4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
For the hypothetical reaction 2 X(g) + Y(g) ⇌ 2 Z(g), Kc = 0.020 at a certain temperature. What is the value of Kc for the reaction 2 Z(g) ⇌ 2 X(g) + Y(g) at the same temperature?
Answer and reasoning
A0.020 A student who thinks K is unchanged when a reaction is reversed picks this. Reversing the equation swaps the numerator and denominator of the expression, so Kc becomes 1/0.020 = 50.
B−0.020 A student who applies the ΔH rule for reversing a reaction to K picks this. K is a ratio of concentrations and cannot be negative; reversing the reaction inverts it, giving 50.
C50Correct The second equation is the reverse of the first, so its equilibrium expression is the reciprocal of the original: Kc = 1/0.020 = 50.
D0.98 A student who thinks K is the fraction of reactant converted, so that the reverse reaction converts the rest, picks this: 1 − 0.020 = 0.98. K is a ratio of equilibrium concentrations, and the reverse reaction has Kc = 1/0.020 = 50.
Working The second equation is the first reversed, so Kc = 1/0.020 = 50. Distractors: K unchanged, 0.020; sign changed as for ΔH, −0.020; 1 − K as if K were the fraction converted, 1 − 0.020 = 0.98.
The table gives Kc values for two hypothetical gas-phase reactions at the same temperature. What is the value of Kc for the reaction 2 E(g) + F(g) ⇌ H(g) at this temperature?
Answer and reasoning
A3.0 A student who multiplies K₁ by 2 when reaction 1 is doubled, as ΔH would be, picks this: (2 × 3.0)(0.50) = 3.0. Doubling the coefficients doubles the exponents, so K₁ must be squared, giving (3.0)² × 0.50 = 4.5.
B9.5 A student who adds the K values of the reactions being combined picks this: (3.0)² + 0.50 = 9.5. When reactions are added, their equilibrium expressions multiply, so Kc = 9.0 × 0.50 = 4.5.
C1.5 A student who thinks K is unaffected by the coefficients used picks this, multiplying 3.0 × 0.50 without squaring K₁. Reaction 1 must be doubled to give the target equation, so its K is squared: 9.0 × 0.50 = 4.5.
D4.5Correct Doubling reaction 1 gives 2 E + 2 F ⇌ 2 G with K = (3.0)² = 9.0. Adding reaction 2 cancels 2 G and one F, giving 2 E + F ⇌ H, so Kc = 9.0 × 0.50 = 4.5.
Working Multiply reaction 1 by 2: 2 E + 2 F ⇌ 2 G, K = (3.0)² = 9.0. Add reaction 2: 2 E + 2 F + 2 G ⇌ 2 G + H + F, which simplifies to 2 E + F ⇌ H. Kc = 9.0 × 0.50 = 4.5. Distractors: K₁ multiplied by 2, (2 × 3.0)(0.50) = 3.0; K values added, 9.0 + 0.50 = 9.5; coefficient change ignored, 3.0 × 0.50 = 1.5.
Reaction 3, A(aq) ⇌ C(aq), is the sum of reaction 1, A(aq) ⇌ B(aq), and reaction 2, B(aq) ⇌ C(aq). A student knows Kc for reaction 1 at 298 K and wants to calculate Kc for reaction 3 at 298 K. Which additional quantity does the student need?
Answer and reasoning
AThe value of Kc for reaction 2Correct Adding reactions 1 and 2 gives reaction 3, so K₃ = K₁ × K₂: ([B]/[A]) × ([C]/[B]) = [C]/[A]. With K₁ known, only K₂ at 298 K is needed.
BThe rate constant for reaction 2 A student who treats a rate constant as if it were an equilibrium constant picks this. The rate constant tells how fast reaction 2 proceeds, not the ratio [C]/[B] at equilibrium, which is what K₂ supplies.
CThe equilibrium concentration of B A student who thinks the species that cancels still appears in the overall expression picks this. [B] cancels when K₁ = [B]/[A] and K₂ = [C]/[B] are multiplied, so K₃ = [C]/[A] needs no value of [B].
DThe initial concentration of A A student who thinks K depends on the starting concentrations picks this. At a given temperature K₃ has the same value whatever the starting mixture; it follows from K₁ × K₂.
Working No numerical data. K₃ = [C]/[A] = ([B]/[A]) × ([C]/[B]) = K₁ × K₂ at the same temperature. The only additional quantity needed is K₂ at 298 K; B cancels, and K does not depend on rates or starting concentrations.
For the hypothetical reaction 2 X(g) ⇌ Y(g), a gas mixture has a reaction quotient Qc = 0.20 at a certain moment. For the same mixture at the same moment, what is the value of Qc for the reaction 2 Y(g) ⇌ 4 X(g)?
Answer and reasoning
A0.20 A student who thinks Q is a fixed measurement of the mixture, unchanged by rewriting the equation, picks this. Q is calculated from the expression for the equation as written, so it changes as K would: (1/0.20)² = 25.
B25Correct 2 Y ⇌ 4 X is the reverse of 2 X ⇌ Y with every coefficient doubled. Q has the same form as K, so it is inverted and squared: Qc = (1/0.20)² = 25. Directly: Qc = [X]⁴/[Y]² = ([X]²/[Y])² = (1/0.20)².
C0.040 A student who squares Q for the doubled coefficients but thinks reversing the equation does not change the value picks this: (0.20)² = 0.040. Reversing also inverts Q, giving (1/0.20)² = 25.
D5.0 A student who inverts Q for the reversed equation but thinks the coefficients do not affect the value picks this: 1/0.20 = 5.0. Doubling the coefficients squares the expression, giving 25.
Working 2 Y ⇌ 4 X is the original equation reversed and multiplied by 2, so Qc = (1/0.20)² = 5.0² = 25. Distractors: Q unchanged, 0.20; squared without inverting, (0.20)² = 0.040; inverted without squaring, 1/0.20 = 5.0.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.6.A.1 K for a reversed reaction Fix
K for a reversed reaction
When a reaction is reversed, products and reactants exchange places in the equilibrium expression, so the new K is the reciprocal of the original: Kreverse = 1/Kforward.
Students often think K describes the equilibrium mixture, which is the same whichever way the equation is written, so reversing a reaction does not change K. In fact No. Reversing the equation puts the former reactants in the numerator and the former products in the denominator, so the new K is the reciprocal of the original.
Students often think Reversing a reaction changes the sign of K, just as reversing a reaction changes the sign of ΔH. In fact No. Reversing a reaction inverts K (K becomes 1/K). K is a ratio of concentrations or pressures and is always positive; changing sign is the rule for ΔH, not for K.
7.6.A.2 K for an equation multiplied by a factor Fix
K for an equation multiplied by a factor
When every coefficient of a reaction is multiplied by a factor c, every exponent in the equilibrium expression is multiplied by c, so the new K is the original K raised to the power c (for example, doubling gives K², halving gives K½).
Dependence of the value of K on the written equation
The numerical value of K belongs to a particular balanced equation as written; the same equilibrium mixture gives different K values for the reversed equation or for an equation with different coefficients, so a K value must always be reported with its equation.
Students often think When the coefficients of an equation are multiplied by a factor c, K is multiplied by the same factor c, as ΔH is. In fact No. Each exponent in the equilibrium expression is multiplied by c, so K is raised to the power c: doubling an equation squares K, and halving it takes the square root of K.
Students often think K depends only on the reaction and the temperature, so it has the same value however the coefficients of the equation are written. In fact Yes. The coefficients are the exponents in the equilibrium expression, so writing the equation with different coefficients gives a different expression and a different numerical value of K for the same mixture.
7.6.A.3 K for a sum of reactions Fix
K for a sum of reactions
When two or more reactions are added to give an overall reaction, the K of the overall reaction is the product of the K values of the reactions added, each written for the reaction exactly as it is used in the sum.
Overall equilibrium expression for a multistep process
A species that is formed in one step and used up in an equal amount in another step cancels completely when the steps are added; its concentration also cancels in the product of the step equilibrium expressions, so it does not appear in the overall K expression. A species that cancels only in part (for example, 2 F on one side and 1 F on the other) stays in the overall equation and in the overall expression with the coefficient that remains.
Students often think The K of an overall reaction is the sum of the K values of the reactions that are added to give it, just as ΔH values are added in Hess's law. In fact No. The overall equilibrium expression is the product of the step expressions, so the overall K is the product of the step K values.
Students often think The overall equilibrium constant of a multistep process equals the K of the step with the smallest K, which limits the overall reaction in the way a slow step limits the rate. In fact No. The overall K is the product of the K values of all the steps as they are used; no single step's K determines it. The idea of one step limiting the whole process applies to reaction rates, not to equilibrium constants.
7.6.A.4 Algebraic manipulations of Q Fix
Algebraic manipulations of Q
Q has the same mathematical form as K, so reversing an equation inverts Q, multiplying its coefficients by c raises Q to the power c, and adding equations multiplies the Q values. A mixture with Q = K for one way of writing a reaction therefore has Q = K for every way of writing it.
Students often think Q is just a measurement of the mixture at that moment, so its value stays the same however the equation is written, even though K must be recalculated. In fact Yes. Q has the same form as K, so reversing the equation inverts Q and multiplying the coefficients by c raises Q to the power c, even though the mixture itself is unchanged.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The table gives Kc values for two hypothetical gas-phase reactions at the same temperature. What is the value of Kc for the reaction A(g) ⇌ D(g) at this temperature?
Answer and reasoning
A3.0 × 10⁻¹ A student who reverses reaction 2 but keeps K₂ unchanged picks this: (1.5 × 10⁻³)(2.0 × 10²) = 3.0 × 10⁻¹. The reversed reaction has K = 1/(2.0 × 10²) = 5.0 × 10⁻³.
B7.5 × 10⁻⁶Correct A ⇌ D is reaction 1 plus reaction 2 reversed (C ⇌ D + B); C and B cancel. Reversing reaction 2 inverts K₂, and adding reactions multiplies their K values: Kc = (1.5 × 10⁻³)/(2.0 × 10²) = 7.5 × 10⁻⁶.
C6.5 × 10⁻³ A student who adds the K values of the combined reactions, as ΔH values are added, picks this: 1.5 × 10⁻³ + 5.0 × 10⁻³ = 6.5 × 10⁻³. The equilibrium expressions multiply, so the K values multiply: 7.5 × 10⁻⁶.
D1.5 × 10⁻³ A student who thinks the overall K equals the K of the least favorable step, as if it limited the process like a slow step limits a rate, picks this. The overall K is the product of the K values of both steps as used, 7.5 × 10⁻⁶.
Working A ⇌ D = reaction 1 + reverse of reaction 2: A + B ⇌ C, then C ⇌ D + B; C and B cancel. K = K₁ × (1/K₂) = (1.5 × 10⁻³)/(2.0 × 10²) = 7.5 × 10⁻⁶. Distractors: reaction 2 used without inverting K₂, (1.5 × 10⁻³)(2.0 × 10²) = 3.0 × 10⁻¹; K values added after inverting K₂, 1.5 × 10⁻³ + 5.0 × 10⁻³ = 6.5 × 10⁻³; smallest K taken as the overall K, 1.5 × 10⁻³.
The diagram represents a gas mixture at equilibrium for the hypothetical reaction 2 A(g) ⇌ A₂(g) at a certain temperature. Each particle shown represents a concentration of 0.010 M. What is the value of Kc for the reaction ½ A₂(g) ⇌ A(g) at this temperature?
Answer and reasoning
A20 A student who halves the coefficients correctly but thinks reversing the reaction leaves K unchanged picks this: (400)½ = 20. Reversing the equation also inverts K, so the value is (1/400)½ = 0.050.
B0.0025 A student who inverts K for the reversed reaction but thinks the coefficients do not affect K picks this: 1/400 = 0.0025. Halving the coefficients takes the square root: (1/400)½ = 0.050.
C0.050Correct The diagram shows one A atom and four A₂ molecules, so [A] = 0.010 M and [A₂] = 0.040 M. For ½ A₂ ⇌ A, Kc = [A]/[A₂]½ = 0.010/0.20 = 0.050, which equals (1/400)½, the reverse of 2 A ⇌ A₂ (K = 400) with coefficients halved.
D0.25 A student who writes the expression for ½ A₂ ⇌ A without the fractional exponent picks this: [A]/[A₂] = 0.010/0.040 = 0.25. The coefficient ½ becomes an exponent: [A]/[A₂]½ = 0.010/0.20 = 0.050.
Working From the diagram: [A] = 1 × 0.010 M = 0.010 M; [A₂] = 4 × 0.010 M = 0.040 M. For ½ A₂ ⇌ A, Kc = [A]/[A₂]½ = 0.010/(0.040)½ = 0.010/0.20 = 0.050. Check: for 2 A ⇌ A₂, K = 0.040/(0.010)² = 400; reversing and halving gives (1/400)½ = 0.050. Distractors: halved but not inverted, 400½ = 20; inverted but not halved, 1/400 = 0.0025; exponent omitted, 0.010/0.040 = 0.25.
Reaction 1 is A(g) ⇌ 2 B(g), and reaction 2 is 2 B(g) ⇌ A(g). At a certain temperature, a mixture of A and B is at equilibrium with respect to reaction 1, so its reaction quotient Q₁ equals K₁. Which statement about the same mixture, judged with respect to reaction 2 (reaction quotient Q₂, equilibrium constant K₂), is correct?
Answer and reasoning
AIt is not at equilibrium, because reversing the equation inverts K but not Q A student who thinks Q describes only the mixture, so its value does not change when the equation is rewritten, picks this. Q₂ = [A]/[B]² = 1/Q₁, just as K₂ = 1/K₁, so Q₂ = K₂.
BIt is not at equilibrium, because K₂ = K₁ but Q₂ is the reciprocal of Q₁ A student who recalculates Q for the reversed equation but thinks reversing a reaction leaves K unchanged picks this. K₂ is also the reciprocal of K₁, so Q₂ = K₂.
CIt is also at equilibrium, because Q₂ and K₂ are both the negatives of Q₁ and K₁ A student who applies the ΔH rule for reversing a reaction to Q and K picks this. Q and K are ratios of concentrations and are never negative; reversing the equation inverts them.
DIt is also at equilibrium, because Q₂ and K₂ are both the reciprocals of Q₁ and K₁Correct Q has the same form as K, so reversing the equation inverts both: Q₂ = 1/Q₁ and K₂ = 1/K₁. Because Q₁ = K₁, Q₂ = K₂, and the mixture is at equilibrium whichever way the reaction is written.
Working No calculation needed. Q₂ = [A]/[B]² = 1/Q₁ and K₂ = 1/K₁. Since Q₁ = K₁, Q₂ = K₂: the mixture is at equilibrium however the equation is written.
A student wants to determine Kc at 25°C for the reaction A(aq) ⇌ C(aq). Without a catalyst the reaction is too slow to reach equilibrium in the laboratory. When catalyst X is present, only reaction 1, A(aq) ⇌ B(aq), takes place, and it reaches equilibrium within minutes; when catalyst Y is present, only reaction 2, B(aq) ⇌ C(aq), takes place, and it reaches equilibrium within minutes. The sum of reactions 1 and 2 is A(aq) ⇌ C(aq); their equilibrium constants at 25°C are K₁ and K₂. Which procedure will allow the student to determine Kc for A(aq) ⇌ C(aq)?
Answer and reasoning
AFind K₁ and K₂ from equilibrium mixtures of reactions 1 and 2 and multiply them togetherCorrect With the appropriate catalyst, reactions 1 and 2 each reach equilibrium quickly, so K₁ and K₂ can be measured at 25°C (a catalyst does not change K). Their sum is A ⇌ C, so Kc = K₁ × K₂: ([B]/[A])([C]/[B]) = [C]/[A].
BFind K₁ and K₂ from equilibrium mixtures of reactions 1 and 2 and add them A student who adds the K values of combined reactions, as ΔH values are added in Hess's law, picks this. The equilibrium expressions multiply, so Kc = K₁ × K₂.
CFind K₁ and K₂ from equilibrium mixtures of reactions 1 and 2 and use the smaller A student who thinks the overall K equals the K of the least favorable step, as a slow step limits a rate, picks this. Every step contributes: Kc = K₁ × K₂.
DMeasure [A] and [C] in an uncatalyzed mixture of A and C one hour after mixing A student who thinks K can be calculated from concentrations at any time picks this. Without a catalyst A ⇌ C is far from equilibrium after one hour, so the concentrations give Q, not Kc.
Working No numerical data. K(A ⇌ C) = K₁ × K₂ at the same temperature. Measuring equilibrium concentrations for reactions 1 and 2 at 25°C gives K₁ and K₂; their product is the required Kc. Adding them, taking the smaller, or using non-equilibrium concentrations of A and C in the uncatalyzed mixture does not give Kc.
Two groups of students each prepared a different starting mixture of the hypothetical gases A and B, allowed it to reach equilibrium at the same temperature, and reported Kc for the equation they chose. The table shows their results. Which statement best evaluates whether the two results are consistent?
Answer and reasoning
AThe results are consistent, because Kc depends on the starting mixture A student who thinks K depends on the starting concentrations picks this. At the same temperature Kc does not depend on the starting mixture; the results agree because Group 2's equation is Group 1's reversed and halved, so its Kc is √(1/0.090) = 3.3.
BThe results are inconsistent, as Group 2's Kc should be ½ × (1/0.090) = 5.6 A student who multiplies K by the factor applied to the coefficients, as ΔH would be, picks this. Halving the coefficients takes the square root of K: √(1/0.090) = 3.3, which matches Group 2's value.
CThe results are inconsistent, because both groups should report Kc = 0.090 A student who thinks K has the same value however the equation is written picks this. Group 2 wrote the reaction reversed with halved coefficients, so its Kc is √(1/0.090) = 3.3.
DThe results are consistent, because Group 2's Kc should equal √(1/0.090) = 3.3Correct Group 2's equation is Group 1's reversed (invert K) and halved (take the square root): √(1/0.090) = √11.1 = 3.3, which is the value Group 2 reported. The different starting mixtures do not affect Kc.
Working Group 2's equation is Group 1's equation reversed and halved, so Kc(2) should be (1/0.090)½ = (11.1)½ = 3.3, which matches. The starting mixtures do not affect Kc. Distractors: K multiplied by ½ instead of raised to the power ½, ½ × 11.1 = 5.6; K independent of the written equation, 0.090.
At a certain temperature, the equilibrium constant for the hypothetical reaction A(g) + B(g) ⇌ C(g) is Kc. A student rewrites the equation as 3 A(g) + 3 B(g) ⇌ 3 C(g). How does Kc for the rewritten equation compare with the original Kc at the same temperature?
Answer and reasoning
AIt is the third power of the original KcCorrect Tripling every coefficient triples every exponent: Kc′ = [C]³/([A]³[B]³) = ([C]/([A][B]))³, the cube of the original Kc.
BIt equals three times the original Kc A student who multiplies K by the factor applied to the coefficients, as ΔH would be, picks this. The coefficients become exponents, so tripling them cubes Kc.
CIt is unchanged from the original Kc A student who thinks K depends only on the reaction and temperature, not on how the equation is written, picks this. The rewritten equation has a different expression, [C]³/([A]³[B]³), whose value is the cube of Kc.
DIt equals the cube root of the original Kc A student who mixes up the rule for multiplying coefficients with the rule for dividing them picks this. Tripling the coefficients raises Kc to the third power; the cube root would apply if the coefficients were divided by 3.
Working Original: Kc = [C]/([A][B]). Rewritten: Kc′ = [C]³/([A]³[B]³) = (Kc)³. Distractors: 3Kc (factor applied as a multiplier), Kc (coefficients thought irrelevant), (Kc)1/3 (rule for dividing coefficients used).
The graph shows how the concentrations of A and B change with time after A is dissolved in water at a constant temperature and reacts according to the equation A(aq) ⇌ 2 B(aq). Based on the graph, what is the value of Kc for the reaction 2 B(aq) ⇌ A(aq) at this temperature?
Answer and reasoning
A12 A student who thinks K is the same whichever way the equation is written picks this, reporting the value for A ⇌ 2 B: (2.2)²/0.40 = 12. The reversed equation has the reciprocal expression, [A]/[B]² = 0.083.
B0.18 A student who writes the expression as a ratio of concentrations with no exponents picks this: [A]/[B] = 0.40/2.2 = 0.18. The coefficient 2 of B is an exponent, so Kc = 0.40/(2.2)² = 0.083.
C1.0 A student who thinks equilibrium is the point where the two concentrations are equal picks this, reading 1.0 M for both where the curves cross: 1.0/(1.0)² = 1.0. The concentrations are still changing there; the equilibrium values are 0.40 M and 2.2 M, giving 0.083.
D0.083Correct The curves level off at [A] = 0.40 M and [B] = 2.2 M. The equation 2 B ⇌ A is the reverse of A ⇌ 2 B, so its expression is the reciprocal: Kc = [A]/[B]² = 0.40/(2.2)² = 0.083, which is 1/12.
Working From the flat parts of the curves, [A] = 0.40 M and [B] = 2.2 M at equilibrium. For 2 B ⇌ A, Kc = [A]/[B]² = 0.40/(2.2)² = 0.40/4.84 = 0.083. (For A ⇌ 2 B, Kc = 4.84/0.40 = 12, and 1/12 = 0.083.) Distractors: K for the equation as originally written, (2.2)²/0.40 = 12; exponent omitted, 0.40/2.2 = 0.18; concentrations read where the curves cross, 1.0/(1.0)² = 1.0.
At a certain temperature, reaction 1, 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), has equilibrium constant K₁, and reaction 2, 2 NO(g) + O₂(g) ⇌ 2 NO₂(g), has equilibrium constant K₂. Which expression gives the equilibrium constant at the same temperature for the reaction SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g)?
Answer and reasoning
AK₁/K₂ A student who inverts K₂ for the reversed reaction but thinks K does not depend on the coefficients picks this. Both reactions must be halved to give the target equation, so each K is raised to the power ½: √(K₁/K₂).
B√(K₁/K₂)Correct The target equation is half of reaction 1 plus half of the reverse of reaction 2 (the ½ O₂ cancels). Halving takes the square root of each K, reversing inverts K₂, and adding the reactions multiplies the results: √K₁ × √(1/K₂) = √(K₁/K₂).
C√(K₁ × K₂) A student who halves both reactions correctly but thinks reversing reaction 2 leaves its K unchanged picks this. The reversed reaction has K = 1/K₂, so the result is √(K₁/K₂).
DK₁ × K₂ A student who multiplies the K values of the given reactions exactly as they are listed picks this. Reaction 1 must be halved and reaction 2 reversed and halved before they add up to the target equation, so the adjusted values √K₁ and √(1/K₂) are multiplied: √(K₁/K₂).
Working No numerical data. Halve reaction 1: SO₂ + ½ O₂ ⇌ SO₃, K = K₁½. Reverse and halve reaction 2: NO₂ ⇌ NO + ½ O₂, K = (1/K₂)½. Adding them cancels ½ O₂ and gives SO₂ + NO₂ ⇌ SO₃ + NO, so K = K₁½ × (1/K₂)½ = √(K₁/K₂). Distractors: coefficients ignored, K₁/K₂; reaction 2 reversed without inverting K₂, √(K₁ × K₂); given K values multiplied as listed, K₁ × K₂.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
