rate
constant: \(0.693 / (4.80 \times 10^{-4}\text{ s}^{-1}) = 1443.75\text{ s}\), which rounds to
\(1.44 \times 10^3\text{ s}\).
Question 3
A reaction has the experimental rate law: \(\text{Rate} = k[A]^2[B]\). If the concentration
of reactant \(A\) is doubled while the concentration of reactant \(B\) is halved, what will
happen to the overall rate of the reaction?
A) The rate will decrease by a factor of 2.
,B) The rate will remain exactly the same.
C) The rate will increase by a factor of 2.
D) The rate will increase by a factor of 4.
Rationale: Let the initial rate be \(\text{Rate}_1 = k[A]^2[B]\). If \([A]\) becomes \(2[A]\)
and \([B]\) becomes \(0.5[B]\), the new rate expression becomes \(\text{Rate}_2 =
k(2[A])^2(0.5[B]) = k(4[A]^2)(0.5[B]) = 2k[A]^2[B]\). Thus, the rate doubles.
Question 4
Consider the following multi-step reaction mechanism:
Step 1: \(NO(g) + O_2(g) \rightleftharpoons NO_3(g)\) (fast, equilibrium)
Step 2: \(NO_3(g) + NO(g) \rightarrow 2 NO_2(g)\) (slow)
Which chemical species acts as a reaction intermediate within this system?
A) \(NO(g)\)
B) \(O_2(g)\)
C) \(NO_3(g)\)
D) \(NO_2(g)\)
Rationale: A reaction intermediate is a chemical species that is generated in an early
step of a mechanism and subsequently consumed in a later step, meaning it does not
appear in the overall balanced chemical equation. Here, \(NO_{3}\) is produced in Step
1 and used up in Step 2.
Question 5
The activation energy (\(E_{a}\)) for a particular forward chemical reaction is 75 kJ/mol,
and the overall enthalpy change (\(\Delta H\)) is -40 kJ/mol. What is the activation
energy for the reverse reaction?
A) 35 kJ/mol
B) 40 kJ/mol
C) 115 kJ/mol
D) -115 kJ/mol
Rationale: A negative enthalpy change (\(\Delta H = -40\text{ kJ/mol}\)) indicates an
exothermic reaction where products are lower in energy than the reactants. To go
backward from products to the transition state, the energy barrier is the sum of the
forward activation energy and the absolute value of enthalpy change: \(75\text{ kJ/mol}
+ 40\text{ kJ/mol} = 115\text{ kJ/mol}\).
Question 6
Which statement regarding catalysts is true?
A) A catalyst increases the yield of products by shifting the equilibrium constant.
B) A catalyst alters the mechanism of a reaction to lower the activation energy.
C) A catalyst increases the collision frequency of molecules by increasing the system
temperature.
D) A catalyst is permanently consumed during the rate-determining step of a reaction.
Rationale: Catalysts provide an alternative pathway or mechanism for a chemical
reaction that possesses a lower activation energy barrier. They speed up both the
,forward and reverse reactions equally, meaning they have no effect on the final position
of equilibrium or the value of K.
Question 7
The rate constant (\(k\)) of a reaction is measured at several different temperatures. A
plot of \(\ln k\) versus \(1/T\) (where \(T\) is in Kelvin) produces a straight line. What can
be determined from the slope of this line?
A) The overall order of the reaction.
B) The pre-exponential collision factor (\(A\)).
C) The activation energy (\(E_{a}\)) of the reaction.
D) The enthalpy change (\(\Delta H\)) of the process.
Rationale: According to the linear form of the Arrhenius equation, \(\ln k = -
\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). When plotting \(\ln k\) vs \(1/T\), the slope of
the line (\(m\)) is equal to \(-\frac{E_{a}}{R}\). Multiplying the slope by the negative gas
constant (\(-R\)) yields the activation energy.
Question 8
For the gas-phase equilibrium reaction \(2 SO_2(g) + O_2(g) \rightleftharpoons 2
SO_3(g)\), how does the numerical value of \(K_{p}\) relate to the value of \(K_{c}\) at a
temperature of 500 K?
A) \(K_p = K_c\)
B) \(K_p = K_c(RT)^{-1}\)
C) \(K_p = K_c(RT)^1\)
D) \(K_p = K_c(RT)^{-3}\)
Rationale: The relationship between the gas-phase equilibrium constants is given by
\(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of
gas (moles of gaseous product minus moles of gaseous reactant). Here, \(\Delta n = 2 -
(2 + 1) = -1\). Therefore, \(K_p = K_c(RT)^{-1}\).
Question 9
At a specific temperature, the equilibrium constant \(K_{c}\) is equal to \(4.0 \times 10^{-
2}\) for the reaction: \(2 HF(g) \rightleftharpoons H_2(g) + F_2(g)\). What is the value of
\(K_{c}\) for the reverse reaction at this same temperature?
A) 25
B) 5.0
C) \(2.0 \times 10^{-1}\)
D) \(4.0 \times 10^2\)
Rationale: When a reversible chemical reaction is written in the reverse direction, its
new equilibrium constant (\(K_{\text{new}}\)) is the reciprocal of the forward equilibrium
constant (\(1/K_{\text{forward}}\)). Calculating this gives: \(1 / (4.0 \times 10^{-2}) = 1 /
0.04 = 25\).
Question 10
The exothermic reaction \(CO(g) + 2 H_2(g) \rightleftharpoons CH_3OH(g)\) is allowed
to reach equilibrium. Which alteration will shift the system toward the right to form more
products?
, A) Increasing the temperature of the reaction vessel.
B) Adding an inert gas like Argon at constant volume.
C) Decreasing the volume of the reaction container.
D) Removing hydrogen (\(H_{2}\)) gas continuously from the mixture.
Rationale: According to Le Chatelier's principle, reducing container volume increases
the total pressure, shifting the equilibrium toward the side with fewer gas molecules.
The reactant side has 3 moles of gas, while the product side has only 1 mole of gas,
driving the reaction to the right.
Question 11
A mixture contains 1.0 M \(A\), 2.0 M \(B\), and 1.5 M \(C\) at a temperature where \(K_c
= 0.50\) for the reaction: \(A(g) + B(g) \rightleftharpoons 2 C(g)\). Which statement
accurately describes the status of the system?
A) The reaction is at equilibrium.
B) The reaction is not at equilibrium; it must shift left to reach it.
C) The reaction is not at equilibrium; it must shift right to reach it.
D) The direction of the shift cannot be determined without knowing the value of \(R\).
Rationale: Calculate the reaction quotient: \(Q_c = \frac{[C]^2}{[A][B]} =
\frac{(1.5)^2}{(1.0)(2.0)} = \frac{2.25}{2.0} = 1.125\). Because \(Q_c (1.125) > K_c
(0.50)\), there are too many products relative to reactants, forcing the system to shift to
the left (toward reactants) to re-establish equilibrium.
Question 12
What is the conjugate base of the amphiprotic ion hydrogen phosphate, \(HPO_{4}^{2-
}\)?
A) \(H_{3}PO_{4}\)
B) \(H_{2}PO_{4}^{-}\)
C) \(PO_{4}^{3-}\)
D) \(OH^{-}\)
Rationale: A conjugate base is formed when a Brønsted-Lowry acid donates a hydrogen
ion (\(H^{+}\)). Removing an \(H^{+}\) ion from \(HPO_{4}^{2-}\) reduces its hydrogen
count by one and drops its electrical charge from \(-2\) to \(-3\), yielding the phosphate
ion, \(PO_{4}^{3-}\).
Question 13
Calculate the pH of a \(0.0050\text{ M}\) aqueous solution of barium hydroxide,
\(Ba(OH)_2\), assuming it behaves as a strong base that dissociates completely.
A) 2.00
B) 2.30
C) 11.70
D) 12.00
Rationale: Barium hydroxide releases two moles of hydroxide ions per mole of formula
unit: \([OH^-] = 2 \times 0.0050\text{ M} = 0.010\text{ M}\). The pOH of this solution is
equal to \(-\log[OH^-] = -\log(0.010) = 2.00\). Since \(\text{pH} + \text{pOH} = 14.00\), the
pH is \(14.00 - 2.00 = 12.00\).
Practice Test Questions & Answers | Complete Study Guide
Prepare for the ACS General Chemistry 2 (Gen Chem 2) Final Exam with this comprehensive
study guide featuring practice questions, verified answers, and detailed rationales. This resource
covers essential topics including chemical kinetics, chemical equilibrium, acids and bases,
buffers, solubility, thermodynamics, electrochemistry, nuclear chemistry, and coordination
compounds. Designed to reinforce fundamental chemistry concepts and improve exam readiness,
the material reflects the key content areas commonly assessed on the ACS Gen Chem 2 Final
Examination. Ideal for college students seeking a reliable resource to strengthen their
understanding, build confidence, and achieve success on the ACS Chemistry final exam.
Question 1
For a first-order reaction, a graph of which variables will yield a straight line with a
negative slope equal to \(-k\)?
A) \([A]\) versus time
B) \(1/[A]\) versus time
C) \(\ln[A]\) versus time
D) \(\ln(1/[A])\) versus time
Rationale: The integrated rate law for a first-order reaction is \(\ln[A]_t = -kt + \ln[A]_0\).
This matches the linear equation format \(y = mx + b\), where graphing \(\ln[A]\) on the
y-axis against time on the x-axis yields a straight line with a slope (\(m\)) of \(-k\).
Question 2
The decomposition of dinitrogen pentoxide is a first-order process with a rate constant
of \(4.80 \times 10^{-4}\text{ s}^{-1}\) at a specific temperature. What is the half-life of
this reaction?
A) \(2.08 \times 10^3\text{ s}\)
B) \(9.60 \times 10^{-4}\text{ s}\)
C) \(1.44 \times 10^3\text{ s}\)
D) \(6.93 \times 10^2\text{ s}\)
Rationale: For any first-order reaction, the half-life is independent of initial concentration
and calculated using the formula \(t_{1/2} = \frac{\ln(2)}{k}\). Plugging in the given rate
constant: \(0.693 / (4.80 \times 10^{-4}\text{ s}^{-1}) = 1443.75\text{ s}\), which rounds to
\(1.44 \times 10^3\text{ s}\).
Question 3
A reaction has the experimental rate law: \(\text{Rate} = k[A]^2[B]\). If the concentration
of reactant \(A\) is doubled while the concentration of reactant \(B\) is halved, what will
happen to the overall rate of the reaction?
A) The rate will decrease by a factor of 2.
,B) The rate will remain exactly the same.
C) The rate will increase by a factor of 2.
D) The rate will increase by a factor of 4.
Rationale: Let the initial rate be \(\text{Rate}_1 = k[A]^2[B]\). If \([A]\) becomes \(2[A]\)
and \([B]\) becomes \(0.5[B]\), the new rate expression becomes \(\text{Rate}_2 =
k(2[A])^2(0.5[B]) = k(4[A]^2)(0.5[B]) = 2k[A]^2[B]\). Thus, the rate doubles.
Question 4
Consider the following multi-step reaction mechanism:
Step 1: \(NO(g) + O_2(g) \rightleftharpoons NO_3(g)\) (fast, equilibrium)
Step 2: \(NO_3(g) + NO(g) \rightarrow 2 NO_2(g)\) (slow)
Which chemical species acts as a reaction intermediate within this system?
A) \(NO(g)\)
B) \(O_2(g)\)
C) \(NO_3(g)\)
D) \(NO_2(g)\)
Rationale: A reaction intermediate is a chemical species that is generated in an early
step of a mechanism and subsequently consumed in a later step, meaning it does not
appear in the overall balanced chemical equation. Here, \(NO_{3}\) is produced in Step
1 and used up in Step 2.
Question 5
The activation energy (\(E_{a}\)) for a particular forward chemical reaction is 75 kJ/mol,
and the overall enthalpy change (\(\Delta H\)) is -40 kJ/mol. What is the activation
energy for the reverse reaction?
A) 35 kJ/mol
B) 40 kJ/mol
C) 115 kJ/mol
D) -115 kJ/mol
Rationale: A negative enthalpy change (\(\Delta H = -40\text{ kJ/mol}\)) indicates an
exothermic reaction where products are lower in energy than the reactants. To go
backward from products to the transition state, the energy barrier is the sum of the
forward activation energy and the absolute value of enthalpy change: \(75\text{ kJ/mol}
+ 40\text{ kJ/mol} = 115\text{ kJ/mol}\).
Question 6
Which statement regarding catalysts is true?
A) A catalyst increases the yield of products by shifting the equilibrium constant.
B) A catalyst alters the mechanism of a reaction to lower the activation energy.
C) A catalyst increases the collision frequency of molecules by increasing the system
temperature.
D) A catalyst is permanently consumed during the rate-determining step of a reaction.
Rationale: Catalysts provide an alternative pathway or mechanism for a chemical
reaction that possesses a lower activation energy barrier. They speed up both the
,forward and reverse reactions equally, meaning they have no effect on the final position
of equilibrium or the value of K.
Question 7
The rate constant (\(k\)) of a reaction is measured at several different temperatures. A
plot of \(\ln k\) versus \(1/T\) (where \(T\) is in Kelvin) produces a straight line. What can
be determined from the slope of this line?
A) The overall order of the reaction.
B) The pre-exponential collision factor (\(A\)).
C) The activation energy (\(E_{a}\)) of the reaction.
D) The enthalpy change (\(\Delta H\)) of the process.
Rationale: According to the linear form of the Arrhenius equation, \(\ln k = -
\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). When plotting \(\ln k\) vs \(1/T\), the slope of
the line (\(m\)) is equal to \(-\frac{E_{a}}{R}\). Multiplying the slope by the negative gas
constant (\(-R\)) yields the activation energy.
Question 8
For the gas-phase equilibrium reaction \(2 SO_2(g) + O_2(g) \rightleftharpoons 2
SO_3(g)\), how does the numerical value of \(K_{p}\) relate to the value of \(K_{c}\) at a
temperature of 500 K?
A) \(K_p = K_c\)
B) \(K_p = K_c(RT)^{-1}\)
C) \(K_p = K_c(RT)^1\)
D) \(K_p = K_c(RT)^{-3}\)
Rationale: The relationship between the gas-phase equilibrium constants is given by
\(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of
gas (moles of gaseous product minus moles of gaseous reactant). Here, \(\Delta n = 2 -
(2 + 1) = -1\). Therefore, \(K_p = K_c(RT)^{-1}\).
Question 9
At a specific temperature, the equilibrium constant \(K_{c}\) is equal to \(4.0 \times 10^{-
2}\) for the reaction: \(2 HF(g) \rightleftharpoons H_2(g) + F_2(g)\). What is the value of
\(K_{c}\) for the reverse reaction at this same temperature?
A) 25
B) 5.0
C) \(2.0 \times 10^{-1}\)
D) \(4.0 \times 10^2\)
Rationale: When a reversible chemical reaction is written in the reverse direction, its
new equilibrium constant (\(K_{\text{new}}\)) is the reciprocal of the forward equilibrium
constant (\(1/K_{\text{forward}}\)). Calculating this gives: \(1 / (4.0 \times 10^{-2}) = 1 /
0.04 = 25\).
Question 10
The exothermic reaction \(CO(g) + 2 H_2(g) \rightleftharpoons CH_3OH(g)\) is allowed
to reach equilibrium. Which alteration will shift the system toward the right to form more
products?
, A) Increasing the temperature of the reaction vessel.
B) Adding an inert gas like Argon at constant volume.
C) Decreasing the volume of the reaction container.
D) Removing hydrogen (\(H_{2}\)) gas continuously from the mixture.
Rationale: According to Le Chatelier's principle, reducing container volume increases
the total pressure, shifting the equilibrium toward the side with fewer gas molecules.
The reactant side has 3 moles of gas, while the product side has only 1 mole of gas,
driving the reaction to the right.
Question 11
A mixture contains 1.0 M \(A\), 2.0 M \(B\), and 1.5 M \(C\) at a temperature where \(K_c
= 0.50\) for the reaction: \(A(g) + B(g) \rightleftharpoons 2 C(g)\). Which statement
accurately describes the status of the system?
A) The reaction is at equilibrium.
B) The reaction is not at equilibrium; it must shift left to reach it.
C) The reaction is not at equilibrium; it must shift right to reach it.
D) The direction of the shift cannot be determined without knowing the value of \(R\).
Rationale: Calculate the reaction quotient: \(Q_c = \frac{[C]^2}{[A][B]} =
\frac{(1.5)^2}{(1.0)(2.0)} = \frac{2.25}{2.0} = 1.125\). Because \(Q_c (1.125) > K_c
(0.50)\), there are too many products relative to reactants, forcing the system to shift to
the left (toward reactants) to re-establish equilibrium.
Question 12
What is the conjugate base of the amphiprotic ion hydrogen phosphate, \(HPO_{4}^{2-
}\)?
A) \(H_{3}PO_{4}\)
B) \(H_{2}PO_{4}^{-}\)
C) \(PO_{4}^{3-}\)
D) \(OH^{-}\)
Rationale: A conjugate base is formed when a Brønsted-Lowry acid donates a hydrogen
ion (\(H^{+}\)). Removing an \(H^{+}\) ion from \(HPO_{4}^{2-}\) reduces its hydrogen
count by one and drops its electrical charge from \(-2\) to \(-3\), yielding the phosphate
ion, \(PO_{4}^{3-}\).
Question 13
Calculate the pH of a \(0.0050\text{ M}\) aqueous solution of barium hydroxide,
\(Ba(OH)_2\), assuming it behaves as a strong base that dissociates completely.
A) 2.00
B) 2.30
C) 11.70
D) 12.00
Rationale: Barium hydroxide releases two moles of hydroxide ions per mole of formula
unit: \([OH^-] = 2 \times 0.0050\text{ M} = 0.010\text{ M}\). The pOH of this solution is
equal to \(-\log[OH^-] = -\log(0.010) = 2.00\). Since \(\text{pH} + \text{pOH} = 14.00\), the
pH is \(14.00 - 2.00 = 12.00\).
