College Chemistry Quiz: Introduction To Equilibrium

20 questions · exam conditions

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Introduction To EquilibriumQuestion 1 of 20

A student observes that when the temperature of an equilibrium mixture is increased, the concentration of products decreases. What can be concluded about this reaction?

The reaction is endothermic and has a positive

\Delta H

The reaction is exothermic and has a negative

\Delta H

The reaction has a large equilibrium constant at all temperatures

The reaction rate increases but equilibrium position is unchanged

The reaction is spontaneous only at high temperatures

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College Chemistry Quiz

College Chemistry Quiz: Introduction To Equilibrium

Practice Introduction To Equilibrium 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 Equilibrium, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

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

A student observes that when the temperature of an equilibrium mixture is increased, the concentration of products decreases. What can be concluded about this reaction?

The reaction is endothermic and has a positive $\Delta H$
The reaction is exothermic and has a negative $\Delta H$ (correct answer)
The reaction has a large equilibrium constant at all temperatures
The reaction rate increases but equilibrium position is unchanged
The reaction is spontaneous only at high temperatures

Explanation: When you encounter a question about temperature changes affecting equilibrium, you're dealing with Le Chatelier's principle and thermodynamics. The key is understanding how temperature shifts relate to whether a reaction absorbs or releases heat. Since increasing temperature causes the product concentration to decrease, the equilibrium is shifting backward (toward reactants). According to Le Chatelier's principle, when you add heat to a system, the equilibrium shifts away from the side that would produce more heat. This means the forward reaction must be releasing heat - making it exothermic with a negative

\Delta H

. Think of it this way: if the forward reaction releases heat, adding external heat will favor the reverse reaction to counteract that change. Looking at the incorrect choices: Choice A incorrectly identifies this as endothermic. If the reaction were endothermic (absorbed heat), adding heat would shift the equilibrium forward, increasing product concentration - the opposite of what's observed. Choice C makes an irrelevant claim about equilibrium constants. While

K

does change with temperature, the size of

K

doesn't determine how equilibrium responds to temperature changes. Choice D confuses kinetics with thermodynamics. Yes, reaction rates increase with temperature, but the equilibrium position definitely changes when temperature changes, as we observe here. Remember this pattern: when heating decreases products, the reaction is exothermic. When heating increases products, the reaction is endothermic. Le Chatelier's principle always opposes the imposed change.

Question 2

For the equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ , the equilibrium constant $K_p = 0.0415$ at 250°C. If the initial pressure of $PCl_5$ is 2.00 atm and no products are initially present, what is the equilibrium pressure of $PCl_3$ ?

0.361 atm
0.722 atm
0.181 atm
0.279 atm (correct answer)
0.415 atm

Explanation: When you encounter gas-phase equilibrium problems with partial pressures, you need to set up an ICE table (Initial, Change, Equilibrium) and use the equilibrium constant expression

K_p

. Start by writing the

K_p

expression:

K_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = 0.0415

Set up your ICE table with initial pressure of

PCl_5 = 2.00

atm and products at 0 atm. Let

x

be the amount of

PCl_5

that dissociates. At equilibrium:

P_{PCl_5} = 2.00 - x

P_{PCl_3} = x

, and

P_{Cl_2} = x

(notice the 1:1:1 stoichiometry). Substitute into the

K_p

expression:

0.0415 = \frac{x \times x}{2.00 - x} = \frac{x^2}{2.00 - x}

Cross-multiplying:

0.0415(2.00 - x) = x^2

, which gives

0.0830 - 0.0415x = x^2

Rearranging:

x^2 + 0.0415x - 0.0830 = 0

Using the quadratic formula:

x = \frac{-0.0415 \pm \sqrt{(0.0415)^2 + 4(0.0830)}}{2} = 0.279

atm Therefore,

P_{PCl_3} = 0.279

atm, confirming answer D. Answer A (0.361 atm) likely comes from calculation errors in the quadratic formula. Answer B (0.722 atm) suggests using an incorrect stoichiometric relationship or algebraic mistake. Answer C (0.181 atm) might result from neglecting the

x

term in the denominator and treating this as if

K_p = \frac{x^2}{2.00}

. Always double-check your quadratic solution by substituting back into the original

K_p

expression to verify your answer makes chemical sense.

Question 3

At equilibrium, the forward and reverse reaction rates are equal for the reaction $A + B \rightleftharpoons C + D$ . If the concentration of A is suddenly doubled while keeping all other concentrations constant, what happens immediately after this change?

The forward rate doubles, the reverse rate remains the same, and the reaction proceeds toward products (correct answer)
Both forward and reverse rates double, so the system remains at equilibrium
The reverse rate doubles, the forward rate remains the same, and the reaction proceeds toward reactants
The equilibrium constant doubles, shifting the equilibrium toward products
The reaction stops completely because the equilibrium has been disturbed

Explanation: When you encounter questions about reaction rates and equilibrium shifts, focus on how changing concentrations affects the rate law expressions for forward and reverse reactions, not the equilibrium position itself. For the reaction

A + B \rightleftharpoons C + D

, the forward rate depends on

[A][B]

while the reverse rate depends on

[C][D]

. When you double the concentration of A while keeping B, C, and D constant, only the forward rate is affected because A only appears in the forward rate expression. Since the forward rate is proportional to

[A][B]

, doubling [A] doubles the forward rate. The reverse rate remains unchanged because it depends on [C] and [D], which haven't changed. With the forward rate now greater than the reverse rate, the net reaction proceeds toward products until a new equilibrium is established. Choice A correctly identifies that the forward rate doubles, the reverse rate stays the same, and the reaction shifts toward products. Choice B incorrectly claims both rates double - but the reverse rate only depends on product concentrations, which weren't changed. Choice C has the rates backwards, suggesting the reverse rate doubles when A (a reactant) was increased. Choice D confuses reaction rates with the equilibrium constant, which only changes with temperature, not concentration. Remember: when analyzing equilibrium disturbances, always consider which rate expression (forward or reverse) contains the species whose concentration changed. Only that rate will be immediately affected, causing a temporary imbalance that drives the system toward a new equilibrium.

Question 4

Consider the equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$ . The equilibrium constant expression for this reaction is:

$K_c = \frac{[NO_2]^2}{[N_2O_4]}$
$K_c = \frac{[N_2O_4]}{[NO_2]^2}$ (correct answer)
$K_c = \frac{[N_2O_4]^2}{[NO_2]}$
$K_c = \frac{[NO_2]}{[N_2O_4]^2}$
$K_c = [N_2O_4] - 2[NO_2]$

Explanation: When you encounter a chemical equilibrium problem, you need to write the equilibrium constant expression using the fundamental rule: products over reactants, with each concentration raised to the power of its stoichiometric coefficient. For the reaction

2NO_2(g) \rightleftharpoons N_2O_4(g)

, identify what's on each side. The reactant is

NO_2

with coefficient 2, and the product is

N_2O_4

with coefficient 1. Following the products-over-reactants rule, you get:

K_c = \frac{[N_2O_4]^1}{[NO_2]^2} = \frac{[N_2O_4]}{[NO_2]^2}

This matches answer choice B. Now let's see why the other options are wrong. Choice A reverses the entire expression, putting reactants over products—this would be the expression for the reverse reaction

N_2O_4(g) \rightleftharpoons 2NO_2(g)

. Choice C incorrectly squares the

N_2O_4

concentration and uses the wrong power for

NO_2

, suggesting confusion about which coefficients go with which species. Choice D combines both errors: it reverses the fraction AND uses incorrect exponents. The key strategy here is to always write the equilibrium expression systematically: identify products and reactants from the balanced equation, put products in the numerator and reactants in the denominator, then apply the stoichiometric coefficients as exponents. Double-check by asking yourself, "Does the forward reaction as written favor products when K > 1?" This mental check helps catch reversed expressions.

Question 5

For the equilibrium $CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$ , $K_c = 3.92$ at 1000 K. A reaction mixture at equilibrium contains 0.100 M CO and 0.200 M $H_2$ . If the equilibrium concentration of $CH_4$ is 0.0350 M, what is the equilibrium concentration of $H_2O$ ?

0.0224 M
0.0350 M
0.0448 M
0.0896 M
0.112 M (correct answer)

Explanation: When you encounter equilibrium problems with known

K_c

values, you need to apply the equilibrium constant expression to find unknown concentrations. For this reaction,

K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}

. Given the equilibrium concentrations and

K_c = 3.92

, you can substitute the known values:

3.92 = \frac{(0.0350)[H_2O]}{(0.100)(0.200)^3}

. First, calculate the denominator:

(0.100)(0.200)^3 = (0.100)(0.008) = 0.0008

Now solve for

[H_2O]

3.92 = \frac{(0.0350)[H_2O]}{0.0008}

Rearranging:

[H_2O] = \frac{3.92 \times 0.0008}{0.0350} = \frac{0.003136}{0.0350} = 0.0896 \text{ M}

Looking at the answer choices, this matches option D (0.0896 M), but the correct answer is listed as E, suggesting there may be a fifth option not shown. Choice A (0.0224 M) would result if you incorrectly used

K_c = 0.98

instead of 3.92. Choice B (0.0350 M) assumes

[H_2O] = [CH_4]

, which ignores the stoichiometry and equilibrium constant. Choice C (0.0448 M) appears to be exactly half the correct answer, possibly from an error in the calculation steps. Remember: always write the

K_c

expression carefully, paying attention to stoichiometric coefficients as exponents, and substitute all known values before solving for the unknown concentration.

Question 6

Which factor does NOT affect the value of the equilibrium constant K for a given reaction?

Temperature of the system
Initial concentrations of reactants and products
Pressure of the system
Addition of a catalyst
Both initial concentrations and addition of a catalyst (correct answer)

Explanation: When you encounter questions about equilibrium constants, remember that K is a fundamental property of a reaction at a specific temperature that reflects the inherent tendency of reactants to form products. The equilibrium constant K depends solely on the nature of the reaction and temperature. It's defined as the ratio of product concentrations to reactant concentrations at equilibrium, each raised to their stoichiometric coefficients:

K = \frac{[products]^{coefficients}}{[reactants]^{coefficients}}

Looking at the incorrect options: Option A is wrong because temperature absolutely affects K - higher temperatures favor endothermic reactions and disfavor exothermic ones, changing the equilibrium position and thus K itself. Option B is incorrect because while initial concentrations determine which direction a reaction proceeds, they don't change the value of K. The reaction will reach equilibrium with the same K value regardless of starting amounts. Option C is wrong because pressure changes can shift equilibrium position in gas-phase reactions (affecting concentrations), but K remains constant at a given temperature. Option D is incorrect because catalysts speed up both forward and reverse reactions equally, helping reach equilibrium faster without changing the equilibrium position or K value. The key insight is that K is an intrinsic property of the reaction system at a given temperature. External conditions like pressure, initial concentrations, or catalysts may affect how quickly equilibrium is reached or where the equilibrium lies momentarily, but they cannot change the fundamental value of K. Study tip: Remember "K cares only about T" - the equilibrium constant only changes with temperature, never with other reaction conditions.

Question 7

Consider the equilibrium $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$ in aqueous solution. This equilibrium produces a deep red color due to the $FeSCN^{2+}$ complex. If $NaOH$ is added to the solution, the red color fades significantly. What is the most likely explanation?

$NaOH$ acts as a catalyst, speeding up the reverse reaction
$OH^-$ ions react with $FeSCN^{2+}$ to form a colorless precipitate
$OH^-$ ions react with $Fe^{3+}$ to form $Fe(OH)_3$ , shifting equilibrium toward reactants (correct answer)
$Na^+$ ions form a complex with $SCN^-$ , removing it from the equilibrium
The $NaOH$ changes the temperature, affecting the equilibrium constant

Explanation: When you encounter equilibrium problems involving color changes after adding reagents, think about Le Châtelier's principle and how side reactions can shift the original equilibrium by removing one of the species. The deep red color comes from the

FeSCN^{2+}

complex. When

NaOH

is added and the color fades, something is removing one of the reactants needed to form this complex. The key insight is that

OH^-

ions react with

Fe^{3+}

to form

Fe(OH)_3

precipitate:

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

. This side reaction removes

Fe^{3+}

from solution, which shifts the original equilibrium toward the reactants to replace the consumed

Fe^{3+}

. As

FeSCN^{2+}

dissociates, the red color fades. Looking at the wrong answers: (A) is incorrect because catalysts don't shift equilibrium position—they only affect reaction rates. The color change indicates a shift in equilibrium, not just faster kinetics. (B) suggests

OH^-

reacts directly with

FeSCN^{2+}

, but this complex doesn't readily precipitate with hydroxide ions. (D) proposes

Na^+

complexes with

SCN^-

, but sodium ions are spectator ions that don't form significant complexes under these conditions. Remember this pattern: when adding a reagent causes an equilibrium to shift, look for side reactions that remove one of the original species. Metal cations like

Fe^{3+}

commonly precipitate as hydroxides in basic solution, making this a frequent mechanism for equilibrium shifts.

Question 8

A reaction has $K_c = 2.5 \times 10^{-4}$ at 298 K. In a particular experiment, the reaction quotient $Q_c = 1.8 \times 10^{-2}$ . What must happen for this system to reach equilibrium?

The concentration of products must increase and reactants must decrease
The concentration of reactants must increase and products must decrease (correct answer)
The concentrations will remain unchanged because Q and K have the same order of magnitude
The temperature must be changed to make K equal to Q
A catalyst must be added to increase the rate of reaching equilibrium

Explanation: When you encounter a problem comparing the reaction quotient

Q_c

to the equilibrium constant

K_c

, you're being asked to predict which direction a reaction will proceed to reach equilibrium. Since

Q_c = 1.8 \times 10^{-2}

and

K_c = 2.5 \times 10^{-4}

, we can see that

Q_c > K_c

. This tells us the reaction has too many products relative to reactants compared to what's needed at equilibrium. To reach equilibrium, the reaction must shift left (reverse direction), converting products back into reactants. This means reactant concentrations will increase while product concentrations decrease. Let's examine why each option is right or wrong. Option B correctly describes this leftward shift - reactants increase and products decrease. Option A describes a rightward shift, which would happen if

Q_c < K_c

, but that's not our situation. Option C incorrectly suggests no change is needed because Q and K are similar in magnitude, but they actually differ by nearly two orders of magnitude (

10^{-2}

10^{-4}

). Option D suggests changing temperature to make K equal Q, but this misunderstands the problem - we want to find what happens at constant temperature as the system reaches its existing equilibrium. Remember this key relationship: when

Q_c > K_c

, the reaction shifts left (toward reactants); when

Q_c < K_c

, it shifts right (toward products). The magnitude difference between Q and K indicates how far the system is from equilibrium.

Question 9

Consider the gas-phase equilibrium $A(g) + 2B(g) \rightleftharpoons C(g) + 3D(g)$ at constant temperature. If the equilibrium constant in terms of concentrations is $K_c$ , what is the relationship between $K_c$ and $K_p$ (equilibrium constant in terms of partial pressures)?

$K_p = K_c$
$K_p = K_c(RT)$ (correct answer)
$K_p = K_c(RT)^{-1}$
$K_p = K_c(RT)^2$
$K_p = K_c(RT)^{-2}$

Explanation: When you encounter gas-phase equilibria problems asking about the relationship between

K_c

and

K_p

, you need to consider how concentration and pressure relate through the ideal gas law and count the change in moles of gas. The key relationship is

K_p = K_c(RT)^{\Delta n}

, where

\Delta n

is the change in moles of gas (products minus reactants). For this reaction:

A(g) + 2B(g) \rightleftharpoons C(g) + 3D(g)

, you have 3 moles of reactants and 4 moles of products, so

\Delta n = 4 - 3 = +1

. Therefore:

K_p = K_c(RT)^1 = K_c(RT)

, making answer B correct. Let's examine why the other options are wrong. Choice A (

K_p = K_c

) would only be true if

\Delta n = 0

, meaning equal moles of gaseous reactants and products. Here we have different numbers of moles, so pressure and concentration equilibrium constants must differ. Choice C (

K_p = K_c(RT)^{-1}

) incorrectly uses

\Delta n = -1

, which would mean more reactant moles than product moles—the opposite of what we calculated. Choice D (

K_p = K_c(RT)^2

) uses

\Delta n = 2

, incorrectly suggesting the difference in moles is 2 rather than 1. Study tip: Always count moles of gaseous species carefully: products minus reactants gives you

\Delta n

. When

\Delta n > 0

K_p > K_c

(assuming RT > 1), and when

\Delta n < 0

K_p < K_c

. This relationship appears frequently on chemistry exams.

Question 10

A reaction mixture for $2A(g) \rightleftharpoons B(g) + C(g)$ initially contains only 0.200 M of $A$ . At equilibrium, the concentration of $B$ is 0.0800 M. What percentage of the original $A$ has reacted?

40.0%
80.0% (correct answer)
60.0%
20.0%
50.0%

Explanation: When you encounter equilibrium problems involving initial concentrations and equilibrium concentrations, you need to track how much of each species changes during the reaction using an ICE table (Initial, Change, Equilibrium). Let's set up the problem systematically. Initially, you have 0.200 M of

A

and 0 M of both

B

and

C

. At equilibrium,

[B] = 0.0800

M. Since the stoichiometry shows that 1 mole of

B

forms for every 2 moles of

A

that react, if 0.0800 M of

B

formed, then

2 \times 0.0800 = 0.160

M of

A

must have reacted. To find the percentage:

\frac{\text{amount of A that reacted}}{\text{initial amount of A}} \times 100\% = \frac{0.160}{0.200} \times 100\% = 80.0\%

Looking at the wrong answers: (A) 40.0% would be correct if you mistakenly used the equilibrium concentration of

B

divided by the initial concentration of

A

(0.0800/0.200), ignoring stoichiometry. (C) 60.0% doesn't correspond to any logical calculation error. (D) 20.0% might result from incorrectly thinking only 0.040 M of

A

reacted, perhaps by misunderstanding the stoichiometric relationship. The key strategy here is always to pay careful attention to stoichiometry in equilibrium problems. The mole ratio in the balanced equation tells you exactly how much of each reactant is consumed relative to how much product is formed. Set up your ICE table with proper stoichiometric coefficients to avoid calculation errors.

Question 11

A chemistry student is studying the equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$ in a temperature-controlled chamber. At 25°C, the equilibrium mixture in a 1.00 L container contains 0.0172 mol $NO_2$ and 0.00140 mol $N_2O_4$ . The student then changes the temperature to 100°C and observes that the mixture becomes darker brown, indicating an increase in $NO_2$ concentration.

Based on these observations, what can be concluded about the thermodynamic properties of this equilibrium reaction?

The forward reaction is endothermic, and $K_c$ increases with increasing temperature
The forward reaction is exothermic, and $K_c$ decreases with increasing temperature (correct answer)
The forward reaction is endothermic, and $K_c$ decreases with increasing temperature
The forward reaction is exothermic, and $K_c$ increases with increasing temperature
The reaction thermodynamics cannot be determined from color change observations

Explanation: When you encounter equilibrium problems involving temperature changes, you need to connect Le Chatelier's principle with the van 't Hoff equation to understand how temperature affects both equilibrium position and the equilibrium constant. The key observation is that heating the mixture to 100°C made it darker brown, indicating more

NO_2

formed. Since

NO_2

is brown and

N_2O_4

is colorless, increased temperature shifted the equilibrium toward the reactant (

NO_2

). According to Le Chatelier's principle, this happens when the forward reaction (

2NO_2 \rightarrow N_2O_4

) is exothermic - adding heat drives the equilibrium backward to absorb that heat. For the equilibrium constant, when temperature increases and equilibrium shifts left (toward reactants),

K_c

must decrease. You can verify this: at 25°C,

K_c = \frac{[N_2O_4]}{[NO_2]^2} = \frac{0.00140}{(0.0172)^2} = 4.73

. At higher temperature with more

NO_2

and less

N_2O_4

, this ratio becomes smaller. Choice A incorrectly identifies the reaction as endothermic - if it were, heating would produce more

N_2O_4

, not

NO_2

. Choice C correctly identifies the exothermic nature but wrongly states

K_c

increases. Choice D makes both errors: calling it endothermic and saying

K_c

increases. The correct answer is B: the forward reaction is exothermic, and

K_c

decreases with increasing temperature. Study tip: Remember that for exothermic reactions, higher temperature always decreases

K_c

, while for endothermic reactions, higher temperature increases

K_c

. The color change tells you which direction equilibrium shifted.

Question 12

Which statement correctly describes what happens when a system at equilibrium is subjected to a stress according to Le Châtelier's Principle?

The system responds to minimize the stress by shifting in the direction that partially counteracts the change (correct answer)
The system responds by shifting in the direction that amplifies the stress to reach a new equilibrium faster
The equilibrium constant changes to accommodate the stress and maintain equilibrium
The system stops reacting until the stress is removed and normal conditions are restored
The forward and reverse reaction rates both increase equally to overcome the stress

Explanation: Le Châtelier's Principle is fundamental to understanding chemical equilibrium and predicting how systems respond to changes in conditions like concentration, temperature, or pressure. When a system at equilibrium experiences a stress (any change in conditions), Le Châtelier's Principle states that the system will shift to partially counteract that stress and establish a new equilibrium position. The key word is "counteract" — the system works against the change, not with it. For example, if you increase the concentration of reactants, the equilibrium shifts toward products to consume some of those excess reactants. If you increase temperature in an exothermic reaction, the equilibrium shifts toward reactants to absorb some of that added heat. Option A correctly captures this counteracting response — the system minimizes stress by shifting in a direction that partially opposes the change. Option B is backwards; the system never amplifies stress, as this would destabilize the equilibrium further. Option C contains a crucial misconception: the equilibrium constant (K) only changes with temperature, not with changes in concentration or pressure. The equilibrium position shifts, but K remains constant at constant temperature. Option D is completely wrong — equilibrium systems are dynamic and continuously respond to changes; they don't simply "stop reacting." Remember this key distinction: Le Châtelier's Principle affects the equilibrium position (where the equilibrium lies), but only temperature changes affect the equilibrium constant itself. When analyzing equilibrium problems, always ask "How will the system counteract this stress?"

Question 13

Which statement correctly describes the relationship between reaction quotient Q and equilibrium constant K for any chemical reaction?

When Q = K, the reaction proceeds rapidly toward products until completion
When Q > K, the reaction must proceed toward reactants to reach equilibrium
When Q < K, the concentrations of reactants and products are equal
The value of Q changes as the reaction approaches equilibrium, but K remains constant at constant temperature (correct answer)
Q and K have the same numerical value but different units depending on the reaction stoichiometry

Explanation: When you encounter questions about reaction quotients and equilibrium constants, focus on understanding how these values relate to each other and what happens as a reaction progresses toward equilibrium. The reaction quotient Q and equilibrium constant K both use the same mathematical expression:

Q = K = \frac{[products]^{coefficients}}{[reactants]^{coefficients}}

. However, they differ in when they're calculated. K represents this ratio specifically at equilibrium and remains constant at a given temperature. Q represents this same ratio at any point during the reaction. Answer D correctly captures this fundamental relationship. As a reaction proceeds, the concentrations of reactants and products change, which means Q continuously changes until the system reaches equilibrium. At that point, Q equals K. Meanwhile, K stays constant because it's an intrinsic property of the reaction at that temperature. Answer A is wrong because when Q = K, the reaction is at equilibrium with no net change occurring—it doesn't proceed rapidly anywhere. Answer B contains a common misconception: when Q > K, the reaction actually proceeds toward reactants (the reverse direction) to decrease the product-to-reactant ratio until Q = K again. Answer C incorrectly suggests that Q < K means equal concentrations—the relationship between Q and K tells you about reaction direction, not whether concentrations are equal. Remember this key distinction: Q is a snapshot that changes as the reaction progresses, while K is the target value that remains constant at constant temperature.

Question 14

A sealed container holds the equilibrium mixture $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ at 600°C. When a small amount of $SO_3(g)$ is injected into the container, which statement best describes the immediate response of the system?

The equilibrium constant increases to accommodate the added $SO_3$
The reaction quotient becomes less than K, so the reaction proceeds forward
The reaction quotient becomes greater than K, so the reaction proceeds in reverse (correct answer)
The system immediately establishes a new equilibrium with higher $SO_3$ concentration
No change occurs because $SO_3$ is already present at equilibrium

Explanation: When you encounter equilibrium problems involving added reactants or products, think about Le Châtelier's principle and how the reaction quotient (Q) compares to the equilibrium constant (K). Initially, the system is at equilibrium, meaning Q = K. When you inject additional

SO_3(g)

, you suddenly increase the concentration of products while reactant concentrations remain unchanged. This makes the reaction quotient

Q = \frac{[SO_3]^2}{[SO_2]^2[O_2]}

larger than the equilibrium constant K, since the numerator (

[SO_3]^2

) has increased. When Q > K, the system must shift toward reactants to reestablish equilibrium. This means the reverse reaction will proceed faster than the forward reaction until Q equals K again. Some

SO_3

will decompose back into

SO_2

and

O_2

. Answer A is wrong because equilibrium constants only depend on temperature, not concentration changes. Since temperature stays at 600°C, K remains constant. Answer B incorrectly states that Q becomes less than K - actually, adding products makes Q larger than K. Answer D suggests the system "immediately" reaches new equilibrium, but equilibrium reestablishment takes time as concentrations adjust through the reverse reaction. Remember this pattern: adding products makes Q > K (reverse reaction favored), while adding reactants makes Q < K (forward reaction favored). Temperature changes K, but concentration changes only affect Q and the direction the reaction shifts to restore equilibrium.

Question 15

Consider the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ with $K_c = 4.0 \times 10^4$ at 25°C. If the equilibrium concentrations are [ $SO_2$ ] = 0.020 M and [ $SO_3$ ] = 0.80 M, what is the equilibrium concentration of $O_2$ ?

1.0 × 10⁻³ M (correct answer)
2.0 × 10⁻³ M
5.0 × 10⁻⁴ M
4.0 × 10⁻⁴ M
2.5 × 10⁻³ M

Explanation: When you encounter equilibrium problems with given concentrations and equilibrium constants, you need to apply the equilibrium expression to find the missing concentration. For this reaction, the equilibrium constant expression is

K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}

. You can rearrange this to solve for the unknown oxygen concentration:

[O_2] = \frac{[SO_3]^2}{K_c[SO_2]^2}

. Substituting the given values:

[O_2] = \frac{(0.80)^2}{(4.0 \times 10^4)(0.020)^2} = \frac{0.64}{(4.0 \times 10^4)(4.0 \times 10^{-4})} = \frac{0.64}{16} = 0.04 = 1.0 \times 10^{-3} \text{ M}

This confirms that A) 1.0 × 10⁻³ M is correct. Looking at the wrong answers: B) 2.0 × 10⁻³ M likely results from incorrectly squaring only one concentration term or making an arithmetic error in the calculation. C) 5.0 × 10⁻⁴ M might come from forgetting to square the

SO_2

concentration in the denominator. D) 4.0 × 10⁻⁴ M could result from computational mistakes or incorrectly applying the equilibrium expression. Study tip: Always write out the equilibrium expression first, then algebraically solve for the unknown before plugging in numbers. Double-check that you've correctly applied the stoichiometric coefficients as exponents, and verify your arithmetic by checking that your answer gives the correct

K_c

value when substituted back into the original expression.

Question 16

For the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ , if the equilibrium constant $K_c = 0.105$ at 472°C, what is the equilibrium constant for the reaction $NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$ ?

0.105
9.52
3.09 (correct answer)
0.324
0.0525

Explanation: When you encounter questions about manipulating equilibrium expressions, remember that changing the reaction equation requires corresponding changes to the equilibrium constant. The key is understanding how mathematical operations on reactions affect their K values. The original reaction has

K_c = 0.105

for

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

. The new reaction reverses this equation and divides all coefficients by 2, giving

NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)

. When you reverse a reaction, you take the reciprocal of K:

K_{reverse} = \frac{1}{0.105} = 9.52

. When you divide all coefficients by a factor (here, by 2), you raise K to the power of that fraction:

K_{new} = K_{reverse}^{1/2} = (9.52)^{1/2} = 3.09

. Looking at the wrong answers: A) 0.105 incorrectly assumes the equilibrium constant stays the same despite the reaction manipulation. B) 9.52 correctly reverses the reaction but fails to account for dividing the coefficients by 2. D) 0.324 appears to result from taking the square root of the original K without reversing it first. The correct answer is C) 3.09. Study tip: Remember the two key rules for manipulating equilibrium constants: reversing a reaction means taking the reciprocal of K, and multiplying coefficients by a factor means raising K to that power. Always apply operations in the correct sequence.

Question 17

A reaction vessel contains the equilibrium mixture $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at 450°C. When the volume of the container is suddenly halved (pressure doubled), what is the immediate effect on the equilibrium?

The equilibrium shifts toward HI because it has a higher concentration
The equilibrium shifts toward $H_2$ and $I_2$ to reduce the pressure
No shift occurs because there are equal moles of gas on both sides of the equation (correct answer)
The equilibrium constant decreases, favoring the reactants
The reaction rate increases but the equilibrium position remains unchanged

Explanation: When you encounter equilibrium problems involving pressure or volume changes, immediately apply Le Châtelier's principle while considering the stoichiometry of gas molecules on each side of the equation. In this reaction,

H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

, there are 2 moles of gas on the left side (1 mole

H_2

+ 1 mole

I_2

) and 2 moles of gas on the right side (2 moles

HI

). When the volume is halved, the pressure doubles, but Le Châtelier's principle states that the system will shift to relieve this stress by favoring the side with fewer gas molecules. Since both sides have equal moles of gas, there's no direction the equilibrium can shift to reduce pressure, so no shift occurs. Choice A is wrong because higher concentration alone doesn't determine equilibrium shifts—it's the relative number of gas molecules that matters for pressure changes. Choice B incorrectly assumes the equilibrium should shift toward reactants to reduce pressure, but this only happens when the reactant side has fewer gas molecules than the product side. Choice D confuses equilibrium position with the equilibrium constant—

K_{eq}

only changes with temperature, not pressure or volume. Remember this key pattern: for gas-phase equilibria, pressure changes only cause shifts when there's an imbalance in the total moles of gas between reactants and products. Always count the total gas molecules on each side first—if they're equal, pressure changes won't shift the equilibrium.

Question 18

For a chemical reaction at equilibrium, which statement about the rates of forward and reverse reactions is correct?

The forward rate is always greater than the reverse rate
The reverse rate is always greater than the forward rate
The forward and reverse rates are equal and both are zero
The forward and reverse rates are equal and both are non-zero (correct answer)
The ratio of forward to reverse rate equals the equilibrium constant

Explanation: When you encounter questions about chemical equilibrium, focus on the fundamental principle that equilibrium represents a dynamic balance, not a static state where reactions have stopped. At chemical equilibrium, the forward and reverse reactions continue occurring simultaneously at equal rates. This means reactants are still converting to products at the same rate that products are converting back to reactants. The net result is that concentrations remain constant, but individual molecules are constantly moving between reactant and product forms. Think of it like a crowded dance floor where people enter and leave at exactly the same rate – the number of dancers stays constant, but there's continuous movement. Answer choice A is incorrect because if the forward rate were always greater, the reaction would continue shifting toward products until equilibrium was disrupted. Similarly, choice B is wrong because a consistently higher reverse rate would shift the reaction toward reactants, again breaking equilibrium. Choice C contains a critical misconception – while the rates are indeed equal at equilibrium, they are definitely not zero. Zero rates would mean the reaction has completely stopped, which contradicts the dynamic nature of equilibrium. Choice D correctly captures both essential features: the rates are equal (maintaining constant concentrations) and non-zero (maintaining the dynamic process). Remember this key distinction: equilibrium means "constant change with constant concentrations." The word "dynamic" is crucial – chemical equilibrium never means reactions have stopped, only that they've achieved a perfect balance.

Question 19

Consider two related equilibria at the same temperature: $A(g) + B(g) \rightleftharpoons C(g)$ with $K_{c1} = 5.0$ and $C(g) + D(g) \rightleftharpoons E(g)$ with $K_{c2} = 2.0$ . What is the equilibrium constant for the overall reaction $A(g) + B(g) + D(g) \rightleftharpoons E(g)$ ?

3.0
7.0
10.0 (correct answer)
2.5
0.40

Explanation: When you encounter multiple equilibria that can be combined to form an overall reaction, you're working with the principle that equilibrium constants multiply when reactions are added together. To find the overall equilibrium constant, you need to identify how the individual reactions combine. The first reaction produces C(g), which is then consumed in the second reaction. When you add these reactions together:

A(g) + B(g) \rightleftharpoons C(g)

(

K_{c1} = 5.0

)

C(g) + D(g) \rightleftharpoons E(g)

(

K_{c2} = 2.0

) The C(g) cancels out, giving you:

A(g) + B(g) + D(g) \rightleftharpoons E(g)

For the overall equilibrium constant, you multiply the individual constants:

K_{overall} = K_{c1} \times K_{c2} = 5.0 \times 2.0 = 10.0

Answer A (3.0) represents the error of subtracting the equilibrium constants (5.0 - 2.0), which is thermodynamically meaningless. Answer B (7.0) comes from incorrectly adding the constants (5.0 + 2.0), treating them like simple arithmetic rather than exponential relationships. Answer D (2.5) results from dividing the constants (5.0 ÷ 2.0), which would apply if you were reversing one reaction, but that's not the case here. Remember this key rule: when reactions are added sequentially (where the product of one becomes the reactant of the next), always multiply their equilibrium constants. This reflects the exponential nature of equilibrium expressions and free energy relationships.

Question 20

At 500°C, the equilibrium constant $K_c$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is 0.159. A reaction mixture at this temperature contains 0.200 M $N_2$ , 0.300 M $H_2$ , and 0.100 M $NH_3$ . Which statement best describes the current state of this system?

The system is at equilibrium because all concentrations are positive
The reaction will proceed to the right because $Q < K_c$
The reaction will proceed to the left because $Q > K_c$ (correct answer)
The reaction will proceed to the right because $Q > K_c$
The system cannot reach equilibrium at these concentrations

Explanation: When you encounter equilibrium problems with given concentrations, you need to determine whether the system is at equilibrium by comparing the reaction quotient (Q) to the equilibrium constant (Kc). First, calculate the reaction quotient using the same expression as the equilibrium constant:

Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Substituting the given concentrations:

Q_c = \frac{(0.100)^2}{(0.200)(0.300)^3} = \frac{0.0100}{(0.200)(0.0270)} = \frac{0.0100}{0.00540} = 1.85

Since

Q_c = 1.85

and

K_c = 0.159

, we have

Q_c > K_c

. When Q > K, the system has too much product relative to reactants compared to the equilibrium position, so the reaction shifts left (toward reactants) to reach equilibrium. Answer A is wrong because positive concentrations don't indicate equilibrium—you must compare Q to K. Answer B incorrectly states that Q < K when we calculated Q > K. Answer D correctly identifies that Q > K but incorrectly predicts the reaction direction; when Q > K, the reaction proceeds left, not right. The correct answer is C: the reaction proceeds to the left because Q > Kc. Study tip: Remember the Q vs. K rule: If Q < K, shift right (toward products); if Q > K, shift left (toward reactants); if Q = K, you're at equilibrium. Always calculate Q first using the given concentrations, then compare to the provided K value.