3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A student wants to determine the average rate, in mol/s, at which Mg(s) is consumed when a strip of magnesium reacts with HCl(aq). Which measurements does the student need?
Answer and reasoning
AThe mass of the Mg strip before and after the reaction A student who thinks the rate is the amount of reactant converted picks this. The two masses give the moles of Mg consumed, but without the time over which it was consumed no rate can be found.
BThe mass of Mg that reacts and the time it takes to reactCorrect The rate is the amount converted per unit time: moles of Mg consumed (from the mass that reacted, divided by 24.30 g/mol) divided by the time taken. Both measurements are needed.
CThe times at which bubbling begins and at which it stops A student who thinks the rate is the time a reaction takes picks this. The two times give the duration, but a rate in mol/s also needs the moles of Mg consumed in that time.
DThe starting mass of Mg and the starting volume of HCl(aq) A student who thinks the rate is set by the total amounts of reactants present picks this. Starting amounts do not say how much Mg was consumed in a measured time, which is what the rate is.
Working Rate in mol/s = (moles of Mg consumed)/(time). Moles of Mg consumed = (mass of Mg that reacted)/(24.30 g/mol), so the mass that reacted and the time over which it reacted are both needed.
The hypothetical reaction 2 X(g) → 5 Y(g) + Z(g) takes place in a sealed rigid container. The graph shows [X] as a function of time. What is the average rate of formation of Y over the first 100 s?
Answer and reasoning
A4.0 × 10⁻⁴ M s⁻¹ A student who thinks every species changes concentration at the same rate picks this, the rate of disappearance of X. The balanced equation gives 5 Y for every 2 X, so [Y] rises 2.5 times as fast as [X] falls.
B2.0 × 10⁻³ M s⁻¹ A student who multiplies the rate for X by the coefficient of Y alone picks this: 5 × 4.0 × 10⁻⁴. The ratio of coefficients is 5 : 2, so the rate for X is multiplied by 5/2, not by 5.
C1.5 × 10⁻³ M s⁻¹ A student who finds a rate by dividing the concentration at one time by that time picks this: (5/2) × (0.060 M/100 s). The average rate uses the change in [X], 0.100 M − 0.060 M, not the value at 100 s.
D1.0 × 10⁻³ M s⁻¹Correct [X] falls from 0.100 M to 0.060 M in 100 s, so X disappears at an average of 0.040 M/100 s = 4.0 × 10⁻⁴ M s⁻¹. Five moles of Y form for every two moles of X used, so Y forms 2.5 times as fast: 1.0 × 10⁻³ M s⁻¹.
Working From the graph, [X] falls from 0.100 M at 0 s to 0.060 M at 100 s. Average rate of disappearance of X = (0.100 − 0.060) M / 100 s = 4.0 × 10⁻⁴ M s⁻¹. Y forms 5 mol per 2 mol X, so rate(Y) = (5/2)(4.0 × 10⁻⁴) = 1.0 × 10⁻³ M s⁻¹. Distractors: equal rates, 4.0 × 10⁻⁴; multiplied by 5 only, 2.0 × 10⁻³; [X] at 100 s divided by time, ×(5/2): 1.5 × 10⁻³.
When a small amount of solid MnO₂ is added to a solution of hydrogen peroxide, H₂O₂(aq), bubbles of O₂(g) form rapidly as the H₂O₂ decomposes into H₂O and O₂. MnO₂ acts as a catalyst. Which statement about the MnO₂ is correct?
Answer and reasoning
AMnO₂ increases the total amount of O₂ that a sample of H₂O₂ can produce. A student who thinks a catalyst increases the amount of product picks this. The amount of O₂ is set by the amount of H₂O₂ that decomposes; MnO₂ only makes that amount form faster.
BMnO₂ speeds up the decomposition and is not used up in the reaction.Correct A catalyst increases the rate of a reaction without being used up overall, so the MnO₂ can be recovered at the end. It does not change the products or how much O₂ the H₂O₂ can give.
CMnO₂ is used up as it reacts, so its atoms end up as part of the products. A student who thinks a catalyst is a reactant that is consumed picks this. A catalyst is not used up in the overall reaction; the products are H₂O and O₂, and the MnO₂ remains.
DMnO₂ causes a decomposition of H₂O₂ that does not occur in its absence. A student who thinks a catalyst makes possible a reaction that cannot otherwise occur picks this. H₂O₂ solutions decompose slowly on their own; the catalyst makes the same reaction go much faster.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.1.A.1 Reaction rate Fix
Reaction rate
The amount of a reactant converted to products (or of a product formed) per unit of time, often expressed as a change in concentration per unit time, in M s⁻¹.
Average rate over a time interval
The change in a concentration divided by the time over which it occurred, Δ[ ]/Δt; for a reactant the change is negative, so the rate of disappearance is reported as −Δ[reactant]/Δt.
Rate versus amount of product
The rate describes how fast products form, not how much product forms in the end; the final amount is set by the amounts of reactants and the stoichiometry.
Students often think The rate of a reaction at a time is the concentration at that time divided by the time. In fact No. An average rate is the CHANGE in concentration divided by the CHANGE in time between two points; the concentration at one time also includes the starting value.
Students often think The rate of a reaction is the amount of reactant converted or product formed, so measuring how much reacts is enough to give the rate. In fact No. The rate is the amount converted PER UNIT TIME; the same amount converted in a shorter time is a faster rate.
5.1.A.2 Stoichiometric relationship between rates Fix
Stoichiometric relationship between rates
The rates of change of the concentrations of the species in a reaction are in the same ratio as their coefficients: for 2 X → 4 Y + Z, [Y] increases twice as fast as [X] decreases, and [Z] increases half as fast as [X] decreases.
Students often think Every reactant and product in a reaction changes concentration at the same rate, whatever the coefficients in the balanced equation. In fact No. The rates are in the ratio of the coefficients: for N₂ + 3 H₂ → 2 NH₃, [H₂] falls three times as fast as [N₂], and [NH₃] rises twice as fast as [N₂] falls.
Students often think A species with a larger coefficient changes concentration more slowly, so the ratio of rates is the inverse of the ratio of coefficients. In fact No. A larger coefficient means more moles of that species react or form for each event of the reaction, so its concentration changes faster.
5.1.A.3 Factors that influence reaction rate Fix
Factors that influence reaction rate
Reactant concentrations, temperature, the surface area of a solid reactant, the presence of a catalyst and other environmental factors all influence how fast a reaction proceeds.
Surface area of a solid reactant
Dividing a solid into smaller pieces exposes more of its particles at the surface, where they can meet particles of the other reactant, so the reaction is faster even though the mass of solid is the same.
Catalyst
A substance that increases the rate of a reaction without being used up in the overall reaction; it does not change which products form or how much product the reactants can give.
Effect of concentration on rate
A more concentrated solution has more reactant particles per unit volume, so they meet the other reactant more often; the concentration, not the total amount of solution, is what matters.
Students often think The rate of a reaction depends on the total amount of each reactant present, so samples containing the same number of moles react at the same rate and a smaller sample reacts more slowly. In fact No. For a dissolved reactant it is the concentration, the amount per unit volume, that affects the rate; a larger volume of the same solution contains more reactant but reacts no faster at the start.
Students often think A higher concentration or a larger surface area makes the reaction faster because each collision between reactant particles is more energetic. In fact No. At the same temperature the particles have the same distribution of energies; a higher concentration or larger surface area makes contacts between reactant particles more frequent, not more energetic.
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
Identical strips of magnesium are placed in the two samples of HCl(aq) shown, at the same temperature. Each box in the figure shows an equal small volume of solution next to the metal surface. Which claim about the initial rate of H₂(g) production, with its reasoning, is correct?
Answer and reasoning
AFaster in the 2.0 M acid, because H₃O⁺ ions in a more concentrated acid strike the Mg surface more energetically. A student who thinks a higher concentration makes collisions more energetic picks this. At the same temperature the ions have the same range of energies; the higher concentration makes contacts with the metal more frequent, not harder.
BThe same in both, because each sample contains 0.10 mol of H₃O⁺ in total, so the same amount of acid is able to react. A student who thinks the rate depends on the total amount of reactant rather than its concentration picks this. Both samples do contain 0.10 mol of acid, but the metal meets only the ions near its surface, and there are twice as many of those in the 2.0 M acid.
CFaster in the 2.0 M acid, because more H₃O⁺ ions are next to the Mg surface, so they collide with it more frequently.Correct The 2.0 M acid has twice as many H₃O⁺ ions per unit volume, as the boxes show, so more ions reach the metal surface each second and react with Mg atoms. The macroscopic result is faster production of H₂.
DFaster in the 1.0 M acid, because H₃O⁺ ions in a more dilute acid have more room to move to the Mg surface. A student who thinks particles move more freely in a dilute solution picks this. The ions move at speeds set by the temperature; with half as many ions per unit volume, fewer reach the metal each second.
A student measures the volume of H₂(g) produced over time when a 0.10 g strip of magnesium ribbon reacts with excess 1.0 M HCl(aq). The experiment is then repeated with 0.10 g of magnesium powder, with everything else the same. In each graph the solid curve is the ribbon trial. Which numbered graph best shows the result of the powder trial as the dashed curve?
Answer and reasoning
AGraph 1Correct The powder exposes far more Mg surface to the acid, so H₂ forms faster at the start (a steeper curve). The mass of Mg, the limiting reactant, is unchanged, so the same final volume of H₂ is produced, just sooner.
BGraph 2 A student who thinks a faster reaction produces more product picks this. The powder does react faster, but Mg is the limiting reactant and its mass is unchanged, so the final volume of H₂ is the same.
CGraph 3 A student who thinks only concentration and temperature affect the rate picks this. Dividing the magnesium into a powder exposes more Mg atoms to the acid, so H₂ forms faster.
DGraph 4 A student who thinks a powder has a smaller surface area than a ribbon of the same mass picks this. Each grain is small, but together the grains expose much more surface, so the powder reacts faster, not more slowly.
A solid reacts with an aqueous solution in an exothermic reaction. The same mass of solid, fully covered by solution, is used in each trial. Which change would decrease the initial rate of the reaction?
Answer and reasoning
ACooling the solution before adding the solidCorrect Lowering the temperature lowers the rate of a reaction such as this one; whether the reaction releases or absorbs heat does not change that, so the initial rate decreases.
BGrinding the solid into a fine powder A student who thinks smaller pieces have a smaller surface area picks this. A powder exposes more surface than the same mass in larger pieces, so grinding increases the rate.
CUsing a smaller volume of the same solution A student who thinks the rate depends on the total amount of reactant picks this. The concentration is unchanged, and the solid is still fully covered, so the initial rate is the same.
DWarming the solution before the solid is added A student who thinks warming slows an exothermic reaction picks this. Raising the temperature increases the rate of a reaction such as this one, whether it releases or absorbs heat.
Nitrogen and hydrogen react according to the equation N₂(g) + 3 H₂(g) → 2 NH₃(g). Which statement correctly relates the rates at which the concentrations change while the reaction proceeds?
Answer and reasoning
A[N₂], [H₂] and [NH₃] change at the same rate. A student who thinks every species changes concentration at the same rate picks this. The rates follow the coefficients 1 : 3 : 2, so H₂ is used up fastest.
B[N₂] decreases three times as fast as [H₂] decreases. A student who inverts the ratio of coefficients picks this. H₂ has the larger coefficient, so more H₂ reacts in each interval, and [H₂] falls three times as fast as [N₂].
C[NH₃] increases two times as fast as [H₂] decreases. A student who multiplies the rate for H₂ by the coefficient of NH₃ alone picks this. The ratio of coefficients is 2 : 3, so [NH₃] increases only two-thirds as fast as [H₂] decreases.
D[H₂] decreases three times as fast as [N₂] does.Correct Three moles of H₂ react for every mole of N₂, so in any time interval the change in [H₂] is three times the change in [N₂]: [H₂] falls three times as fast.
Working Rates of change are in the ratio of the coefficients: −Δ[N₂]/Δt : −Δ[H₂]/Δt : Δ[NH₃]/Δt = 1 : 3 : 2. So [H₂] falls three times as fast as [N₂]; [NH₃] rises 2/3 as fast as [H₂] falls.
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
