2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
For the hypothetical reaction X₂(g) ⇌ 2 X(g), Kc = 3.6 × 10⁻⁶ at a certain temperature. A rigid container at this temperature is filled with X₂ to an initial concentration of 0.10 M, with no X present. What is the equilibrium concentration of X?
Answer and reasoning
A6.0 × 10⁻⁴ MCorrect With [X] = 2x and [X₂] = 0.10 − x ≈ 0.10 M (Kc is very small), Kc = [X]²/[X₂] gives [X]² = (3.6 × 10⁻⁶)(0.10) = 3.6 × 10⁻⁷, so [X] = 6.0 × 10⁻⁴ M. Only 0.3% of the X₂ dissociates, so the approximation is valid.
B3.6 × 10⁻⁷ M A student who writes the equilibrium expression without the exponent, Kc = [X]/[X₂], picks this: [X] = (3.6 × 10⁻⁶)(0.10). The coefficient 2 makes [X] squared, so [X] = √((3.6 × 10⁻⁶)(0.10)) = 6.0 × 10⁻⁴ M.
C3.6 × 10⁻⁶ M A student who thinks K equals the equilibrium concentration of the product picks this. Kc is the ratio [X]²/[X₂]; solving it with [X₂] ≈ 0.10 M gives [X] = 6.0 × 10⁻⁴ M.
D6.7 × 10⁻² M A student who thinks reactant and product concentrations become equal at equilibrium picks this, setting 0.10 − x = 2x so that [X] = [X₂] = 0.067 M. That would make Kc about 0.067, not 3.6 × 10⁻⁶; with so small a Kc, very little X₂ dissociates.
Working X₂ ⇌ 2 X: [X₂] = 0.10 − x, [X] = 2x. Kc = (2x)²/(0.10 − x) = 3.6 × 10⁻⁶. K is very small, so 0.10 − x ≈ 0.10: 4x² = 3.6 × 10⁻⁷, x = 3.0 × 10⁻⁴, [X] = 2x = 6.0 × 10⁻⁴ M (x is 0.3% of 0.10, so the approximation holds; the quadratic gives 5.99 × 10⁻⁴ M). Equivalently [X] = √(Kc × 0.10). Distractors: exponent omitted, [X] = Kc × 0.10 = 3.6 × 10⁻⁷ M; [X] = Kc = 3.6 × 10⁻⁶ M; equal concentrations, 0.10 − x = 2x, [X] = 6.7 × 10⁻² M.
For the hypothetical reaction A(g) + B(g) ⇌ C(g), Kc = 4.0 at a certain temperature. At one moment, a reaction mixture at this temperature has Qc = 25. Which statement correctly describes the mixture at that moment?
Answer and reasoning
AThe forward reaction is faster than the reverse reaction, so [C] rises A student who reverses the meaning of the Q–K comparison picks this. When Q > K, there is already too much product relative to reactants, so the net reaction consumes C, not forms it.
BThe forward and reverse reactions are equally fast, so [C] stays the same A student who thinks any mixture containing reactants and products is at equilibrium picks this. The rates are equal only when Q = K; here Qc = 25 is not equal to Kc = 4.0.
COnly the reverse reaction occurs, and it continues until Qc equals Kc A student who thinks only one reaction occurs while a system moves toward equilibrium picks this. Both reactions occur; the reverse reaction is faster, so the net change is toward A and B until the two rates become equal.
DThe reverse reaction is faster than the forward reaction, so [C] decreasesCorrect Qc = 25 is greater than Kc = 4.0, so the ratio [C]/([A][B]) is too large. The reverse reaction is faster than the forward reaction, so there is a net consumption of C and formation of A and B until Qc = Kc.
Working No calculation needed beyond the comparison. Qc = 25 > Kc = 4.0, so the mixture contains too much C relative to A and B: the reverse reaction is faster than the forward reaction, and the net change consumes C and forms A and B until Qc falls to 4.0. Both reactions occur throughout.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.7.A.1 Predicting equilibrium concentrations Fix
Predicting equilibrium concentrations
Given the balanced equation, the initial concentrations or partial pressures, and K, the equilibrium values are found by letting the amounts change in the mole ratio of the equation (an ICE table), substituting the equilibrium expressions into K and solving for the change.
Small-change approximation
When K is very small, the amount of reactant that reacts is negligible compared with its initial concentration, so the equilibrium reactant concentration can be replaced by the initial value when solving for the change; the result should be checked to confirm that the change is small.
Students often think The equilibrium expression is the ratio of product to reactant concentrations without exponents, so [X] = Kc × [X₂]. In fact No. Each coefficient is an exponent in the expression: for X₂ ⇌ 2 X, Kc = [X]²/[X₂], so [X] is the square root of Kc[X₂].
Students often think A reaction reaches equilibrium when the reactant and product concentrations (or partial pressures) have become equal. In fact No. At equilibrium the forward and reverse rates are equal; the concentrations take whatever values make Q equal K, and these are generally unequal.
7.7.A.2 Reaction quotient compared with K Fix
Reaction quotient compared with K
Comparing Q with K predicts the direction of net change: Q < K, net consumption of reactants and formation of products; Q > K, net consumption of products and formation of reactants; Q = K, the system is at equilibrium.
Dynamic equilibrium (Q = K)
When Q = K, the forward and reverse reactions continue at equal rates, so the proportions of reactants and products remain constant even though both reactions are still occurring.
Approach to equilibrium from the product side
A mixture containing only products has Q larger than any finite K, so the net reaction forms reactants until Q falls to K, whatever the value of K.
Students often think When Q is less than K the reaction proceeds toward the reactants, and when Q is greater than K it proceeds toward the products. In fact No. When Q < K, the ratio of products to reactants is too small, so the net reaction consumes reactants and forms products until Q increases to K.
Students often think A reaction mixture that contains reactants and products together is at equilibrium, so it will not change. In fact No. Any mixture contains reactants and products once the reaction has started; it is at equilibrium only when its composition gives Q = K.
5 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 5
For the hypothetical reaction A(g) ⇌ B(g), Kc = 3.0 at a certain temperature. A rigid container at this temperature is filled so that [A] = 0.30 M and [B] = 0.10 M initially. Which of the numbered graphs best represents how [A] and [B] change until equilibrium is reached?
Answer and reasoning
AGraph 1 A student who thinks a mixture containing both reactant and product is already at equilibrium picks Graph 1. Equilibrium requires Qc = Kc; here Qc = 0.10/0.30 = 0.33 is less than 3.0, so the net reaction converts A to B.
BGraph 2 A student who thinks reactant and product concentrations become equal at equilibrium picks Graph 2. Equal concentrations of 0.20 M would make Qc = 1.0, not 3.0; at equilibrium [B]/[A] must equal 3.0, which gives [A] = 0.10 M and [B] = 0.30 M.
CGraph 3Correct Qc = 0.10/0.30 = 0.33, less than Kc = 3.0, so the net reaction forms B. With x M of A converted, (0.10 + x)/(0.30 − x) = 3.0, so x = 0.20: [A] levels off at 0.10 M and [B] at 0.30 M, as shown in Graph 3.
DGraph 4 A student who thinks a reaction with K greater than 1 goes to completion picks Graph 4. With Kc = 3.0 the equilibrium mixture still contains A: [B]/[A] = 3.0 gives [A] = 0.10 M, not zero.
Working Qc = 0.10/0.30 = 0.33 < 3.0, so net forward reaction. Let x M of A react: Kc = (0.10 + x)/(0.30 − x) = 3.0 → 0.10 + x = 0.90 − 3x → x = 0.20. Equilibrium: [A] = 0.10 M, [B] = 0.30 M. The other graphs show equal final concentrations (0.20 M each), complete conversion ([A] = 0, [B] = 0.40 M), or no change.
In the hypothetical reaction X(aq) + Y(aq) ⇌ Z(aq), X and Y are colorless and Z is red; Kc = 50 at 25°C. A student plans to prepare a solution at 25°C in which, at the moment of mixing, [X] = 0.10 M, [Y] = 0.10 M, and [Z] = 0.20 M, and then to watch its color. Which prediction is correct?
Answer and reasoning
AThe red color fades, because some Z must decompose to reach equilibrium A student who reverses the meaning of the Q–K comparison picks this. Qc = 20 is less than Kc = 50, so there is too little Z relative to X and Y, and more Z forms.
BThe red color becomes more intense, because too little Z is presentCorrect Qc = 0.20/((0.10)(0.10)) = 20, which is less than Kc = 50. The net reaction consumes X and Y and forms red Z until Qc reaches 50, so the color deepens.
CThe color does not change, because X, Y and Z are all already present A student who thinks a mixture containing reactants and products is already at equilibrium picks this. The mixture is at equilibrium only if Qc = Kc; here Qc = 20, not 50, so its composition changes.
DThe red color deepens until no X or Y remains, because Kc is greater than 1 A student who thinks any reaction with K > 1 goes to completion picks this. The color does deepen, but only until Qc = 50; with Kc = 50, X and Y remain in the equilibrium mixture.
Working Qc = [Z]/([X][Y]) = 0.20/((0.10)(0.10)) = 20. Qc = 20 < Kc = 50, so the net reaction forms Z: the red color becomes more intense until Qc = 50. With Kc = 50 the reaction does not consume all the X and Y.
For the hypothetical reaction X(g) + Y(g) ⇌ 2 Z(g), Kp = 4.0 at a certain temperature. A rigid container at this temperature is filled with the gases at the initial partial pressures shown in the table. What is the partial pressure of Z once equilibrium is reached?
Answer and reasoning
A0.12 atm A student who writes Kp with the reactants on top, x²/(0.60 − 2x)² = 4.0, picks this: x = 0.24 and PZ = 0.12 atm. Kp has the product, (PZ)², in the numerator, which gives PZ = 0.30 atm.
B0.20 atm A student who thinks the partial pressures become equal at equilibrium picks this, setting 0.60 − 2x = x. Equal pressures of 0.20 atm would make Qp = 1.0, not 4.0.
C0.30 atmCorrect Only Z is present at first, so the net reaction forms X and Y. With PX = PY = x and PZ = 0.60 − 2x, Kp = (0.60 − 2x)²/x² = 4.0, so (0.60 − 2x)/x = 2.0 and x = 0.15. PZ = 0.60 − 0.30 = 0.30 atm.
D0.40 atm A student who lets every species change by the same amount, writing PZ = 0.60 − x, picks this: (0.60 − x)/x = 2.0 gives x = 0.20 and PZ = 0.40 atm. Two moles of Z react for each mole of X formed, so PZ = 0.60 − 2x.
Working Only Z is present, so Qp is infinitely large > Kp: the net reaction forms X and Y. Let x atm of X form: PX = PY = x, PZ = 0.60 − 2x. Kp = (0.60 − 2x)²/x² = 4.0 → (0.60 − 2x)/x = 2.0 → x = 0.15. PZ = 0.60 − 0.30 = 0.30 atm. Distractors: inverted expression, x²/(0.60 − 2x)² = 4.0 → x = 0.24, PZ = 0.12 atm; equal pressures, 0.60 − 2x = x → PZ = 0.20 atm; Z changes by x instead of 2x, (0.60 − x)/x = 2.0 → x = 0.20, PZ = 0.40 atm.
The diagram represents a mixture of A₂(g), B₂(g) and AB(g) in a rigid container at one moment; the number of molecules of each kind shown is proportional to its concentration. The gases react according to the equation A₂(g) + B₂(g) ⇌ 2 AB(g), for which Kc = 6.0 at the temperature of the mixture. Which statement correctly describes the mixture shown?
Answer and reasoning
AQc = 18, so [AB] will rise A student who reverses the meaning of the Q–K comparison picks this. Qc = 18 is correct, but Qc greater than Kc means the ratio of AB to A₂ and B₂ is too large, so AB is consumed, not formed.
BQc = 18, so [AB] will decreaseCorrect The diagram shows 6 AB, 2 A₂ and 1 B₂, so Qc = 6²/(2 × 1) = 18 (the scale factor between numbers and concentrations cancels). Qc is greater than Kc = 6.0, so the mixture has too much AB: the net reaction consumes AB and forms A₂ and B₂.
CQc = 3.0, so [AB] will rise A student who writes the expression without exponents picks this: 6/(2 × 1) = 3.0, which is less than 6.0. The coefficient 2 of AB is an exponent, so Qc = 6²/(2 × 1) = 18, which is greater than Kc.
DQc = 6.0, so [AB] will not change A student who thinks any mixture that contains reactants and products together is at equilibrium picks this, taking Qc to equal Kc. The composition shown gives Qc = 6²/(2 × 1) = 18, not 6.0, so the mixture is not at equilibrium.
Working The diagram shows 2 A₂, 1 B₂ and 6 AB. Qc = [AB]²/([A₂][B₂]); the same number of molecules appears on each side of the equation, so the factor that converts numbers of molecules to concentrations cancels: Qc = 6²/(2 × 1) = 18. Qc = 18 > Kc = 6.0, so there is a net consumption of AB (and formation of A₂ and B₂) until Qc falls to 6.0. Distractors: direction reversed for Q > K; exponent omitted, 6/(2 × 1) = 3.0 < 6.0; mixture assumed to be at equilibrium, Qc = Kc = 6.0.
For the hypothetical reaction 2 A(g) ⇌ B(g), mixture 1 is at equilibrium at a certain temperature, so its reaction quotient equals Kc. Mixture 2, at the same temperature, has twice the [A] and three times the [B] of mixture 1. How does the reaction quotient, Qc, of mixture 2 compare with Kc?
Answer and reasoning
AIt is three-fourths the value of KcCorrect Qc = [B]/[A]². Doubling [A] multiplies the denominator by 2² = 4 and tripling [B] multiplies the numerator by 3, so Qc = (3/4)Kc. Because Qc is less than Kc, mixture 2 undergoes a net consumption of A and formation of B.
BIt is three-halves the value of Kc A student who writes the expression without exponents, [B]/[A], picks this: tripling [B] and doubling [A] would multiply the ratio by 3/2. The coefficient 2 of A is an exponent, so the denominator becomes four times as large and Qc = (3/4)Kc.
CIt is still equal to the value of Kc A student who thinks any mixture containing reactants and products together is at equilibrium picks this. Mixture 2 has a different composition from mixture 1: Qc = 3[B]/(2[A])² = (3/4)Kc, so it is not at equilibrium.
DIt is four-thirds the value of Kc A student who writes the expression with the reactant on top, [A]²/[B], picks this: doubling [A] and tripling [B] would multiply it by 4/3. The product B is in the numerator and [A]² in the denominator, so Qc = (3/4)Kc.
Working Qc = [B]/[A]². Mixture 1: [B]/[A]² = Kc. Mixture 2: 3[B]/(2[A])² = 3[B]/(4[A]²) = (3/4)Kc. Qc < Kc, so in mixture 2 there is a net consumption of A and formation of B. Distractors: exponent omitted, 3[B]/(2[A]) = (3/2) times the value for mixture 1; mixture assumed to be at equilibrium, Qc = Kc; reactant term written in the numerator, (2[A])²/(3[B]) = (4/3) times the value for mixture 1.
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
