2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
The hypothetical reaction 2 X(g) ⇌ Y(g) reached equilibrium in a rigid container at constant temperature. Additional Y(g) was then injected into the container. The table shows the concentrations of X and Y at equilibrium and immediately after the injection. What is the value of the reaction quotient, Q, immediately after the injection?
Answer and reasoning
A8.00 A student who thinks Q equals K at every moment picks this, the value of K = 0.500/(0.250)². The injection changed [Y], so Q immediately afterward is 1.25/(0.250)² = 20.0, not K.
B20.0Correct Q has the same form as K, Q = [Y]/[X]², but uses the concentrations present at the moment in question. Immediately after the injection, Q = 1.25/(0.250)² = 20.0, which is greater than K = 0.500/(0.250)² = 8.00, so the system is no longer at equilibrium.
C5.00 A student who writes Q without powers picks this: 1.25/0.250 = 5.00. The coefficient 2 for X makes the denominator [X]², so Q = 1.25/(0.250)² = 20.0.
D2.50 A student who uses the coefficient as a multiplier picks this: 1.25/(2 × 0.250) = 2.50. In Q the coefficient is an exponent, so the denominator is (0.250)² and Q = 20.0.
Working Q = [Y]/[X]² for 2 X ⇌ Y. Immediately after the injection, Q = 1.25/(0.250)² = 1.25/0.0625 = 20.0. (For comparison, K = 0.500/(0.250)² = 8.00, so Q > K and the net reaction will proceed toward X.)
A student studies the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) at 25°C. The concentration of the red FeSCN²⁺ ion can be measured with a spectrophotometer, and [Fe³⁺] and [SCN⁻] can then be calculated from the amounts mixed. The student wants to find out whether adding more SCN⁻ to an equilibrium mixture changes the value of K or changes only Q. Which procedure would answer the student's question?
Answer and reasoning
AImmediately after adding the SCN⁻, find all three concentrations and compare [FeSCN²⁺]/([Fe³⁺][SCN⁻]) with its value before the addition A student who thinks the ratio equals K at every moment picks this. Immediately after the addition the mixture is not at equilibrium, so the ratio measured then is Q, not K; it must be measured after the color has stopped changing.
BAfter the color stops changing, measure [FeSCN²⁺] alone and compare it with the [FeSCN²⁺] in the mixture before the addition A student who thinks a larger equilibrium concentration of product means a larger K picks this. [FeSCN²⁺] does rise, but [Fe³⁺] and [SCN⁻] also change, and only the full ratio shows whether K changed.
CAfter the color stops changing, find all three concentrations and compare [FeSCN²⁺]/([Fe³⁺][SCN⁻]) with its value before the additionCorrect K is the value of [FeSCN²⁺]/([Fe³⁺][SCN⁻]) at equilibrium. Once the color is steady the mixture is again at equilibrium, so comparing the ratio with its value before the addition shows whether K changed. (At constant temperature it does not: the addition changes Q, and the system returns to the same K.)
DImmediately after adding the SCN⁻, measure how quickly [FeSCN²⁺] rises and compare it with how quickly it rose in the original mixture A student who thinks a larger K means a faster reaction picks this. How quickly the color develops depends on the reactant concentrations and the kinetics, not on K, so a rate measurement cannot show whether K changed.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.10.A.1 Reaction quotient, Q Fix
Reaction quotient, Q
An expression with the same form as the equilibrium-constant expression (product concentrations or partial pressures over reactant concentrations or partial pressures, each raised to the power of its coefficient), evaluated with the values present at any moment. Q equals K only when the system is at equilibrium.
Disturbance (stress) to an equilibrium
A change in conditions, such as adding or removing a reactant or product, changing the volume, or changing the temperature, that makes Q differ from K and so takes the system out of equilibrium.
Direction of net reaction from Q and K
If Q < K, the net reaction proceeds toward products (Q rises toward K); if Q > K, the net reaction proceeds toward reactants (Q falls toward K); if Q = K, the system is at equilibrium and there is no net reaction.
New equilibrium state
The state reached after a disturbed system has responded: the concentrations or partial pressures have redistributed so that Q again equals K. The new concentrations generally differ from those before the disturbance.
Students often think Q is just another name for K, so the reaction quotient of a system equals K at every moment, including immediately after the system is disturbed. In fact No. Adding a gaseous or dissolved reactant or product changes one of the concentrations in the Q expression, so Q no longer equals K and the system is no longer at equilibrium. Q returns to K only as the net reaction proceeds.
Students often think The reaction quotient is simply the concentration of the product divided by the concentration of the reactant, with no powers, whatever the coefficients in the equation. In fact Yes. Q has the same form as K: each concentration or partial pressure is raised to the power of its coefficient in the balanced equation, so for 2 X ⇌ Y, Q = [Y]/[X]².
7.10.A.2 Stress that changes Q only Fix
Stress that changes Q only
Adding or removing a reactant or product at constant temperature changes Q but leaves K unchanged. A change of volume or a dilution, which changes several concentrations at once, also leaves K unchanged; it changes Q unless the exponents of the species affected add up to the same total in the numerator and the denominator of Q.
Temperature dependence of K
K has a fixed value only at a given temperature. A change in temperature changes the value of K, so a system that was at equilibrium is left with Q ≠ K, and the concentrations or partial pressures redistribute until Q equals the new K.
Le Châtelier's principle
A qualitative rule: when a system at equilibrium is disturbed, the net reaction proceeds in the direction that partly counteracts the disturbance. The comparison of Q with K gives the same prediction and shows why.
Students often think K changes when the concentrations change, so a larger equilibrium concentration of product after a reactant is added shows that K has increased. In fact No. At constant temperature K keeps its value. Adding a reactant makes Q smaller than K, the net reaction forms more product, and at the new equilibrium all the concentrations have adjusted so that the ratio in the Q expression equals the same K.
Students often think K is a constant for a reaction under all conditions, so a temperature change, like a concentration change, alters only Q. In fact Yes. Each value of K applies at one temperature. A change in temperature changes K; concentration and volume changes at constant temperature change Q only.
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
The hypothetical reaction R(g) ⇌ 2 P(g) was at equilibrium in a rigid container. At time t₁ a single change was made to the system. The graph shows [R] and [P] before and after t₁. Which statement identifies the change and correctly compares the value of K after t₁ with its value before t₁?
Answer and reasoning
AThe temperature was changed, and K has the same value after t₁ as before A student who thinks K does not depend on temperature picks this. A temperature change does change K: from the graph, K = [P]²/[R] falls from 0.10 before t₁ to 0.022 after t₁.
BA catalyst was added to the mixture, and K is smaller after t₁ than before A student who thinks a catalyst shifts an equilibrium picks this. A catalyst speeds up the forward and reverse reactions alike and does not change K, so it cannot move a system already at equilibrium to new concentrations.
CSome P was removed from the container, and K is the same after t₁ as before A student who reads the falling [P] curve as removal of P picks this. Removing P would make [P] drop suddenly at t₁ and then partly recover as R reacts; instead [P] falls smoothly while [R] rises, and K = [P]²/[R] changes from 0.10 to 0.022.
DThe temperature was changed, and K is smaller after t₁ than beforeCorrect Neither concentration jumps at t₁, so nothing was added or removed; instead both drift to new constant values, which happens when K itself changes, as it does with temperature. K = [P]²/[R] is (0.20)²/0.40 = 0.10 before t₁ and (0.10)²/0.45 = 0.022 after, so K is smaller.
Working No sudden jump in either concentration at t₁, so no R or P was added or removed and the volume was not changed; both concentrations drift to new constant values, so K itself changed: a temperature change. K = [P]²/[R]: before t₁, (0.20)²/0.40 = 0.10; after, (0.10)²/0.45 = 0.022. K is smaller after t₁. (Changes are consistent: R +0.05 M, P −0.10 M, a 1 : 2 ratio.)
The hypothetical reaction A(aq) + B(aq) ⇌ AB(aq) is at equilibrium in a solution at 25°C. The solution is then diluted with water to twice its original volume at 25°C. Which describes Q immediately after the dilution and the net reaction that follows?
Answer and reasoning
AQ = 2K; the net reaction then forms more A(aq) and B(aq)Correct Halving every concentration halves the numerator of Q = [AB]/([A][B]) but divides the denominator by 4, so Q = 2K. With Q > K the net reaction proceeds toward reactants, forming A and B until Q falls back to K.
BQ = K; the mixture stays at equilibrium after dilution A student who thinks diluting every species by the same factor leaves Q unchanged picks this. Q has one concentration factor in the numerator and two in the denominator, so halving them all doubles Q.
CQ = K/2; the net reaction then forms more AB(aq) A student who writes Q as reactants over products, [A][B]/[AB], finds that dilution halves it and so expects more AB to form. Q is products over reactants, [AB]/([A][B]), which doubles on dilution, so the net reaction forms A and B.
DQ = 2K; the net reaction then forms more AB(aq) A student who thinks Q > K drives the reaction toward products picks this. Q = 2K is correct, but when Q > K the net reaction proceeds toward reactants, lowering [AB] and raising [A] and [B] until Q = K.
Working Q = [AB]/([A][B]). After dilution every concentration is halved: Q = (½[AB])/((½[A])(½[B])) = 2[AB]/([A][B]) = 2K. Q > K, so the net reaction proceeds toward reactants, forming more A and B until Q again equals K.
A sealed glass flask contains an equilibrium mixture of brown NO₂(g) and colorless N₂O₄(g): 2 NO₂(g) ⇌ N₂O₄(g). The flask is moved from a water bath at 25°C to an ice-water bath at 0°C. The brown color becomes paler and then stays the same. Which claim about K = [N₂O₄]/[NO₂]² is supported by this observation?
Answer and reasoning
AK is the same at 0°C, because a temperature change alters only Q, as a concentration change does A student who thinks K does not depend on temperature picks this. Cooling the flask did not change any concentration directly, yet the mixture shifted to a new composition; that can happen only if K changed.
BK is greater at 0°C, because the new equilibrium mixture contains less NO₂ and more N₂O₄Correct The volume of the sealed flask is fixed, so the paler color means [NO₂] fell and [N₂O₄] rose as the new equilibrium was reached. Nothing was added or removed, so the shift shows that cooling changed K: [N₂O₄]/[NO₂]² is larger at the new equilibrium, so K is greater at 0°C.
CK is smaller at 0°C, because a reaction proceeds less far toward products at lower temperature A student who thinks cooling lowers K for every reaction picks this. Here cooling shifted the mixture toward the product, N₂O₄, as the paler color shows, so K is greater at 0°C.
DK is smaller at 0°C, because the forward and reverse reactions are both slower at 0°C A student who thinks K measures how fast a reaction is picks this. Both reactions are indeed slower at 0°C, but K describes the equilibrium composition, and the paler color shows a larger fraction of N₂O₄, so K is greater.
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
