1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
The reversible reaction A(g) ⇌ B(g) takes place in a sealed, rigid container at constant temperature. The particulate diagrams show the same small region of the container at four successive times. Which statement correctly describes the reactions in the container between time 3 and time 4?
Answer and reasoning
AThe forward reaction is faster than the reverse reaction A student who reads 'more B than A' as 'the forward reaction is faster' picks this. If the forward reaction were faster, the number of B particles would still be rising; it stays at 8, so the rates are equal.
BThe reverse reaction is now faster than the forward reaction A student who thinks the substance present in the larger amount must react faster picks this, because there are twice as many B particles as A particles. If the reverse reaction were faster, the number of A particles would rise; it stays at 4, so the rates are equal.
CThe forward and reverse reactions have both come to a stop A student who thinks reactions stop at equilibrium picks this, because the numbers of A and B no longer change. A and B are both still present, so both reactions continue; the numbers stay constant because the two reactions occur at equal rates.
DThe forward and reverse reactions are occurring at equal ratesCorrect From time 1 to time 3 the number of A particles falls (10, 7, 4) and the number of B particles rises (2, 5, 8): a net conversion of A to B, so the forward rate is the greater. Between time 3 and time 4 the numbers stay at 4 A and 8 B, so there is no net conversion: the forward and reverse reactions are occurring at equal rates.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.2.A.1 Forward and reverse reactions Fix
Forward and reverse reactions
In a reversible reaction written as reactants ⇌ products, the forward reaction converts reactants to products and the reverse reaction converts products to reactants. Both take place at the same time in the same mixture.
Net conversion
The overall change that results from the forward and reverse reactions together. If the forward rate is greater than the reverse rate there is a net conversion of reactants to products; if the reverse rate is greater there is a net conversion of products to reactants.
Direction of a reversible reaction
The direction of net conversion at a given moment. It is decided by which of the two rates, forward or reverse, is greater at that moment, not by which substance is present in the larger amount.
Equilibrium state
The state reached when the rates of the forward and reverse reactions are equal. Both reactions continue, there is no net conversion, and the amounts of reactants and products stay constant; the amounts need not be equal to one another.
Students often think When more product than reactant is present, the forward reaction is the faster one, because it is the reaction that has made the most. In fact No. The amounts present do not show which rate is greater. In a mixture at equilibrium that contains more product than reactant, the forward and reverse rates are equal; what the larger amount of product shows is how far the reaction went before the rates became equal.
Students often think Whichever substance is present in the larger amount reacts faster, so the reaction that consumes it is the faster of the two. In fact No. A rate depends on the rate constant of that reaction as well as on the amount of the substance reacting. For a reaction A ⇌ B in which each direction is a single step (rates kf[A] and kr[B]), the substance present in the larger amount at equilibrium is the one consumed by the reaction with the smaller rate constant, and the two rates are equal.
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
The reversible reaction A(g) ⇌ B(g) takes place in a rigid container at constant temperature. Each direction is a single elementary step, so the rate of the forward reaction is kf[A] and the rate of the reverse reaction is kr[B]. The mixture is at equilibrium when half of the B is suddenly removed from the container. How do the rates immediately after the removal compare with the rates at equilibrium?
Answer and reasoning
AThe forward rate is unchanged, and the reverse rate is halvedCorrect The forward rate, kf[A], depends on [A], which has not changed, so it is unchanged. The reverse rate, kr[B], is halved because [B] is halved. The forward rate is now the greater, so there is a net conversion of A to B until the two rates are equal again.
BThe forward rate is doubled, and the reverse rate is unchanged A student who thinks the forward reaction speeds up to replace the B that was removed picks this. The forward rate depends only on [A], which has not changed; it is the reverse rate that changes, because [B] is halved.
CThe forward rate is halved, and the reverse rate is halved A student who thinks the forward and reverse rates are always equal picks this, so that a change in one rate is matched by the other. The rates are equal only at equilibrium; removing B halves the reverse rate and leaves the forward rate as it was.
DThe forward rate is unchanged, and the reverse rate is unchanged A student who thinks rates are fixed at a given temperature picks this. The rate constants are unchanged, but the reverse rate, kr[B], also depends on [B], which has been halved.
Working Forward rate = kf[A]: [A] and kf are unchanged at that moment, so the forward rate is unchanged. Reverse rate = kr[B]: [B] is halved and kr is unchanged, so the reverse rate is halved. The forward rate is now greater than the reverse rate, so there is a net conversion of A to B until the rates are equal again.
A student models the reaction A(g) ⇌ B(g), which takes place in a sealed container that at first holds mostly A, with the diagrams shown. The model accounts for the observation that the amount of B in the container increases until time 3 and then stays constant. Which statement correctly describes a further feature or a limitation of the model?
Answer and reasoning
AIt does not show either the forward or the reverse reaction occurring at time 3 A student who thinks the reactions stop, or cancel, at equilibrium picks this. The arrows at time 3 are equal in length but not zero: the model shows both reactions occurring at equal rates.
BIt does not show how the amounts of A and B in the container compare at time 3Correct The arrows represent rates. A longer forward arrow at times 1 and 2 means a net conversion of A to B, and equal arrows at time 3 mean no net conversion, which accounts for the observation. But equal rates do not mean equal amounts, and nothing in the model shows how much A or B is present.
CIt shows that the amounts of A and B present in the container are equal at time 3 A student who thinks equal rates mean equal amounts picks this. The equal arrows at time 3 show equal rates; the amounts of A and B are constant from then on but need not be equal.
DIt shows that there is a net conversion of B to A between time 2 and time 3 A student who reads a growing reverse arrow and a shrinking forward arrow as the reverse reaction winning picks this. Between time 2 and time 3 the forward arrow is still the longer one, so there is still a net conversion of A to B.
A sample of colorless N₂O₄(g) is placed in a sealed, rigid flask at constant temperature, where the reaction N₂O₄(g) ⇌ 2 NO₂(g) occurs. NO₂(g) is brown. The brown color of the mixture deepens for several minutes and then stops changing. Which claim about the period during which the color is deepening is best supported?
Answer and reasoning
ANO₂ forms but is not yet consumed, so the reverse reaction has not yet begun A student who thinks the reverse reaction starts only once equilibrium is reached picks this. As soon as some NO₂ is present it begins to react to form N₂O₄; the color deepens because NO₂ forms faster than it is consumed.
BNO₂ forms faster and faster, so the forward rate rises until the color stops changing A student who thinks the forward reaction speeds up as a reaction approaches equilibrium picks this. The forward rate falls as N₂O₄ is used up; the color stops changing when the falling forward rate and the rising reverse rate become equal.
CNO₂ forms faster than it is used up, so the forward reaction is the faster oneCorrect A deepening brown color shows that the amount of NO₂ is increasing: there is a net conversion of N₂O₄ to NO₂. That happens when the forward reaction, which forms NO₂, is faster than the reverse reaction, which consumes it.
DNO₂ is used up at a rising rate, so the reverse rate exceeds the forward rate A student who takes a rising reverse rate to mean that the reverse reaction is the faster one picks this. The reverse rate does rise as NO₂ accumulates, but while the color is deepening it is still smaller than the forward rate.
Hydrogen iodide is formed in the reversible reaction H₂(g) + I₂(g) ⇌ 2 HI(g). A sealed, rigid container at 700 K initially contains only HI(g). Which prediction about the system before it reaches equilibrium is correct?
Answer and reasoning
AThe forward reaction is faster than the reverse reaction, and the amount of HI increases A student who thinks a reaction always approaches equilibrium in the direction in which the equation is written picks this. With no H₂ or I₂ present at the start, the forward reaction cannot be the faster one.
BThe forward and reverse reactions are equally fast, and the amount of HI stays the same A student who thinks the forward and reverse rates of a reversible reaction are always equal picks this. The rates are equal only at equilibrium; at the start the forward rate is zero and the reverse rate is not.
CThe forward and reverse reactions do not take place, and the amount of HI is unchanged A student who thinks the reverse reaction can only follow the forward reaction picks this, since there are no reactants for the forward reaction to use. HI molecules react to form H₂ and I₂ whenever HI is present.
DThe reverse reaction is faster than the forward one, and the amount of HI decreasesCorrect At first no H₂ or I₂ is present, so the forward rate is zero, while HI is reacting to form H₂ and I₂. The reverse rate is greater than the forward rate, so there is a net conversion of products to reactants and the amount of HI decreases until the rates become equal.
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
