AP Chemistry Unit 5 Review

• Rate = k[X][Y₂]

• Rate = k[X]²[Y₂]

• Rate = k[X][Y₂]²

• Rate = k[X]²[Y₂]²

Answer explanation

Comparing experiment 1 with experiment 2, a doubling of [Y₂]_i while keeping [X]_i constant resulted in a doubling of the reaction rate. Thus, the reaction is first order with respect to Y₂. A similar comparison of experiment 2 with experiment 3 reveals that the reaction is first order with respect to X. Thus, the exponent of the concentrations of reactants in the rate law are both equal to 1.

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NGSS.HS-PS1-5

• 16 M/s

• 32 M/s

• 64 M/s

• 128 M/s

Answer explanation

The initial rate of disappearance of X is equal to twice the initial rate of appearance of X₂Y₂ and given by the following: (Δ[X]/Δt)=−2 × (Δ[X₂Y₂]/Δt) = 2 × (initial rate of reaction)= 2× 32 M/s = 64 M/s. The negative sign indicates that the concentration of X is decreasing.

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NGSS.HS-PS1-5

• The pressures of the gases used by the second chemist must have been lower than those used by the first chemist, thus the collisions between reacting particles were less frequent than they were in the first chemist's experiments.

• The pressures of the gases used by the second chemist must have been lower than those used by the first chemist, thus the number of collisions with sufficient energy to cause reaction was lower than it was in the first chemist's experiments.

• The second chemist must have added a catalyst for the reaction, thus providing a different reaction pathway for the reactant particles to react with an activation energy that was lower than that of the uncatalyzed reaction in the first chemist's experiments.

• The second chemist must have added a catalyst for the reaction, thus providing energy to reactant particles to increase their rate of reaction compared to their rate of reaction in the first chemist's experiments.

Answer explanation

The presence of a catalyst is the most likely cause of the increased rates measured by the second chemist. An alternative reaction pathway with a lower activation energy would explain the greater reaction rates compared to the rates measured by the first chemist.

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• Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.

• 
  Decreasing the temperature will increase the frequency of the collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.

• Increasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.

• Decreasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.

Answer explanation

The rate changes proportionally to the change in the concentration of sucrose and the increase in the rate of hydrolysis can be explained by an increase in the frequency of the collisions between the reactant molecules.

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NGSS.HS-PS1-5

• The activation energy of the reaction is smaller in trial 2 than it is in trial 1.

• The frequency of collisions between reactant molecules is greater in trial 2 than it is in trial 1.

• The value of the rate constant for the reaction is smaller in trial 2 than it is in trial 1.

• The value of the rate constant for the reaction is greater in trial 2 than it is in trial 1.

Answer explanation

The concentration of H₂ in trial 2 has been doubled. As reactant concentration increases, the frequency of collisions between reactant molecules increases. This leads to an increase in reaction rate.

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• Trial 1, because there is a higher concentration of Zn(s) in the reaction mixture.

• Trial 1, because the sample of Zn(s) has less surface area for the reaction to take place.

• Trial 2, because there is a higher concentration of HCl(aq) in the reaction mixture.

• Trial 2, because the sample of Zn(s) has a greater surface area for the reaction to take place.

Answer explanation

The sample of powdered Zn(s) in trial 2 has a greater surface area than the piece of Zn(s) in trial 1; therefore, more Zn atoms are exposed to HCl(aq) in trial 2 than in trial 1, leading to a faster initial reaction rate.

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NGSS.HS-PS1-5

• The reaction will proceed at a slower rate with increasing temperature.

• The rate of the reaction will double when the concentrations of both CH₃I and NaOH are doubled.

• The rate of the reaction will double if the concentration of CH₃I is doubled while keeping the concentration of NaOH constant.

• 
  A larger amount of CH₃OH will be produced if the concentrations of CH₃I and NaOH are halved.

Answer explanation

Based on the information given, doubling the concentration of CH₃I while keeping the concentration of NaOH constant will double the rate of the reaction. Based on collision theory, the rate is faster as a result of the increased frequency of collisions when the concentration is increased.

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• 
  Zeroth order

• First order

• Second order

• Third order

Answer explanation

For a first-order reaction, graphing ln[N₂O₅] versus time produces a straight line.

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• n_Bi

• n_Po

• ln(n_Bi)

• 1/n_Bi

Answer explanation

Because the reaction is a first-order radioactive decay, a plot of n_Bi versus time would produce a straight line.

Tags

NGSS.HS-PS1-8

• Rate = k

• Rate = k[NH₃]

• Rate = k[NH₃]₂

• Rate = k[N₂][H₂]

Answer explanation

The plots of [NH₃] versus time for the two different initial concentrations of ammonia are straight lines with identical slopes. The slope is equal to Δ[NH₃]/Δt, which is constant and equal to the rate of reaction. In other words, the rate is independent of [NH₃], and the reaction is zeroth order with Rate=k.

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NGSS.HS-PS1-5