College Chemistry Quiz: Introduction To Le Chateliers Principle

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Introduction To Le Chateliers PrincipleQuestion 1 of 20

For SO₂(g) + O₂(g) ⇌ SO₃(g), what is true at equilibrium in a closed container?

Concentrations are equal

Forward and reverse rates match

Only reactants remain

Reaction quotient is zero

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College Chemistry Quiz

College Chemistry Quiz: Introduction To Le Chateliers Principle

Practice Introduction To Le Chateliers Principle in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Introduction To Le Chateliers Principle, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For SO₂(g) + O₂(g) ⇌ SO₃(g), what is true at equilibrium in a closed container?

Concentrations are equal
Forward and reverse rates match (correct answer)
Only reactants remain
Reaction quotient is zero

Explanation: This question tests introductory college chemistry skills focused on understanding Le Chatelier's Principle and its impact on equilibrium. Le Chatelier's Principle states that a system at equilibrium will adjust to counteract any imposed changes, such as concentration, temperature, or pressure. At equilibrium in a closed container for SO₂(g) + O₂(g) ⇌ SO₃(g), forward and reverse rates are equal, keeping concentrations constant. Choice B is correct as it states the rates match, defining equilibrium. Choice A is incorrect because concentrations are not necessarily equal, just constant. To help students, debunk myths like reactions stopping at equilibrium. Use examples to show dynamic nature through rates.

Question 2

For SO₂(g) + O₂(g) ⇌ SO₃(g), what happens to Q right after removing SO₃(g)?

Q decreases (correct answer)
Q increases
Q becomes equal to K instantly
Q becomes 1

Explanation: This question tests introductory college chemistry skills focused on understanding Le Chatelier's Principle and its impact on equilibrium. Le Chatelier's Principle states that a system at equilibrium will adjust to counteract any imposed changes, such as concentration, temperature, or pressure. Right after removing SO₃(g) from SO₂(g) + O₂(g) ⇌ SO₃(g), Q decreases due to a smaller numerator. Choice A is correct as it states Q decreases, prompting a right shift. Choice B is incorrect because removing product decreases Q, not increases it. To help students, use the Q expression to show immediate effects. Practice with removal scenarios to reinforce concepts.

Question 3

A student places solid $CaCO_3(s)$ in a closed container and allows the following equilibrium to establish at 800°C: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ . After equilibrium is reached, the student doubles the volume of the container while keeping temperature constant. Which of the following best describes the immediate effect on the system?

The equilibrium will shift left because decreasing pressure favors the side with fewer gas molecules
The equilibrium will shift right because decreasing pressure favors the side with more gas molecules
The equilibrium will shift left because the concentration of $CO_2$ increases
The equilibrium will shift right because the partial pressure of $CO_2$ decreases (correct answer)
No shift will occur because the equilibrium involves only solid reactants and products

Explanation: When you encounter equilibrium problems involving volume or pressure changes, focus on Le Châtelier's principle and how gas molecules respond to pressure changes. This reaction involves solids and one gas (

CO_2

), so only the gas phase matters for pressure effects. When the container volume doubles at constant temperature, the partial pressure of

CO_2

immediately drops to half its original value. This creates a stress on the equilibrium system. According to Le Châtelier's principle, the system will respond by shifting in the direction that relieves this stress. Since the

CO_2

pressure decreased, the equilibrium shifts right to produce more

CO_2

gas and restore equilibrium. Choice A incorrectly states that decreasing pressure favors fewer gas molecules, which is backwards. Lower pressure actually favors the side with more gas molecules (or in this case, the side that produces gas). Choice B has the right direction but wrong reasoning—it's not about "more gas molecules" since there's only one gas species, but about the pressure decrease of that gas. Choice C is completely wrong because when volume doubles, the concentration of

CO_2

decreases, not increases. Choice D correctly identifies both the cause (partial pressure of

CO_2

decreases) and the effect (equilibrium shifts right). Study tip: For equilibrium problems involving volume changes, remember that increasing volume decreases gas pressure, which shifts equilibrium toward the side that produces more gas molecules to counteract the pressure drop.

Question 4

For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , a chemist observes that adding a catalyst increases the rate at which equilibrium is reached. A student claims that the catalyst must shift the equilibrium position toward products because it makes the reaction go faster. Which statement best evaluates this claim?

The claim is correct because catalysts always favor product formation in equilibrium reactions
The claim is correct because increasing the forward reaction rate shifts equilibrium toward products
The claim is incorrect because catalysts affect both forward and reverse reaction rates equally (correct answer)
The claim is incorrect because catalysts only affect endothermic reactions, not exothermic ones
The claim is partially correct because catalysts shift equilibrium position but don't affect reaction rates

Explanation: When you encounter questions about catalysts and chemical equilibrium, focus on understanding what catalysts actually do versus common misconceptions about their effects. A catalyst works by providing an alternative reaction pathway with lower activation energy. Crucially, it lowers the activation energy barrier for both the forward AND reverse reactions equally. This means while the catalyst helps the system reach equilibrium faster, it doesn't change where that equilibrium position lies. The equilibrium constant

K_{eq}

remains unchanged because it depends only on temperature, not on the presence of a catalyst. The correct answer is C because catalysts affect both forward and reverse reaction rates equally. The student's reasoning contains a fundamental error: faster reaction rate does not mean shifted equilibrium position. The catalyst speeds up both directions of the reaction proportionally. Answer A is wrong because catalysts never favor products or reactants—they're completely neutral regarding equilibrium position. Answer B reflects the student's flawed logic and misunderstands that equal rate increases in both directions cancel out any potential shift effect. Answer D incorrectly suggests catalysts only work for certain types of reactions based on thermodynamics, when actually catalysts affect reaction kinetics regardless of whether a reaction is endothermic or exothermic. Remember this key distinction: catalysts affect kinetics (how fast equilibrium is reached) but not thermodynamics (where equilibrium lies). When you see catalyst questions, always ask yourself whether the question is about reaction rate or equilibrium position—they're completely different concepts.

Question 5

A mixture containing $H_2(g)$ , $I_2(g)$ , and $HI(g)$ is at equilibrium according to: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ . When additional $H_2(g)$ is injected into the container, the concentration of $I_2$ is observed to decrease. Which of the following best explains this observation using Le Châtelier's principle?

Adding $H_2$ increases the total pressure, which favors the side with fewer gas molecules
Adding $H_2$ shifts the equilibrium right, consuming $I_2$ to form more $HI$ (correct answer)
Adding $H_2$ dilutes the $I_2$ , causing its concentration to decrease directly
Adding $H_2$ increases temperature due to collision frequency, shifting equilibrium left
Adding $H_2$ changes the equilibrium constant, requiring $I_2$ concentration to decrease

Explanation: When you encounter equilibrium problems involving Le Châtelier's principle, focus on how the system responds to stress by shifting to counteract the change. Adding

H_2

to this equilibrium system creates a stress by increasing the concentration of a reactant. According to Le Châtelier's principle, the equilibrium will shift to relieve this stress by consuming the excess

H_2

. This means the reaction shifts right (toward products), consuming both

H_2

and

I_2

to produce more

HI

. As the reaction proceeds in the forward direction,

I_2

concentration decreases because it's being consumed to form additional

HI

. This perfectly explains the observation and confirms answer B. Looking at the incorrect options: Answer A incorrectly focuses on pressure effects. While adding gas does increase total pressure, this reaction has equal moles of gas on both sides (3 moles total), so pressure changes don't favor either direction. Answer C suggests simple dilution, but this misses the point entirely—we're told

I_2

concentration specifically decreases due to the equilibrium shift, not dilution effects. Answer D incorrectly claims that adding

H_2

increases temperature through collision frequency, which isn't how temperature changes occur in chemical systems, and even if it did, we're not given information about whether this reaction is exothermic or endothermic. Remember: Le Châtelier's principle always involves the system shifting to oppose the applied stress. When you add a reactant, the equilibrium shifts toward products, consuming other reactants in the process.

Question 6

Consider the equilibrium: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ . A chemist measures the equilibrium concentrations at 250°C and finds $[PCl_5] = 0.10 \text{ M}$ , $[PCl_3] = 0.050 \text{ M}$ , and $[Cl_2] = 0.050 \text{ M}$ . If the chemist then adds enough $Cl_2$ to increase its concentration to 0.10 M, which prediction is most consistent with Le Châtelier's principle?

$[PCl_5]$ will increase to exactly 0.125 M when the new equilibrium is established
$[PCl_5]$ will increase to some value greater than 0.10 M when the new equilibrium is established (correct answer)
$[PCl_5]$ will decrease because adding $Cl_2$ shifts the equilibrium toward more products
$[PCl_5]$ will remain at 0.10 M because the equilibrium constant doesn't change
$[PCl_5]$ will change unpredictably because the system is no longer at equilibrium

Explanation: When you encounter equilibrium problems involving concentration changes, Le Châtelier's principle is your roadmap: the system will shift to counteract any disturbance you impose on it. Let's first establish the baseline. The initial equilibrium constant

K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.050)(0.050)}{0.10} = 0.025

. This value remains constant at constant temperature. When you suddenly increase

[Cl_2]

from 0.050 M to 0.10 M, you're adding more product to the system. According to Le Châtelier's principle, the equilibrium shifts left (toward reactants) to consume some of this excess

Cl_2

. This shift converts some

PCl_3

and

Cl_2

back into

PCl_5

, increasing

[PCl_5]

above its original 0.10 M value. Answer A incorrectly assumes you can calculate an exact final concentration without solving the complete ICE table equilibrium problem. The 0.125 M value has no theoretical basis. Answer C reflects a fundamental misunderstanding of Le Châtelier's principle. Adding a product shifts equilibrium toward reactants, not toward more products. Answer D ignores that while

K_c

stays constant, the individual concentrations must adjust to maintain this constant ratio after the disturbance. Study tip: Remember that Le Châtelier's principle predicts the direction of shift, but calculating exact final concentrations requires setting up an ICE table with the equilibrium constant. For qualitative predictions, focus on which direction counteracts your imposed change.

Question 7

For the equilibrium $CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$ with $\Delta H = -206 \text{ kJ/mol}$ , an industrial chemist wants to maximize $CH_4$ production. The chemist considers three changes: (I) increasing temperature, (II) increasing pressure, (III) adding a catalyst. Which combination of changes will achieve the desired goal?

I and II only, because both shift equilibrium toward products
II and III only, because high pressure favors fewer gas molecules and catalysts speed up product formation
II only, because pressure increases favor the side with fewer gas molecules (correct answer)
I and III only, because temperature and catalysts both increase reaction rates
III only, because only catalysts increase both forward and reverse reaction rates equally

Explanation: When you encounter equilibrium problems involving industrial production, apply Le Chatelier's principle systematically to each proposed change, considering both the reaction stoichiometry and thermodynamics. Let's analyze each change for maximizing

CH_4

production. For temperature: Since

\Delta H = -206 \text{ kJ/mol}

, this reaction is highly exothermic. Le Chatelier's principle tells us that increasing temperature shifts equilibrium toward the endothermic direction (reactants), decreasing

CH_4

yield. For pressure: Count the gas molecules on each side. Reactants have 4 moles of gas

(1 CO + 3 H_2)

, while products have 2 moles

(1 CH_4 + 1 H_2O)

. Higher pressure favors the side with fewer gas molecules, so increasing pressure shifts equilibrium toward products, increasing

CH_4

production. For catalysts: Catalysts increase reaction rates equally in both directions but don't shift equilibrium position—they help reach equilibrium faster without changing the final concentrations. Option A is wrong because increasing temperature actually decreases

CH_4

production for this exothermic reaction. Option B incorrectly assumes catalysts shift equilibrium toward products—they only affect reaction speed. Option D makes the same catalyst error and incorrectly suggests temperature increases help, when higher temperatures reduce yield for exothermic reactions. Only increasing pressure (option C) will maximize

CH_4

production by shifting equilibrium toward the side with fewer gas molecules. Study tip: For industrial equilibrium problems, always check: Is the reaction exothermic or endothermic? How many gas molecules are on each side? Remember that catalysts affect kinetics, not equilibrium position.

Question 8

A chemist studies the equilibrium $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ in a closed container. When the chemist adds more solid $CaCO_3$ to the container, the equilibrium position does not change. A student suggests this violates Le Châtelier's principle because adding more reactant should shift the equilibrium toward products. Which statement best addresses the student's concern?

The student is correct; Le Châtelier's principle predicts the equilibrium should shift right when more $CaCO_3$ is added
The student is incorrect because Le Châtelier's principle only applies to gas-phase reactions, not solid-gas equilibria
The student is incorrect because the concentration of pure solids doesn't appear in the equilibrium expression and remains constant (correct answer)
The student is correct, but the shift is too small to detect experimentally in solid-gas systems
The student is incorrect because adding solids changes the total pressure, which counteracts any equilibrium shift

Explanation: When you encounter equilibrium problems involving solids and gases, remember that Le Châtelier's principle depends on changes in concentration, and pure solids have constant concentration regardless of the amount present. The key insight is understanding what appears in the equilibrium expression. For this reaction, the equilibrium constant expression is

K = [CO_2]

because the concentrations of pure solids (

CaCO_3

and

CaO

) are constants that get incorporated into the equilibrium constant itself. Adding more solid

CaCO_3

doesn't change its "concentration" in the thermodynamic sense—there's still the same amount of

CaCO_3

available at the surface for reaction. Therefore, no shift occurs, and Le Châtelier's principle is not violated. Option A incorrectly assumes that adding solid reactant should shift equilibrium like adding a dissolved or gaseous reactant would. Option B wrongly limits Le Châtelier's principle to gas-phase reactions—the principle applies to all equilibria, but you must correctly identify what can actually change concentration. Option D suggests the shift is too small to detect, but there is genuinely no shift because the driving force (concentration change) doesn't exist for pure solids. The correct answer is C because it identifies that pure solid concentrations are constant and don't appear as variables in equilibrium expressions. Study tip: When analyzing equilibrium shifts, always write out the equilibrium expression first. Only species whose concentrations can actually change (gases, aqueous solutions) will cause shifts when their amounts are altered.

Question 9

Consider the equilibrium: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$ . The $FeSCN^{2+}$ complex is deep red in color. When $NaOH$ is added to this equilibrium mixture, the red color fades significantly. Given that $Fe^{3+}$ forms a precipitate with $OH^-$ , which explanation best accounts for the color change?

$NaOH$ neutralizes $FeSCN^{2+}$ , destroying the colored complex directly
$NaOH$ reacts with $SCN^-$ to form a colorless compound, shifting equilibrium left
$NaOH$ removes $Fe^{3+}$ by precipitation, shifting the equilibrium left according to Le Châtelier's principle (correct answer)
$NaOH$ increases the pH, which destabilizes the $FeSCN^{2+}$ complex thermodynamically
$NaOH$ dilutes the solution, decreasing all concentrations and shifting equilibrium toward reactants

Explanation: When you encounter equilibrium problems involving color changes and added reagents, think about Le Châtelier's principle: how does the added substance affect the concentration of species in the equilibrium? In this equilibrium, the deep red color comes from

FeSCN^{2+}

. When

NaOH

is added, the

OH^-

ions react with

Fe^{3+}

to form

Fe(OH)_3

precipitate:

Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3(s)

. This precipitation removes

Fe^{3+}

from solution, decreasing its concentration. According to Le Châtelier's principle, the equilibrium shifts left to replace the consumed

Fe^{3+}

, which breaks down

FeSCN^{2+}

complexes and causes the red color to fade. This makes option C correct. Option A is wrong because

NaOH

doesn't directly neutralize or react with the

FeSCN^{2+}

complex—it's a basic compound, not an acid-base reaction with the complex. Option B incorrectly suggests

NaOH

reacts with

SCN^-

, but

SCN^-

(thiocyanate) doesn't react with hydroxide ions under these conditions. Option D mentions pH destabilization, but this is vague and incorrect—the

FeSCN^{2+}

complex itself isn't particularly pH-sensitive; the real issue is the removal of

Fe^{3+}

. Remember: when analyzing equilibrium shifts, always identify which component is being removed or added, then predict the direction of shift needed to counteract that change. Precipitation reactions are common ways to remove ionic species from equilibrium systems.

Question 10

A chemist studies the equilibrium $H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)$ and finds that at 500 K, the equilibrium lies far to the right ( $K_{eq} = 7.2 \times 10^4$ ). When the temperature is increased to 700 K, the equilibrium constant decreases to $K_{eq} = 3.8 \times 10^2$ . If a student heats an equilibrium mixture from 500 K to 700 K, which change will be observed?

The concentration of $HBr$ will increase because higher temperature increases reaction rates
The concentration of $HBr$ will decrease because the equilibrium shifts toward reactants (correct answer)
The concentration of $HBr$ will remain constant because the equilibrium constant is always large
The concentration of $H_2$ will decrease because increasing temperature favors bond formation
No change will occur because changing temperature doesn't affect equilibrium position

Explanation: When you encounter equilibrium problems involving temperature changes, focus on how the equilibrium constant changes and what that reveals about the reaction's thermodynamics. The key insight here is that

K_{eq}

decreases dramatically as temperature increases (from

7.2 \times 10^4

at 500 K to

3.8 \times 10^2

at 700 K). Since a smaller equilibrium constant means the equilibrium position shifts toward reactants, heating the mixture will decrease

HBr

concentration and increase

H_2

and

Br_2

concentrations. This decrease in

K_{eq}

with increasing temperature also tells us this is an exothermic reaction—heat is a "product" that gets relieved when temperature increases. Option A incorrectly confuses reaction rates with equilibrium position. While higher temperatures do increase reaction rates, this doesn't determine which direction the equilibrium shifts. Option C misses the fundamental point that equilibrium constants are temperature-dependent, not constant values. The magnitude of

K_{eq}

doesn't determine whether it changes with temperature. Option D contradicts the data—if

H_2

concentration were decreasing,

K_{eq}

would be increasing, not decreasing as observed. Therefore, B correctly identifies that

HBr

concentration decreases because the equilibrium shifts toward reactants when heated. Remember: when

K_{eq}

decreases with increasing temperature, the reaction is exothermic, and heating shifts the equilibrium toward reactants. When

K_{eq}

increases with temperature, the reaction is endothermic, and heating favors products.

Question 11

A student observes that for the equilibrium $COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$ , adding an inert gas at constant pressure causes the equilibrium to shift right. The student is surprised because they expected inert gases to have no effect on equilibrium. Which explanation best accounts for this observation?

The inert gas reacts slightly with $COCl_2$ , removing it from the equilibrium
Adding gas at constant pressure requires the volume to increase, effectively diluting all species (correct answer)
The inert gas acts as a catalyst, speeding up the forward reaction more than the reverse
The inert gas increases the temperature through molecular collisions, favoring the endothermic direction
The observation is incorrect; inert gases never affect equilibrium position under any conditions

Explanation: When you encounter questions about how inert gases affect chemical equilibrium, the key is understanding Le Châtelier's principle and how pressure and volume changes influence gas-phase reactions. The correct explanation is B. When you add an inert gas at constant pressure, the container must expand to accommodate the additional gas molecules while maintaining the same pressure. This volume increase effectively dilutes all the reacting species, lowering their partial pressures. According to Le Châtelier's principle, the equilibrium responds by shifting toward the side with more gas molecules to counteract this change. Since

COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)

has one molecule on the left and two on the right, the equilibrium shifts right. Choice A is wrong because inert gases are chemically unreactive by definition—they don't participate in chemical reactions. Choice C incorrectly suggests inert gases can catalyze reactions, but catalysts change reaction rates without affecting equilibrium position, and inert gases don't catalyze anything. Choice D assumes adding inert gas increases temperature, but the question doesn't specify this condition, and the effect described occurs purely due to pressure-volume relationships. Remember this distinction: adding inert gas at constant volume has no effect on equilibrium (partial pressures unchanged), but adding inert gas at constant pressure shifts equilibrium toward the side with more gas molecules due to the resulting volume increase and dilution effect.

Question 12

Consider the equilibrium: $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ with $\Delta H = +53 \text{ kJ/mol}$ . A chemist wants to determine whether a mixture is at equilibrium by applying a small stress and observing the response. The chemist adds a tiny amount of $H_2$ and observes that $[HI]$ increases slightly. What can the chemist conclude about the original state of the system?

The system was at equilibrium, and Le Châtelier's principle explains the $HI$ increase (correct answer)
The system was not at equilibrium, with $Q > K$ before $H_2$ addition
The system was not at equilibrium, with $Q < K$ before $H_2$ addition
The system was at equilibrium, but the $H_2$ addition changed the equilibrium constant
The observation is inconsistent with Le Châtelier's principle and indicates experimental error

Explanation: When you encounter equilibrium problems involving stress tests, you're dealing with Le Châtelier's principle and the distinction between systems at equilibrium versus those approaching equilibrium. Let's analyze what happened: adding

H_2

caused

[HI]

to increase. If the system was at equilibrium before the addition, Le Châtelier's principle predicts exactly this response. Adding

H_2

(a product) shifts the equilibrium left toward reactants, increasing

[HI]

. This is the expected behavior for an equilibrium system under stress. Answer A correctly identifies this scenario - the system was at equilibrium, and Le Châtelier's principle explains the observed increase in

[HI]

. Answer B suggests

Q > K

initially, meaning the system had excess products and would naturally shift left to reach equilibrium. However, if this were true, adding more

H_2

would push the system even further from equilibrium in the wrong direction, making the leftward shift more dramatic than Le Châtelier's principle alone would predict. Answer C proposes

Q < K

, indicating excess reactants. A system in this state would naturally shift right toward products, so adding

H_2

would create competing effects that wouldn't simply result in increased

[HI]

. Answer D incorrectly suggests the equilibrium constant changed. The equilibrium constant

K

only changes with temperature, not with concentration changes. Study tip: Remember that Le Châtelier's principle only applies to systems already at equilibrium. If you observe the predicted response to a small stress, the system was likely at equilibrium initially.

Question 13

A student studies the equilibrium $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$ in a closed container at 25°C. The student observes that when the volume of the container is doubled at constant temperature, the partial pressure of $NH_3$ decreases but then gradually returns to a value higher than the original pressure. Which sequence of effects best explains this observation?

Volume increase → pressure decrease → equilibrium shift left → $NH_3$ pressure increases above original
Volume increase → pressure decrease → equilibrium shift right → $NH_3$ pressure increases above original (correct answer)
Volume increase → no immediate change → equilibrium shift right → $NH_3$ pressure equals original
Volume increase → pressure decrease → no equilibrium shift → $NH_3$ pressure returns to original
Volume increase → concentration decrease → equilibrium shift right → $NH_3$ pressure increases above original

Explanation: When you encounter equilibrium problems involving volume changes, focus on how Le Châtelier's principle applies to gaseous systems. This reaction involves a solid decomposing into two gases, so volume changes will affect the gas-phase equilibrium. When the container volume doubles at constant temperature, the immediate effect is that both

NH_3

and

H_2S

partial pressures drop by half due to the increased space (pressure decrease). However, this disturbs the equilibrium. Since the reaction produces 2 moles of gas from 1 mole of solid, increasing volume favors the forward reaction (more gas particles). The equilibrium shifts right, causing more

NH_4HS

to decompose and produce additional

NH_3

and

H_2S

. The final

NH_3

pressure ends up higher than the original because the system has more total gas molecules in the larger volume. Option A incorrectly suggests the equilibrium shifts left. A leftward shift would decrease gas production, which contradicts Le Châtelier's principle when volume increases favor more gas formation. Option C wrongly states there's no immediate pressure change when volume doubles - this violates basic gas laws. Option D claims no equilibrium shift occurs, but volume changes always affect equilibria involving different numbers of gas molecules on each side. Remember: for gas-phase equilibria, increasing volume always favors the side with more gas molecules. Count the gas particles on each side of the equation to predict the shift direction.

Question 14

Consider the equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$ with $\Delta H = -58 \text{ kJ/mol}$ . A student places this system in a water bath and observes that when ice is added to the bath, the gas mixture becomes less brown over time. However, when the ice melts and the bath warms back to room temperature, the mixture becomes brown again. Which statement best explains this reversible color change?

Cooling shifts equilibrium toward $N_2O_4$ (colorless), warming shifts back toward $NO_2$ (brown) (correct answer)
Cooling increases the solubility of $NO_2$ in water, removing it from the gas phase
Cooling decreases molecular motion, allowing $NO_2$ molecules to associate into colorless dimers
Cooling shifts equilibrium toward $NO_2$ formation, but the lower temperature makes brown color less visible
The color change is due to thermal expansion and contraction of the container, not chemical equilibrium

Explanation: When you encounter a chemical equilibrium problem involving temperature changes, think about Le Châtelier's principle and how temperature affects the position of equilibrium based on whether the reaction is exothermic or endothermic. This reaction has

\Delta H = -58 \text{ kJ/mol}

, meaning it's exothermic in the forward direction (forming

N_2O_4

). According to Le Châtelier's principle, decreasing temperature favors the exothermic direction, while increasing temperature favors the endothermic direction. When ice cools the system, equilibrium shifts toward

N_2O_4

formation. Since

NO_2

is brown and

N_2O_4

is colorless, this shift reduces the brown color. When the bath warms back up, equilibrium shifts back toward

NO_2

, restoring the brown color. Choice A correctly identifies this temperature-equilibrium relationship. Choice B incorrectly suggests dissolution in water, but the problem involves gas-phase equilibrium, not solubility. Choice C mentions molecular association into dimers, which is actually what's happening chemically, but fails to recognize this as an equilibrium shift driven by temperature change according to Le Châtelier's principle. Choice D has the equilibrium direction backwards—cooling favors

N_2O_4

formation, not

NO_2

formation. Study tip: For equilibrium problems involving temperature, always identify whether

\Delta H

is positive or negative first. Then remember: cooling favors the exothermic direction (negative

\Delta H

), while heating favors the endothermic direction (positive

\Delta H

). This pattern appears frequently on chemistry exams.

Question 15

A chemist studies the equilibrium $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$ at high temperature. The chemist finds that increasing the amount of solid carbon in the container does not change the equilibrium concentrations of $CO_2$ and $CO$ . However, a student argues that this contradicts Le Châtelier's principle because adding more reactant should shift the equilibrium toward products. Which response best addresses the student's argument?

The student is correct; the chemist must have made an experimental error in measuring the gas concentrations
The student is incorrect because Le Châtelier's principle only applies when all reactants and products are in the same phase
The student is incorrect because the activity of pure solid carbon remains constant regardless of the amount present (correct answer)
The student is correct, but the equilibrium shift is too small to detect when dealing with solid-gas equilibria
The student is incorrect because carbon acts as a catalyst in this reaction, not a true reactant

Explanation: When you encounter equilibrium problems involving solids and gases, remember that the physical state of reactants and products significantly affects how Le Châtelier's principle applies. The key insight here is understanding chemical activity. For pure solids and liquids, the activity (effective concentration) remains constant regardless of the amount present. In this equilibrium, solid carbon

C(s)

has a constant activity of 1, so adding more carbon doesn't change the "effective concentration" that drives the equilibrium position. The equilibrium expression is

K = \frac{[CO]^2}{[CO_2]}

- notice that

[C]

doesn't appear because its activity is constant. Therefore, adding more solid carbon cannot shift the equilibrium, and the gas concentrations remain unchanged. Option A incorrectly assumes the student's reasoning is valid and blames experimental error. The chemist's observations are entirely correct. Option B makes a false claim about Le Châtelier's principle - the principle does apply to multi-phase systems, but you must consider how each phase behaves. Option D accepts the student's flawed premise and incorrectly suggests the effect is just too small to measure, when in reality there is no effect at all. Option C correctly identifies that solid carbon's activity remains constant regardless of amount present, which is why adding more doesn't affect the equilibrium. Study tip: For equilibrium problems, always check the phases involved. Pure solids and liquids have constant activities, so changing their amounts won't shift equilibria - only gases and aqueous solutions have variable concentrations that can drive equilibrium shifts.

Question 16

For the equilibrium $PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g)$ , a student observes that at 250°C, increasing pressure favors $PCl_5$ formation. The student then heats the system to 400°C and finds that increasing pressure still favors $PCl_5$ formation, but to a lesser extent. If the student continues heating to 600°C, which observation would be most surprising?

Increasing pressure still favors $PCl_5$ formation, but even less than at 400°C
Increasing pressure has no effect on the equilibrium position
Increasing pressure favors $PCl_3$ and $Cl_2$ formation instead of $PCl_5$ (correct answer)
The equilibrium constant decreases compared to its value at 400°C
The reaction reaches equilibrium more slowly due to higher temperature

Explanation: This question tests your understanding of Le Châtelier's principle and how temperature affects equilibrium behavior. When analyzing gas-phase equilibria, you need to consider both the effect of pressure (which depends on the number of gas molecules) and how temperature changes the equilibrium constant. Looking at the reaction

PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g)

, there are 2 moles of gas on the left and 1 mole on the right. According to Le Châtelier's principle, increasing pressure should always favor the side with fewer gas molecules - in this case,

PCl_5

formation. This fundamental relationship between pressure and equilibrium position cannot reverse, regardless of temperature. The student's observations at 250°C and 400°C show the expected behavior: pressure favors

PCl_5

, but the effect weakens as temperature increases. This suggests the forward reaction is exothermic, so higher temperatures favor the reverse reaction. Choice A correctly predicts continued weakening of the pressure effect at higher temperature. Choice B is possible if the equilibrium constant becomes very unfavorable, making pressure changes less noticeable. Choice D makes perfect sense since this appears to be an exothermic reaction where higher temperatures decrease the equilibrium constant. Choice C is impossible because it violates Le Châtelier's principle. The stoichiometry doesn't change with temperature - pressure must always favor the side with fewer gas molecules. Study tip: Remember that temperature affects the equilibrium constant, but pressure effects depend solely on gas molecule counts, which don't change with temperature.

Question 17

A student investigates the equilibrium $2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)$ and calculates that $Q = 0.025$ for a particular mixture. The student then adds a catalyst to speed up the reaction and recalculates $Q = 0.025$ for the same mixture. The student concludes that catalysts don't affect equilibrium position. However, the student's lab partner suggests measuring Q again after waiting longer. Why might this suggestion be important?

Catalysts gradually change the equilibrium constant over time, affecting Q
The mixture wasn't at equilibrium initially, and the catalyst helps it reach equilibrium faster (correct answer)
Catalysts temporarily change Q values before returning them to original values
The catalyst might decompose over time, changing its effect on the equilibrium
Q calculations become more accurate after catalysts have been present longer

Explanation: This question tests your understanding of the relationship between reaction quotients, equilibrium, and catalysts. The key insight is distinguishing between a system that has reached equilibrium versus one that's still approaching it. The reaction quotient

Q

represents the ratio of products to reactants at any given moment, while the equilibrium constant

K

represents this ratio when the system has reached equilibrium. If

Q = K

, the system is at equilibrium. If

Q \neq K

, the system will shift toward equilibrium over time. Answer B is correct because the student likely measured

Q

for a mixture that wasn't yet at equilibrium. When they added the catalyst and immediately remeasured,

Q

was still the same because catalysts don't change equilibrium positions—they only speed up the rate at which equilibrium is reached. The lab partner's suggestion to wait longer is crucial because it allows time for the reaction to actually reach equilibrium. With the catalyst present, this happens faster, so after waiting, they would observe

Q

changing toward its equilibrium value. Answer A is wrong because catalysts never change equilibrium constants—this is a fundamental principle. Answer C incorrectly suggests catalysts temporarily alter

Q

values, but catalysts only affect reaction rates, not the instantaneous concentration ratios. Answer D focuses on catalyst decomposition, which isn't relevant to understanding equilibrium concepts and doesn't address why waiting would reveal important information about the equilibrium position. Remember: catalysts speed up both forward and reverse reactions equally, helping systems reach equilibrium faster without changing where that equilibrium lies.

Question 18

For SO₂(g) + O₂(g) ⇌ SO₃(g), what happens to SO₂(g) as equilibrium shifts right?

SO₂(g) is consumed (correct answer)
SO₂(g) is produced
SO₂(g) stays constant instantly
SO₂(g) becomes a catalyst

Explanation: This question tests introductory college chemistry skills focused on understanding Le Chatelier's Principle and its impact on equilibrium. Le Chatelier's Principle states that a system at equilibrium will adjust to counteract any imposed changes, such as concentration, temperature, or pressure. As the equilibrium SO₂(g) + O₂(g) ⇌ SO₃(g) shifts right, reactants like SO₂(g) are consumed to form products. Choice A is correct as it states SO₂(g) is consumed during a right shift. Choice B is incorrect because a right shift consumes, not produces, SO₂. To help students, link shift directions to consumption and production. Use balanced equations to track species changes.

Question 19

For SO₂(g) + O₂(g) ⇌ SO₃(g), what happens when SO₂(g) and O₂(g) are both added?

Shifts right, forming more SO₃(g) (correct answer)
Shifts left, forming more SO₂(g)
No shift; only K changes
Shifts left, forming more SO₃(g)

Explanation: This question tests introductory college chemistry skills focused on understanding Le Chatelier's Principle and its impact on equilibrium. Le Chatelier's Principle states that a system at equilibrium will adjust to counteract any imposed changes, such as concentration, temperature, or pressure. Adding both SO₂(g) and O₂(g) to SO₂(g) + O₂(g) ⇌ SO₃(g) typically decreases Q, causing a right shift. Choice A is correct as it describes the right shift, forming more SO₃(g). Choice B is incorrect because the combined addition favors products. To help students, discuss how multiple changes interact via Q. Practice with varying addition amounts to see effects.

Question 20

For SO₂(g) + O₂(g) ⇌ SO₃(g), which change decreases SO₃(g) at the new equilibrium?

Add O₂(g)
Remove SO₃(g)
Add SO₃(g)
Remove SO₂(g) (correct answer)

Explanation: This question tests introductory college chemistry skills focused on understanding Le Chatelier's Principle and its impact on equilibrium. Le Chatelier's Principle states that a system at equilibrium will adjust to counteract any imposed changes, such as concentration, temperature, or pressure. Removing SO₂(g) from SO₂(g) + O₂(g) ⇌ SO₃(g) shifts left, decreasing SO₃(g) at the new equilibrium. Choice D is correct as it results in net consumption of SO₃(g). Choice B is incorrect because removing SO₃(g) shifts right, producing some back. To help students, emphasize distinguishing direct changes from shift-induced effects. Practice with scenarios where multiple factors could influence outcomes.