1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
The table gives data for four hypothetical reactions at 25°C. Each reaction has the form R(aq) ⇌ P(aq), and each is started with R only. Which reaction proceeds essentially to completion?
Answer and reasoning
AReaction 1 A student who thinks K is the fraction of reactant converted, so that Kc = 1.0 means complete conversion, picks this. Kc = [P]/[R] = 1.0 means equal concentrations of P and R at equilibrium, so only about half of the R has reacted.
BReaction 2Correct Each reaction has the form R ⇌ P, so Kc = [P]/[R] at equilibrium. Reaction 2 has Kc = 6 × 10¹², so at equilibrium [P] is about 10¹² times [R]: essentially all of the R has been converted to P. Its long time to reach equilibrium describes the rate, not the extent, of the reaction.
CReaction 3 A student who thinks a reaction that reaches equilibrium quickly must go far toward products picks this, because Reaction 3 is the fastest. Its Kc of 0.20 means [P] is only one-fifth of [R] at equilibrium; how fast equilibrium is reached does not tell how far the reaction goes.
DReaction 4 A student who reads a large negative exponent as a large number picks this, because 10⁻¹⁴ has the largest exponent in the table. 3 × 10⁻¹⁴ is a very small number, so in Reaction 4 [P] is only about 3 × 10⁻¹⁴ times [R]: the reaction barely proceeds.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.5.A.1 Magnitude of K and extent of reaction Fix
Magnitude of K and extent of reaction
The size of K describes the relative amounts of products and reactants at equilibrium for the reaction as written: when K is much greater than 1, products predominate at equilibrium; when K is much less than 1, reactants predominate.
Reaction that proceeds essentially to completion
A reaction with a very large K: at equilibrium the limiting reactant is almost entirely converted to products, so its remaining concentration is very small (though not exactly zero) compared with the concentrations of the products.
Reaction that barely proceeds
A reaction with a very small K: at equilibrium only a tiny fraction of the reactants has been converted, so the product concentrations are very small (though not zero) and the reactant concentrations are almost unchanged from their initial values.
Approximations for very large or very small K
Because a very small K means almost no reactant is used, the equilibrium reactant concentration can be taken as its initial value; because a very large K means the limiting reactant is almost used up, the product concentration can be found from the stoichiometry of a complete reaction.
Extent versus rate
K describes how far a reaction proceeds before equilibrium is reached, not how fast it gets there; a reaction with a very large K can be slow, and one with a small K can reach equilibrium quickly.
Students often think A reaction with a very small K does not occur at all, so the equilibrium mixture contains no product. In fact No. A very small K means the reaction barely proceeds: at equilibrium the products are present at very low concentrations, but they are present, because the forward and reverse reactions are both occurring.
Students often think A reaction with a very large K goes completely to completion, so at equilibrium the concentration of the limiting reactant is exactly zero. In fact Almost, but not completely. A very large K means the reaction proceeds essentially to completion: the remaining concentration of the limiting reactant is very small but not zero, as the K expression requires.
4 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 4
For the reaction N₂(g) + O₂(g) ⇌ 2 NO(g), the value of Kc at 25°C is of the order of 10⁻³⁰. A sealed container of N₂(g) and O₂(g) is kept at 25°C until equilibrium is reached. Which claim about the contents of the container at equilibrium, and its justification, is correct?
Answer and reasoning
AN₂, O₂ and NO are present in equal amounts, because the system is at equilibrium A student who thinks reactants and products are present in equal amounts at equilibrium picks this. At equilibrium the forward and reverse rates are equal; with Kc about 10⁻³⁰ the amount of NO is tiny compared with the amounts of N₂ and O₂.
BAlmost all of the N₂ and O₂ has become NO, because the system is at equilibrium A student who thinks that reaching equilibrium means the reaction has gone to completion picks this. Being at equilibrium means only that there is no further net change; the very small Kc shows that the equilibrium lies far toward N₂ and O₂.
CNo NO at all is present, because a reaction with so small a Kc does not occur A student who thinks a reaction with a very small K does not occur at all picks this. A very small K means the reaction barely proceeds: NO is present at equilibrium, at a very low concentration, because both the forward and reverse reactions occur.
DOnly a very small amount of NO is present, because Kc is extremely smallCorrect Kc = [NO]²/([N₂][O₂]) is about 10⁻³⁰, so at equilibrium [NO]² is about 10⁻³⁰ times [N₂][O₂]. The reaction barely proceeds: the mixture is almost entirely N₂ and O₂, with a very small but nonzero concentration of NO.
Working No calculation needed. Kc = [NO]²/([N₂][O₂]) ≈ 10⁻³⁰, far less than 1, so at equilibrium [NO]² is about 10⁻³⁰ times [N₂][O₂]: the reaction barely proceeds and NO is present only at a very low concentration, but not zero.
A solution of the hypothetical substance A is prepared at 25°C, and A reacts according to the equation A(aq) ⇌ B(aq). The graph shows [A] and [B] as functions of time. Which statement about Kc for the reaction at 25°C is best supported by the graph?
Answer and reasoning
AKc is much greater than 1Correct At equilibrium [B] = 0.40 M while [A] is too small to see on the graph, so Kc = [B]/[A] is 0.40 M divided by a very small concentration: much greater than 1. Essentially all of the A has been converted to B, as expected for a very large K.
BKc is much less than 1 A student who thinks a large K favors reactants, reading K as reactants over products, picks this. Kc = [B]/[A] has the product in the numerator, and here [B] is far greater than [A] at equilibrium, so Kc is much greater than 1.
CKc is approximately 1 A student who thinks K is the fraction of reactant converted, so that complete conversion means K = 1, picks this. Kc = 1 would mean [B] = [A] = 0.20 M at equilibrium; complete conversion corresponds to a very large Kc.
DKc is roughly equal to 0.40 A student who thinks K equals the equilibrium concentration of the product picks this, reading 0.40 M off the [B] curve. Kc is the ratio [B]/[A], and [A] at equilibrium is very small, so Kc is much greater than 1.
Working No calculation needed beyond reading the graph. Both curves level off after about 25 min: [B] = 0.40 M and [A] is too small to distinguish from zero. Kc = [B]/[A] = 0.40/(a very small number), so Kc is much greater than 1; the reaction proceeds essentially to completion.
A student plans to mix equal volumes of 0.20 M X(aq) and 0.10 M Y(aq) at 25°C, where they react according to the hypothetical equation X(aq) + Y(aq) ⇌ Z(aq), Kc = 5 × 10¹¹ at 25°C. After the mixture reaches equilibrium, the student will measure the concentrations of the three species. Which prediction about the equilibrium [Y] is correct?
Answer and reasoning
A[Y] will be exactly zero, because all of the Y is used up A student who thinks a reaction with a very large K uses up the limiting reactant completely picks this. The reaction proceeds essentially to completion, but at equilibrium Kc = [Z]/([X][Y]) requires a small nonzero [Y], about 2 × 10⁻¹² M.
B[Y] will remain near 0.050 M, because Y barely reacts with X A student who reads a very large K as favoring the reactants picks this. Kc has the product in the numerator, so a value of 5 × 10¹¹ means Z is strongly favored and almost all of the Y reacts.
C[Y] will be very small but not zero, because Kc is very largeCorrect After mixing, [X] = 0.10 M and [Y] = 0.050 M. With Kc = 5 × 10¹¹ the reaction proceeds essentially to completion, so [Z] ≈ 0.050 M and [X] ≈ 0.050 M, and [Y] = [Z]/(Kc[X]) ≈ 2 × 10⁻¹² M: tiny, but not zero.
D[Y] will be 0.025 M, because [Y] and [Z] become equal at equilibrium A student who thinks reactants and products have equal concentrations at equilibrium picks this, setting 0.050 − x = x. At equilibrium the forward and reverse rates are equal, not the concentrations; with Kc = 5 × 10¹¹ almost all of the Y is converted to Z.
Working After mixing, [X]₀ = 0.10 M and [Y]₀ = 0.050 M. Kc is very large, so the reaction proceeds essentially to completion with Y limiting: [Z] ≈ 0.050 M and [X] ≈ 0.050 M. Then [Y] = [Z]/(Kc[X]) = 0.050/((5 × 10¹¹)(0.050)) = 2 × 10⁻¹² M: very small, not zero. Distractors: complete consumption gives exactly 0; reading a large K as favoring reactants gives [Y] ≈ 0.050 M; equal [Y] and [Z] gives 0.050 − x = x, x = 0.025 M.
A solution initially contains the hypothetical substance R at concentration [R]₀ and no P. R reacts according to the equation R(aq) ⇌ P(aq), for which Kc = 1 × 10⁻⁹ at 25°C. Which expression gives the best estimate of the equilibrium [P]?
Answer and reasoning
A[R]₀/Kc A student who writes K as reactants over products, Kc = [R]/[P], picks this. With Kc = [P]/[R], dividing by a very small Kc would give a [P] far larger than [R]₀, which is impossible here.
BKc × [R]₀Correct Kc = [P]/[R]. Because Kc is very small, the reaction barely proceeds and the equilibrium [R] is almost exactly [R]₀, so [P] = Kc[R] ≈ Kc[R]₀, a very small concentration.
CKc A student who thinks K equals the equilibrium concentration of the product picks this. Kc is the ratio [P]/[R], so [P] also depends on [R], which is about [R]₀.
D½[R]₀ A student who thinks reactants and products are present in equal amounts at equilibrium picks this, setting [P] = [R] so that half of the R has reacted. With Kc = 1 × 10⁻⁹, [P] is about 10⁻⁹ times [R], not equal to it.
Working Kc = [P]/[R] = 1 × 10⁻⁹. The reaction barely proceeds, so [R] at equilibrium ≈ [R]₀ (only about one R in 10⁹ reacts). Then [P] = Kc[R] ≈ Kc[R]₀. Distractors: inverted expression gives [P] = [R]₀/Kc; K read as [P] gives [P] = Kc; equal amounts gives [P] = [R] = ½[R]₀.
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
