1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
The hypothetical reaction X(g) + Y(g) ⇌ 2 Z(g) is allowed to reach equilibrium in a rigid container at a constant temperature. The table shows the concentrations measured in the equilibrium mixture. What is the value of Kc for the reaction at this temperature?
Answer and reasoning
A0.092Correct Kc = [Z]²/([X][Y]) = (0.15)²/((0.35)(0.70)) = 0.0225/0.245 = 0.092. The coefficient 2 of Z becomes the exponent on [Z].
B0.61 A student who leaves the coefficients out of the equilibrium expression picks this: 0.15/((0.35)(0.70)) = 0.61. Because the coefficient of Z is 2, [Z] must be squared.
C1.2 A student who uses the coefficient of Z as a multiplier picks this: (2 × 0.15)/((0.35)(0.70)) = 1.2. In an equilibrium expression a coefficient of 2 becomes an exponent: [Z]², not 2[Z].
D11 A student who puts the reactant concentrations over the product concentration picks this: (0.35)(0.70)/(0.15)² = 11. Kc has the products of the reaction as written in the numerator, so the value is the reciprocal, 0.092.
Working Kc = [Z]²/([X][Y]) = (0.15)²/((0.35)(0.70)) = 0.0225/0.245 = 0.092. Distractors: exponent omitted, 0.15/0.245 = 0.61; coefficient used as a multiplier, (2 × 0.15)/0.245 = 1.2; expression inverted, 0.245/0.0225 = 11.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.4.A.1 Equilibrium constant, KcFix
Equilibrium constant, Kc
The value of the equilibrium expression, written from the balanced equation with molar concentrations of products over reactants, each raised to the power of its coefficient, when the concentrations are those of the system at equilibrium. For a given reaction at a given temperature it has one value.
Equilibrium constant, Kp
For a reaction involving gases, the value of the equilibrium expression written with the equilibrium partial pressures of the gases (in atm), each raised to the power of its coefficient, products over reactants.
Experimental determination of K
K is found by measuring the concentrations or partial pressures of the reactants and products once the system has reached equilibrium and substituting them into the equilibrium expression. When only some equilibrium values are measured, the others can be found from the known starting amounts and the mole ratios of the balanced equation.
Recognizing equilibrium in experimental data
A closed system has reached equilibrium when the measured concentrations or partial pressures (or a property that depends on them, such as color intensity or total pressure) no longer change with time. Values measured before that point describe the reaction quotient, Q, not K.
Stoichiometric changes during approach to equilibrium
As a system moves toward equilibrium, the changes in the amounts of the reactants and products are in the mole ratio given by the coefficients of the balanced equation; for example, in A(g) ⇌ 2 B(g), each mole of A that reacts forms two moles of B.
Independence of K from starting amounts
Experiments on the same reaction at the same temperature that start from different amounts of reactants or products reach different equilibrium concentrations, but substituting each set of equilibrium values into the equilibrium expression gives the same K.
Students often think The equilibrium expression is simply the product of the product concentrations divided by the product of the reactant concentrations, with no exponents, whatever the coefficients are. In fact Yes. Each concentration or partial pressure is raised to a power equal to its coefficient in the balanced equation; for X + Y ⇌ 2 Z, Kc = [Z]²/([X][Y]).
Students often think A coefficient multiplies the concentration of that species in the equilibrium expression, so a coefficient of 2 in front of Z gives 2[Z]. In fact As an exponent: a species with coefficient 2 appears as its concentration squared, for example [Z]², not as 2[Z].
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
A sample of the hypothetical gas A is placed in an evacuated rigid container at constant temperature, where it reacts according to the equation 2 A(g) ⇌ B(g). The graph shows [A] and [B] as functions of time. Based on the graph, what is the value of Kc for the reaction at this temperature?
Answer and reasoning
A4.0 A student who thinks equilibrium is reached where the reactant and product concentrations are equal picks this, using the crossing point: 0.25/(0.25)² = 4.0. At the crossing point both curves are still changing; equilibrium is where they level off, at [A] = 0.15 M and [B] = 0.30 M.
B13Correct The curves level off after about 40 s, so the equilibrium concentrations are [A] = 0.15 M and [B] = 0.30 M. Kc = [B]/[A]² = 0.30/(0.15)² = 13.
C0.53 A student who puts the starting concentration of A into the expression picks this: 0.30/(0.75)² = 0.53. Kc uses the [A] remaining at equilibrium, 0.15 M, because 0.60 M of A has reacted.
D0.075 A student who puts the reactant term over the product term picks this: (0.15)²/0.30 = 0.075. For 2 A ⇌ B, Kc = [B]/[A]², the reciprocal of this value.
Working Both curves are level after about 40 s, so the equilibrium values are [A] = 0.15 M and [B] = 0.30 M (check: 0.75 − 0.15 = 0.60 M of A reacted, forming 0.30 M B, ratio 2:1). Kc = [B]/[A]² = 0.30/(0.15)² = 0.30/0.0225 = 13. Distractors: values at the crossing point (both 0.25 M), 0.25/(0.25)² = 4.0; starting [A] used, 0.30/(0.75)² = 0.53; expression inverted, (0.15)²/0.30 = 0.075.
A student determines Kc for the hypothetical reaction J(aq) ⇌ 2 L(aq) by measuring [J] and [L] in an equilibrium mixture and substituting them into the equilibrium expression. The true value of Kc at the temperature of the experiment is 0.25. The student's value of [J] is correct, but because of a calibration error, the student's value of [L] is 10 times the true equilibrium [L]. What value of Kc does the student calculate?
Answer and reasoning
A2.5 A student who leaves the coefficient out of the expression, writing Kc = [L]/[J], picks this: the value would then be 10 × 0.25 = 2.5. The coefficient 2 makes [L] squared, so a tenfold error in [L] multiplies the calculated Kc by 100.
B0.0025 A student who writes the expression with the reactant on top, [J]/[L]², picks this: a value of [L] that is 10 times too large would then divide the value by 100, giving 0.25/100 = 0.0025. Kc has the product, [L]², in the numerator, so the calculated value is 100 times too large.
C0.25 A student who thinks a calculated K cannot be affected by a measurement error, because K is constant at a given temperature, picks this. The true Kc is fixed at 0.25, but the student's value is calculated from the faulty [L] and is 100 times too large.
D25Correct Kc = [L]²/[J]. A value of [L] that is 10 times too large makes [L]² 10² = 100 times too large with [J] unchanged, so the calculated value is 100 × 0.25 = 25.
A student wants to determine Kp at 400 K for the hypothetical reaction A(g) ⇌ 2 B(g). The student fills an evacuated rigid flask held at 400 K with pure A(g) to a measured pressure and then follows the total pressure in the flask with a pressure gauge. In a trial run, the total pressure rose for about 5 minutes and then stayed constant. Using the measured starting pressure of A, which additional measurement will allow the student to calculate Kp?
Answer and reasoning
AThe total pressure 1 minute after the flask is filled A student who thinks K can be calculated from pressures measured at any time picks this. One minute after filling, the pressure is still rising, so values from that reading give Q for that moment, not Kp.
BHow quickly the total pressure rises in the first minute A student who thinks K can be found from how fast a reaction proceeds picks this. The initial rate of pressure rise is kinetic information; Kp depends on the partial pressures once equilibrium is reached, not on how quickly they get there.
CThe total pressure once it has stopped changingCorrect When the total pressure stops changing, the system is at equilibrium. Then x = Ptotal − P₀ is the pressure of A that reacted, so PA = P₀ − x and PB = 2x at equilibrium, and Kp = (PB)²/PA.
DThe total pressure when the pressures of A and B are equal A student who thinks equilibrium is reached when reactant and product amounts are equal picks this. Equal partial pressures of A and B, if they occur at all, occur at one moment during the approach to equilibrium; a reading at that moment gives Q, and at equilibrium the pressures are constant but need not be equal.
Working No numerical data are given. If x atm of A reacts, PA = P₀ − x and PB = 2x, so Ptotal = P₀ + x. Once the total pressure has stopped changing, the system is at equilibrium: x = Ptotal − P₀, PA = P₀ − x, PB = 2x, and Kp = (PB)²/PA. A reading taken earlier gives Q, an initial rate gives kinetic information, and the moment when PA = PB is not equilibrium in general.
A student determines Kc for the hypothetical reaction A(aq) ⇌ 2 B(aq) at 25°C in two trials. In trial 1, the initial [A] is 0.10 M; in trial 2, the initial [A] is 0.20 M. Neither solution contains B initially, and both are kept at 25°C until equilibrium is reached. Which prediction about the two calculated Kc values is correct?
Answer and reasoning
ATrial 2 gives a larger Kc, because more B is present at equilibrium A student who thinks K depends on the amount of reactant used picks this. Trial 2 does contain more B at equilibrium, but it also contains more A, and the ratio [B]²/[A] is the same as in trial 1 at the same temperature.
BBoth trials give the same Kc, because Kc does not depend on initial [A]Correct At the same temperature the equilibrium expression [B]²/[A] has one value. Trial 2 reaches equilibrium with more A and more B than trial 1, but its equilibrium concentrations give the same Kc.
CBoth trials give the same Kc, because they reach the same equilibrium [B] A student who thinks every equilibrium mixture of a reaction has the same concentrations at a given temperature picks this. Starting with twice as much A gives a higher equilibrium [B]; it is the value of [B]²/[A], not [B] itself, that is the same in both trials.
DTrial 2 gives a smaller Kc, because a smaller fraction of its A reacts A student who thinks K is the fraction of reactant converted picks this. A smaller fraction of A does react in trial 2, but Kc is the ratio [B]²/[A] at equilibrium, which is the same in both trials.
Working No calculation needed. Kc = [B]²/[A]; at the same temperature both trials give the same Kc. Trial 2 contains more B at equilibrium, and a smaller fraction of its A reacts (for A ⇌ 2 B, x²·4/(c − x) = Kc gives a smaller x/c for larger c), but the value of [B]²/[A] is the same.
A sample of the hypothetical gas X₂ is placed in an evacuated rigid container at an initial concentration of 0.80 M, and the reaction X₂(g) ⇌ 2 X(g) occurs at constant temperature. When the concentrations have stopped changing, [X] = 0.40 M. What is the value of Kc for the reaction at this temperature?
Answer and reasoning
A0.27Correct Two moles of X form for each mole of X₂ that reacts, so [X₂] falls by 0.20 M to 0.60 M. Kc = [X]²/[X₂] = (0.40)²/(0.60) = 0.27.
B0.67 A student who writes the equilibrium expression without exponents picks this: 0.40/0.60. The coefficient 2 of X makes the expression [X]²/[X₂].
C0.20 A student who uses the concentration of X₂ that was put in at the start picks this: (0.40)²/(0.80). The equilibrium expression needs the equilibrium concentration of X₂, 0.60 M.
D0.40 A student who thinks [X₂] falls by the same amount that [X] rises picks this, using [X₂] = 0.80 − 0.40 = 0.40 M and (0.40)²/(0.40). Only 0.20 M of X₂ reacts to form 0.40 M of X, so [X₂] = 0.60 M.
Working Forming 0.40 M of X consumes 0.40/2 = 0.20 M of X₂, so at equilibrium [X₂] = 0.80 − 0.20 = 0.60 M. Kc = [X]²/[X₂] = (0.40)²/(0.60) = 0.27.
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
