Question
\(C_{12}H_{22}O_{11}(aq)+H_2O(l)→2C_6H_{12}O_6(aq)\)
The chemical equation shown above represents the hydrolysis of sucrose. Under certain conditions, the rate is directly proportional to the concentration of sucrose. Which statement supports how a change in conditions can increase the rate of this reaction?
A 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.
B 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.
C 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.
D 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
Ans: C
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.
Question
The information in the data table above represents two different trials for an experiment to study the rate of the reaction between NO(g) and \(H_2\)(g), as represented by the balanced equation above the table. Which of the following statements provides the correct explanation for why the initial rate of formation of \(N_2\) is greater in trial 2 than in trial 1? Assume that each trial is carried out at the same constant temperature.
A The activation energy of the reaction is smaller in trial 2 than it is in trial 1.
B The frequency of collisions between reactant molecules is greater in trial 2 than it is in trial 1.
C The value of the rate constant for the reaction is smaller in trial 2 than it is in trial 1.
D The value of the rate constant for the reaction is greater in trial 2 than it is in trial 1.
▶️Answer/Explanation
Ans: B
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.
Question
\(Zn(s)+2 HCl(aq)→ZnCl_2(aq)+H_2(g)\)
Zn(s) reacts with HCl(aq) according to the equation shown above. In trial 1 of a kinetics experiment, a 5.0g piece of Zn(s) is added to 100mL of 0.10MHCl(aq). The rate of reaction between Zn(s) and HCl(aq) is determined by measuring the volume of H2(g) produced over time. In trial 2 of the experiment, 5.0g of powdered Zn(s) is added to 100mL of 0.10M HCl(aq). Which trial will have a faster initial rate of reaction and why?
A Trial 1, because there is a higher concentration of Zn(s) in the reaction mixture.
B Trial 1, because the sample of Zn(s)has less surface area for the reaction to take place.
C Trial 2, because there is a higher concentration of HCl(aq) in the reaction mixture.
D Trial 2, because the sample of Zn(s) has a greater surface area for the reaction to take place.
▶️Answer/Explanation
Ans:D
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.
Question
Consider the reaction represented by the equation \(2X + 2Z → X_2Z_2\). During a reaction in which a large excess of reactant X was present, the concentration of reactant Z was monitored over time. A plot of the natural logarithm of the concentration of Z versus time is shown in the figure above. The order of the reaction with respect to reactant Z is
(A) zero order
(B) first order
(C) second order
(D) third order
▶️Answer/Explanation
Ans:B