2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A rigid 2.0 L container holds a mixture of the gases N₂, H₂ and NH₃, which react according to the equation N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g). The table shows the amount of each gas in the container at a particular moment. What is the value of Qc at that moment?
Answer and reasoning
A0.80 A student who writes the quotient without exponents picks this: 0.30/((0.75)(0.50)) = 0.80. Each concentration is raised to the power of its coefficient, which gives (0.30)²/((0.75)(0.50)³) = 0.96.
B0.96Correct The concentrations are [N₂] = 0.75 M, [H₂] = 0.50 M and [NH₃] = 0.30 M. Qc = [NH₃]²/([N₂][H₂]³) = (0.30)²/((0.75)(0.50)³) = 0.96.
C0.53 A student who multiplies each concentration by its coefficient picks this: (2 × 0.30)/((0.75)(3 × 0.50)) = 0.53. The coefficients are exponents, not multipliers, which gives 0.96.
D0.24 A student who puts the amounts in moles into the expression picks this: (0.60)²/((1.5)(1.0)³) = 0.24. Qc uses concentrations, so each amount must first be divided by 2.0 L, which gives 0.96.
Working Concentrations: [N₂] = 1.5 mol/2.0 L = 0.75 M; [H₂] = 1.0/2.0 = 0.50 M; [NH₃] = 0.60/2.0 = 0.30 M. Qc = [NH₃]²/([N₂][H₂]³) = (0.30)²/((0.75)(0.50)³) = 0.090/0.09375 = 0.96.
The hypothetical reaction X(s) + 2 Y(g) ⇌ Z(g) occurs in a rigid 2.0 L container at constant temperature. At a particular moment the container holds 2.5 mol of X(s), and the gas concentrations are [Y] = 0.40 M and [Z] = 0.80 M. What is the value of Qc at that moment?
Answer and reasoning
A4.0 A student who includes the solid in the quotient, taking its concentration as 2.5 mol/2.0 L = 1.25 M, picks this: 0.80/((1.25)(0.40)²) = 4.0. Pure solids are left out, which gives 5.0.
B2.0 A student who writes the quotient without exponents picks this: 0.80/0.40 = 2.0. The coefficient 2 of Y is the exponent on [Y], which gives 0.80/(0.40)² = 5.0.
C5.0Correct The solid X does not appear in the reaction quotient, because its concentration does not depend on the amount present. Qc = [Z]/[Y]² = 0.80/(0.40)² = 5.0.
D1.0 A student who multiplies [Y] by its coefficient picks this: 0.80/(2 × 0.40) = 1.0. The coefficient is an exponent, not a multiplier, which gives 0.80/(0.40)² = 5.0.
Working X is a pure solid, so it does not appear in the reaction quotient. Qc = [Z]/[Y]² = 0.80/(0.40)² = 0.80/0.16 = 5.0.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.3.A.1 Reaction quotient, QcFix
Reaction quotient, Qc
For a A + b B ⇌ c C + d D, Qc = ([C]c [D]d)/([A]a [B]b): the concentrations of the products, each raised to its coefficient, divided by those of the reactants, each raised to its coefficient. Qc can be calculated for a reaction mixture at any time, whether or not it is at equilibrium.
Reaction quotient, Qp
For a gas-phase reaction, the reaction quotient written with partial pressures in place of concentrations: Qp = ((PC)c (PD)d)/((PA)a (PB)b).
Equilibrium constant, Kc or Kp
The value that the reaction quotient has when the system is at equilibrium at a given temperature: Kc = Qc and Kp = Qp at equilibrium. The equilibrium expression for K has the same form as the expression for Q, with equilibrium concentrations or partial pressures.
Law of mass action
The principle that gives the form of the expression for Q and K from the balanced equation: products over reactants, with each concentration or partial pressure raised to the power of its coefficient.
Q tends toward K
In a mixture that is not at equilibrium, the composition changes in the direction that brings the reaction quotient closer to the equilibrium constant; once Q equals K, Q stays constant.
Kc compared with Kp
Kc is written with molar concentrations and Kp with partial pressures. For the same gas-phase reaction at the same temperature the two generally have different numerical values, so concentrations must be compared with Kc and partial pressures with Kp.
Students often think Q and K are the product concentrations divided by the reactant concentrations, with no exponents, whatever the coefficients are. In fact Yes. Each concentration or partial pressure is raised to the power of that species' coefficient: for N₂ + 3 H₂ ⇌ 2 NH₃, Qc = [NH₃]²/([N₂][H₂]³).
Students often think Each concentration in the expression for Q or K is multiplied by its coefficient, so 2 NH₃ contributes 2[NH₃] and 3 H₂ contributes 3[H₂]. In fact No. A coefficient appears as an exponent on the concentration, not as a multiplier: for 2 NH₃ the term is [NH₃]², not 2[NH₃].
7.3.A.2 Pure solids and pure liquids in Q Fix
Pure solids and pure liquids in Q
The concentration of a pure solid or a pure liquid does not depend on how much of it is present, so these substances do not appear in the expression for Q or K. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Qp = PCO₂.
Students often think Every substance in the balanced equation appears in the expression for Q or K, with a solid's concentration found from its amount in moles and the volume of the container, so that more solid means a larger concentration… In fact No. The concentration of a pure solid or pure liquid does not depend on the amount present, so it is not included; only gases and dissolved species appear.
Students often think Solids are left out of the expression for Q or K because they are not involved in the equilibrium reaction. In fact No. The solids are reactants or products and are formed or used up as the system approaches equilibrium. They are left out of Q because their concentrations do not depend on the amounts present.
3 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 3
A student mixes solutions of A and B, with no C present, to study the hypothetical reaction A(aq) + B(aq) ⇌ C(aq) at constant temperature. The student measures the concentrations at regular intervals and calculates Qc each time. Which of the numbered graphs best represents how the calculated values of Qc are expected to change with time?
Answer and reasoning
AGraph 1 A student who thinks the reaction quotient always has the value of the equilibrium constant picks this graph. Qc depends on the concentrations at each moment: it is zero when no C is present and equals Kc only once equilibrium is reached.
BGraph 2 A student who writes the quotient as reactants over products picks this graph, expecting a very large value when only A and B are present. With products in the numerator, Qc starts at zero and increases toward Kc.
CGraph 3 A student who thinks the reaction continues until the reactants are used up picks this graph, in which Qc rises past Kc. The net reaction stops when Qc reaches Kc, so Qc levels off at that value.
DGraph 4Correct With no C present at the start, Qc = [C]/([A][B]) is zero. As C forms and A and B are used up, Qc increases; the reaction quotient tends toward the equilibrium constant, and once Qc = Kc the concentrations, and so Qc, stay constant.
Two identical rigid, evacuated containers are held at the same high temperature. Container 1 is loaded with 10.0 g of CaCO₃(s) and container 2 with 20.0 g of CaCO₃(s). In each container the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) reaches equilibrium with both solids present. Which comparison of the equilibrium pressures of CO₂ in the two containers is correct, and why?
Answer and reasoning
AEqual in both, because a solid's concentration does not depend on its amountCorrect The concentrations of CaCO₃(s) and CaO(s) are independent of the amounts present, so they do not appear in the reaction quotient: Qp = PCO₂. At equilibrium Qp = Kp, which has one value at this temperature, so the CO₂ pressure is the same in both containers.
BEqual in both, because the solids are not involved in the reaction at equilibrium A student who thinks solids are left out of the reaction quotient because they take no part in the reaction picks this. The pressures are equal, but the reason is wrong: CaCO₃ decomposes and CaO reacts with CO₂ continuously; the solids are omitted because their concentrations do not depend on their amounts.
CGreater in container 2, because its larger [CaCO₃] is balanced by extra CO₂ A student who includes the solid in the reaction quotient, with a concentration that grows with its amount, picks this. The concentration of solid CaCO₃ is the same whatever its mass, so Qp = PCO₂ and the equilibrium pressure is the same in both containers.
DGreater in container 2, because twice as much CaCO₃ decomposes to form CO₂ A student who treats the system as a mole-ratio calculation, with product proportional to the reactant supplied, picks this. Decomposition stops in each container when PCO₂ reaches Kp, so the same amount of CaCO₃ decomposes in each and the extra solid in container 2 remains.
For the hypothetical reaction A(g) ⇌ B(g), Kc = 2.5 at a certain temperature. Three mixtures of A and B are prepared at this temperature. At the moment of mixing, Qc is 0.10 in mixture 1, 2.5 in mixture 2 and 40 in mixture 3. Which statement about the mixtures is correct?
Answer and reasoning
AMixture 2 is at equilibrium, and Qc in mixtures 1 and 3 will change toward 2.5.Correct Mixture 2 has Qc = Kc, so it is at equilibrium. In mixtures 1 and 3, Qc differs from Kc; the reaction quotient tends toward the equilibrium constant, so their compositions change until Qc = 2.5.
BMixtures 1, 2 and 3 are each at equilibrium, with Kc values of 0.10, 2.5 and 40. A student who thinks the equilibrium constant changes with the concentrations picks this, giving each mixture its own Kc. At one temperature Kc has one value, 2.5; mixtures 1 and 3 have Qc values that differ from it, so they are not at equilibrium.
CMixture 2 is at equilibrium, and mixture 1 has the highest ratio of [B] to [A]. A student who reads the quotient as reactants over products picks this, taking the smallest Qc to mean the most product. Qc = [B]/[A], so mixture 1 has the lowest ratio of [B] to [A] and mixture 3 the highest.
DMixtures 1, 2 and 3 will each react until A is used up, with Qc rising beyond 40. A student who thinks a reaction continues until the reactant is used up picks this. The net reaction stops when Qc reaches Kc = 2.5: mixture 2 is already there, and mixtures 1 and 3 change until they reach it.
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
