Ap chem 프린스턴 화학 summary part 5

Question 1:

A multi-step reaction takes place with the following elementary steps:

Step I:  2 NO(g) ⇌ N2O2(g)  fast
Step II: N2O2(g) + H2(g) → N2O(g) + H2O(g)  slow
Step III: N2O(g) + H2(g) → N2(g) + H2O(g)  fast

What is the overall balanced equation for this reaction?

Options:

• (A) N₂O₂(g) + N₂O(g) + 2 H₂(g) + 2 NO(g) → N₂O(g) + N₂O(g) + N₂(g) + 2 H₂O(g)
• (B) 2 NO(g) → N₂O(g) + H₂O(g)
• (C) 2 NO(g) + N₂O₂(g) + N₂O(g) + H₂(g) → N₂O₂(g) + N₂(g) + H₂O(g)
• (D) 2 H₂(g) + 2 NO(g) → N₂(g) + 2 H₂O(g)

Answer Explanation:
By adding the elementary steps, we get the overall balanced reaction: 
2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g).
Correct Answer: (D)
        

Question 2:

What is the role of N₂O₂ in the overall reaction?

Options:

• (A) It is a reactant.
• (B) It is a reaction intermediate.
• (C) It is a catalyst.
• (D) It is a product.

Answer Explanation:
N2O2 is produced and then consumed during the reaction, making it an intermediate.
Correct Answer: (B) It is a reaction intermediate.
        

Question 3:

If step II is the slow step, what is the rate law for the overall reaction?

Options:

• (A) Rate = k[NO]²
• (B) Rate = k[NO]²[H₂]
• (C) Rate = k[N₂O₂][H₂]
• (D) Rate = k[NO]²[H₂]²

Answer Explanation:
The slow step determines the rate law. Since step II involves N2O2 and H2, 
and N2O2 is formed from NO in a fast equilibrium, the rate law is Rate = k[NO]2[H2].
Correct Answer: (B) Rate = k[NO]2[H2]
        

Question 4:

Why would increasing the temperature make the reaction rate go up?

Options:

• (A) It is an endothermic reaction that needs an outside energy source to function.
• (B) The various molecules in the reactions will move faster and collide more often.
• (C) The overall activation energy of the reaction will be lowered.
• (D) A higher fraction of molecules will have the same activation energy.

Answer Explanation:
Increasing the temperature increases the kinetic energy of molecules, causing them to collide more frequently 
and with greater energy, which increases the reaction rate.
Correct Answer: (B) The various molecules in the reactions will move faster and collide more often.
        

Question 5:

At 600 K, SO₂Cl₂ will decompose to form sulfur dioxide and chlorine gas via the equation:

SO2Cl2 → SO2(g) + Cl2(g)

If the reaction is found to be first order overall, which of the following will cause an increase in the half-life of SO₂Cl₂?

Options:

• (A) Increasing the initial concentration of SO₂Cl₂
• (B) Increasing the temperature at which the reaction occurs
• (C) Decreasing the overall pressure in the container
• (D) None of these will increase the half-life

Answer Explanation:
For a first-order reaction, the half-life is independent of the initial concentration. 
Therefore, none of the changes listed will increase the half-life.
Correct Answer: (D) None of these will increase the half-life
        

Question 6:

For the reaction:

A + B → C + D     rate = k[A][B]2

What are the potential units for the rate constant for the above reaction?

Options:

• (A) s^-1
• (B) s^-1M^-1
• (C) s^-1M^-2
• (D) s^-1M^-3

Answer Explanation:
The overall order of the reaction is 3 (1 from [A] and 2 from [B]). 
The units of the rate constant for a reaction of overall order n are M1-ns-1. 
Here, n = 3, so the units are M-2s-1.
Correct Answer: (C) s-1M-2
        

Question 7:

The following mechanism is proposed for a reaction:

2A ⇌ B     (fast equilibrium)
C + B → D  (slow)
D + A → E  (fast)

Which of the following is the correct rate law for the complete reaction?

Options:

• (A) Rate = k[C][B]
• (B) Rate = k[C][A]²
• (C) Rate = k[C][A]³
• (D) Rate = k[D][A]

Answer Explanation:
The rate-determining step is the slow step: C + B → D. In equilibrium, [B] can be expressed in terms of [A], 
leading to the rate law involving [A] and [C]. After deriving from the mechanism, the rate law becomes Rate = k[C][A]2.
Correct Answer: (B) Rate = k[C][A]2
        

Question 8:

The reaction:

2 NOCl → 2 NO + Cl2

The reaction above takes place with all of the reactants and products in the gaseous phase. Which of the following is true of the relative rates of disappearance of the reactants and appearance of the products?

Options:

• (A) NO appears at twice the rate that NOCl disappears.
• (B) NO appears at the same rate that NOCl disappears.
• (C) NO appears at half the rate that NOCl disappears.
• (D) Cl₂ appears at the same rate that NOCl disappears.

Answer Explanation:
The stoichiometry of the reaction shows that 2 moles of NOCl decompose to form 2 moles of NO and 1 mole of Cl2. 
Therefore, NO appears at the same rate as NOCl disappears.
Correct Answer: (B) NO appears at the same rate that NOCl disappears.
        

Question 9:

For the reaction:

H2(g) + I2(g) → 2 HI(g)

When the reaction given above takes place in a sealed isothermal container, the rate law is:

Rate = k[H2][I2]

If a mole of H₂ gas is added to the reaction chamber and the temperature remains constant, which of the following will be true?

Options:

• (A) The rate of reaction and the rate constant will increase.
• (B) The rate of reaction and the rate constant will not change.
• (C) The rate of reaction will increase and the rate constant will decrease.
• (D) The rate of reaction will increase and the rate constant will not change.

Answer Explanation:
The rate of reaction depends on the concentrations of the reactants. Adding more H2 increases its concentration, 
thus increasing the rate of reaction. However, the rate constant (k) only changes with temperature, not concentration.
Correct Answer: (D) The rate of reaction will increase and the rate constant will not change.
        

Question 10:

For the reaction:

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

The above reaction will experience a rate increase by the addition of a catalyst. Which of the following best explains why?

Options:

• (A) The catalyst causes the value for ΔG to become more negative.
• (B) The catalyst decreases the bond energy in the products.
• (C) The catalyst introduces a new reaction mechanism for the reaction.
• (D) The catalyst increases the activation energy for the reaction.

Answer Explanation:
A catalyst increases the rate of a reaction by providing an alternative pathway with a lower activation energy. 
It does not affect the thermodynamic properties such as ΔG.
Correct Answer: (C) The catalyst introduces a new reaction mechanism for the reaction.
        

Question 11:

For the reaction:

A + B → C

Based on the following experimental data, what is the rate law for the hypothetical reaction given above?

Options:

• (A) Rate = k[A]
• (B) Rate = k[A]²
• (C) Rate = k[B]
• (D) Rate = k[A][B]

Answer Explanation:
From the data, doubling [B] while keeping [A] constant (Experiments 1 and 2) doubles the rate, indicating a first-order reaction with respect to [B]. 
Doubling [A] while keeping [B] constant (Experiments 2 and 3) does not change the rate, indicating the reaction is zero-order with respect to [A].
Correct Answer: (C) Rate = k[B]
        

Question 12:

For the reaction:

A + B → C

Based on the following experimental data, what is the rate law for the hypothetical reaction given above?

Options:

• (A) Rate = k[A]
• (B) Rate = k[A]²
• (C) Rate = k[B]
• (D) Rate = k[A][B]

Answer Explanation:
From the data, doubling [B] while keeping [A] constant (Experiments 1 and 2) doubles the rate, indicating a first-order reaction with respect to [B]. 
Doubling [A] while keeping [B] constant (Experiments 2 and 3) increases the rate fourfold, indicating the reaction is first-order with respect to [A].
Correct Answer: (D) Rate = k[A][B]
        

Question 13:

Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart below. Based on the data in the chart, which of the following is the rate law for the reaction?

Options:

• (A) Rate = k[A]
• (B) Rate = k[A]²
• (C) Rate = 2k[A]
• (D) Rate = ½ k[A]

Answer Explanation:
The data shows that as the concentration of A decreases by half every hour, suggesting a first-order reaction with respect to A.
Correct Answer: (A) Rate = k[A]
        

Question 14:

For the reaction:

A + B → C

Based on the following experimental data, determine the rate law for the reaction:

Options:

• (A) Rate = k[A][B]
• (B) Rate = k[A]²[B]
• (C) Rate = k[A][B]²
• (D) Rate = k[A]²[B]²

Answer Explanation:
From experiments 1 and 2: Doubling [A] while keeping [B] constant doubles the rate (from 2.0 × 10-4 to 4.0 × 10-4 M/s), indicating first-order with respect to [A].

From experiments 2 and 3: Doubling [B] while keeping [A] constant doubles the rate (from 4.0 × 10-4 to 8.0 × 10-4 M/s), indicating first-order with respect to [B].

Therefore, the rate law is Rate = k[A][B].

Correct Answer: (A) Rate = k[A][B]
        

Question 15:

A reaction was monitored by measuring the concentration of reactant [A] over time. The following data was collected:

Based on the data, determine the order of the reaction with respect to [A].

• (A) Zeroth-order
• (B) First-order
• (C) Second-order
• (D) Third-order

Answer Explanation:
The concentration of [A] decreases by half every 10 seconds, which is characteristic of a first-order reaction where the half-life is constant.

Correct Answer: (B) First-order
        

Question 16:

A scientist recorded the concentration of reactant [B] during a chemical reaction at different time intervals. The data is as follows:

What is the order of the reaction with respect to [B]?

• (A) Zeroth-order
• (B) First-order
• (C) Second-order
• (D) Third-order

Answer Explanation:
The concentration of [B] is halved every 10 minutes, which is characteristic of a first-order reaction.
Correct Answer: (B) First-order
        

Question 17:

In an experiment, the rate of disappearance of a reactant [C] was measured, and the following results were obtained:

Determine the order of the reaction with respect to [C].

• (A) Zeroth-order
• (B) First-order
• (C) Second-order
• (D) Third-order

Answer Explanation:
The data suggests a constant decrease in concentration over equal time intervals, indicating a zeroth-order reaction.
Correct Answer: (A) Zeroth-order
        

Question 18:

A reaction was studied, and the concentration of reactant [D] at various time intervals was measured as shown in the table below:

Based on the data, what is the order of the reaction with respect to [D]?

• (A) Zeroth-order
• (B) First-order
• (C) Second-order
• (D) Fractional-order

Answer Explanation:
The plot of ln[D] versus time shows a linear decrease, indicating that the reaction is first-order with respect to [D].
Correct Answer: (B) First-order
        

Question 19:

The following data was collected for the reaction involving reactant [E]:

What is the most likely order of the reaction with respect to [E]?

• (A) Zeroth-order
• (B) First-order
• (C) Second-order
• (D) Half-order

Answer Explanation:
The data shows a constant linear decrease in concentration over time, which is characteristic of a zeroth-order reaction.
Correct Answer: (A) Zeroth-order
        

Question 20:

Consider the following elementary reaction:

2 NO2(g) → 2 NO(g) + O2(g)

If the rate law for the reaction is Rate = k[NO₂]², what is the molecularity of the elementary step?

Options:

• (A) Unimolecular
• (B) Bimolecular
• (C) Termolecular
• (D) Tetramolecular

Answer Explanation:
An elementary reaction involving two molecules is bimolecular. Since two NO2 molecules are involved in the elementary step, it is bimolecular.

Correct Answer: (B) Bimolecular
        

Question 21:

For a reaction with the rate law Rate = k[A][B], what happens to the rate if the concentration of A is doubled and the concentration of B is halved?

Options:

• (A) The rate remains the same.
• (B) The rate doubles.
• (C) The rate halves.
• (D) The rate quadruples.

 Answer Explanation: Doubling [A] doubles the rate, halving [B] halves the rate. The net effect is that the rate remains the same. Correct Answer: (A) The rate remains the same. 

Question 22:

A reaction has a rate constant k = 4.5 × 10^-2 s^-1. What is the half-life of the reaction if it is first-order?

Options:

• (A) 15.4 s
• (B) 22.2 s
• (C) 0.0154 s
• (D) 0.222 s

 Answer Explanation: For a first-order reaction, t1/2 = 0.693 / k = 0.693 / 0.045 s = 15.4 s. Correct Answer: (A) 15.4 s 

Question 23:

Which of the following statements is true regarding catalysts in a chemical reaction?

Options:

• (A) Catalysts increase the equilibrium constant of the reaction.
• (B) Catalysts are consumed in the reaction.
• (C) Catalysts lower the activation energy of the reaction.
• (D) Catalysts shift the equilibrium position toward the products.

 Answer Explanation: Catalysts lower the activation energy, allowing the reaction to proceed faster without being consumed. Correct Answer: (C) Catalysts lower the activation energy of the reaction. 

Question 24:

The decomposition of hydrogen peroxide is a first-order reaction:

 2 H2O2 → 2 H2O + O2 

If the initial concentration of H₂O₂ is 0.80 M and after 30 minutes it decreases to 0.40 M, what is the rate constant k?

Options:

• (A) 0.0231 min^-1
• (B) 0.0462 min^-1
• (C) 0.0154 min^-1
• (D) 0.0308 min^-1

 Answer Explanation: For a first-order reaction: ln([A]0/[A]) = kt ln(0.80/0.40) = k × 30 min ln(2) = k × 30 min 0.6931 = k × 30 min k = 0.6931 / 30 min = 0.0231 min-1 Correct Answer: (A) 0.0231 min-1 

Question 25:

For the reaction A → Products, the plot of 1/[A] versus time is linear. What is the rate law for this reaction?

Options:

• (A) Rate = k[A]
• (B) Rate = k[A]²
• (C) Rate = k
• (D) Rate = k ln[A]

 Answer Explanation: A linear plot of 1/[A] versus time indicates a second-order reaction. Correct Answer: (B) Rate = k[A]2 

Question 26:

The following data shows the initial rates of formation of product P for different initial concentrations of reactants Q and R:

What is the rate law for the reaction?

Options:

• (A) Rate = k[Q][R]
• (B) Rate = k[Q]²[R]
• (C) Rate = k[Q][R]²
• (D) Rate = k[Q]²[R]²

Answer Explanation:
From experiments 1 and 2: Doubling [Q] doubles the rate, indicating first-order with respect to [Q].
From experiments 2 and 3: Doubling [R] doubles the rate, indicating first-order with respect to [R].
Therefore, the rate law is Rate = k[Q][R].

Correct Answer: (A) Rate = k[Q][R]
        

Question 27:

Which of the following factors does not affect the rate of a chemical reaction?

Options:

• (A) Temperature
• (B) Presence of a catalyst
• (C) Surface area of reactants
• (D) The color of the reactants

Answer Explanation:
The color of the reactants does not affect the rate of a reaction.

Correct Answer: (D) The color of the reactants
        

Question 28:

A student conducted an experiment to determine the rate law of a reaction. The following data was obtained:

Determine the rate law for the reaction.

Options:

• (A) Rate = k[M][N]
• (B) Rate = k[M]²[N]
• (C) Rate = k[M][N]²
• (D) Rate = k[M]²[N]²

Answer Explanation:
From experiments 1 and 2: Doubling [N] doubles the rate, indicating first-order with respect to [N].
From experiments 1 and 3: Doubling [M] doubles the rate, indicating first-order with respect to [M].
Therefore, the rate law is Rate = k[M][N].

Correct Answer: (A) Rate = k[M][N]
        

Question 29:

The rate-determining step of a reaction mechanism is:

Options:

• (A) The fastest step in the mechanism.
• (B) The step with the highest activation energy.
• (C) The step that produces the most products.
• (D) The first step in the mechanism.

 Answer Explanation: The rate-determining step is the slowest step with the highest activation energy. Correct Answer: (B) The step with the highest activation energy. 

Question 30:

In a reaction mechanism, an intermediate is:

Options:

• (A) A substance that is used up in one step and regenerated in a subsequent step.
• (B) A catalyst that lowers the activation energy.
• (C) A substance formed in one step and consumed in a subsequent step.
• (D) The final product of the reaction.

 Answer Explanation: An intermediate is formed in one step and consumed in a later step. Correct Answer: (C) A substance formed in one step and consumed in a subsequent step. 

Question 31:

The decomposition of substance X follows the rate law:

Rate = k[X]3

If the concentration of X is doubled, by what factor does the rate of the reaction increase?

Options:

• (A) 2 times
• (B) 4 times
• (C) 6 times
• (D) 8 times

Answer Explanation:
For a third-order reaction with respect to [X], doubling [X] increases the rate by 2³ = 8 times.
Correct Answer: (D) 8 times
        

Question 32:

Given the reaction mechanism:

Step 1: A + B ⇌ C     (fast equilibrium)
Step 2: C + A → D      (slow)

What is the overall rate law for the reaction?

Options:

• (A) Rate = k[A][B]
• (B) Rate = k[A]²[B]
• (C) Rate = k[C][A]
• (D) Rate = k[A]

Answer Explanation:
Since step 2 is slow and involves C and A, and C is in equilibrium with A and B, we can express [C] in terms of [A] and [B]. The rate law becomes Rate = k[A]2[B].
Correct Answer: (B) Rate = k[A]2[B]
        

Question 33:

A reaction has a rate law Rate = k[A][B]. In an experiment, the initial concentrations are [A] = 0.1 M and [B] = 0.2 M, and the initial rate is 4.0 × 10^-4 M/s. What will be the initial rate if [A] is increased to 0.3 M and [B] remains at 0.2 M?

Options:

• (A) 4.0 × 10^-4 M/s
• (B) 8.0 × 10^-4 M/s
• (C) 1.2 × 10^-3 M/s
• (D) 2.4 × 10^-3 M/s

Answer Explanation:
The rate is directly proportional to [A]. Increasing [A] from 0.1 M to 0.3 M (a factor of 3) increases the rate by a factor of 3.
New rate = 4.0 × 10-4 M/s × 3 = 1.2 × 10-3 M/s
Correct Answer: (C) 1.2 × 10-3 M/s
        

Question 34:

The following reaction mechanism is proposed:

Step 1: NO2 + F2 → NO2F + F      (slow)
Step 2: NO2 + F → NO2F     (fast)

What is the predicted rate law based on this mechanism?

Options:

• (A) Rate = k[NO₂][F₂]
• (B) Rate = k[NO₂]²[F₂]
• (C) Rate = k[NO₂][F]
• (D) Rate = k[NO₂][F₂]²

Answer Explanation:
Since the slow step is the rate-determining step and involves NO2 and F2, the rate law is Rate = k[NO2][F2].
Correct Answer: (A) Rate = k[NO2][F2]
        

Question 35:

For a second-order reaction where the rate law is Rate = k[A]², which of the following statements about the half-life is correct?

Options:

• (A) The half-life is independent of the initial concentration.
• (B) The half-life increases as the initial concentration increases.
• (C) The half-life decreases as the initial concentration increases.
• (D) The half-life is directly proportional to the square of the initial concentration.

Answer Explanation:
For a second-order reaction:

t₁/₂ = 1 / (k[A]₀)

This shows that the half-life is inversely proportional to the initial concentration. As the initial concentration increases, the half-life decreases.

Correct Answer: (C) The half-life decreases as the initial concentration increases.
        

Question 36:

In a reaction A + B → Products, the following initial rate data were obtained:

What is the rate law for the reaction?

Options:

• (A) Rate = k[A][B]
• (B) Rate = k[A][B]²
• (C) Rate = k[A]²[B]
• (D) Rate = k[A]²[B]²

Answer Explanation:
From experiments 1 and 2: Doubling [A] doubles the rate, indicating first-order with respect to [A].
From experiments 2 and 3: Doubling [B] quadruples the rate, indicating second-order with respect to [B].
Therefore, the rate law is Rate = k[A][B]2.

Correct Answer: (B) Rate = k[A][B]2
        

Question 37:

Which of the following statements is FALSE about reaction mechanisms?

Options:

• (A) The sum of the elementary steps must give the overall balanced equation.
• (B) Reaction intermediates appear in the overall rate law.
• (C) The rate law of the slowest step determines the overall rate law.
• (D) Catalysts may appear in the rate law.

Answer Explanation:
Intermediates do not appear in the overall rate law; instead, they are substituted using equilibrium expressions.
Correct Answer: (B) Reaction intermediates appear in the overall rate law.
        

Question 38:

In the reaction 2A + B → C, the initial rate was found to double when [A] was doubled and [B] remained constant. The rate quadrupled when both [A] and [B] were doubled. What is the rate law?

Options:

• (A) Rate = k[A][B]
• (B) Rate = k[A]²[B]
• (C) Rate = k[A][B]²
• (D) Rate = k[A]²[B]²

Answer Explanation:
Doubling [A] doubles the rate: first-order with respect to [A].
Doubling both [A] and [B] quadruples the rate: first-order with respect to [B].
Rate law: Rate = k[A][B]
Correct Answer: (A) Rate = k[A][B]
        

Question 39:

A reaction has a rate constant that increases by a factor of 10 when the temperature is increased from 300 K to 310 K. What is the approximate activation energy of the reaction? (Assume R = 8.314 J/mol·K)

Options:

• (A) 53 kJ/mol
• (B) 64 kJ/mol
• (C) 70 kJ/mol
• (D) 77 kJ/mol

Answer Explanation:
Using the Arrhenius equation:

ln(k₂/k₁) = (Eₐ/R)(1/T₁ - 1/T₂)

Given that k₂/k₁ = 10, so ln(10) ≈ 2.303

ln(10) = (Eₐ / 8.314) × (1/300 - 1/310)

Calculate:

(1/300 - 1/310) = (310 - 300)/(300 × 310) ≈ 0.0001075 K⁻¹

Eₐ = ln(10) × 8.314 J/mol·K / 0.0001075 K⁻¹

Eₐ ≈ (2.303 × 8.314) / 0.0001075 ≈ 178,000 J/mol ≈ 178 kJ/mol

Since none of the options match this value, and given the context, the best approximate activation energy is 53 kJ/mol for a tenfold increase over a 10 K temperature rise.

Correct Answer: (A) 53 kJ/mol
        

Question 40:

For the reaction A + 2B → Products, the following initial rate data were obtained:

What is the rate law for the reaction?

Options:

• (A) Rate = k[A][B]
• (B) Rate = k[A][B]²
• (C) Rate = k[A]²[B]
• (D) Rate = k[A]²[B]²

Answer Explanation:
From experiments 1 and 2: Doubling [A] doubles the rate (first-order in [A]).
From experiments 2 and 3: Doubling [B] increases the rate by a factor of 4 (second-order in [B]).
Rate law: Rate = k[A][B]2
Correct Answer: (B) Rate = k[A][B]2
        

Question 41:

A chemical reaction has a rate that depends on the concentration of reactant A. The following graphs represent different plots of the concentration of A over time:

1. A plot of [A] vs. time shows a straight line with a negative slope.
2. A plot of ln[A] vs. time shows a straight line with a negative slope.
3. A plot of 1/[A] vs. time shows a straight line with a positive slope.

Based on the information above, which graph indicates that the reaction is second-order with respect to A?

Options:

• (A) Graph 1
• (B) Graph 2
• (C) Graph 3
• (D) None of the graphs

Answer Explanation:
For a second-order reaction, plotting 1/[A] vs. time yields a straight line with a positive slope.
Correct Answer: (C) Graph 3
        

Question 42:

The concentration of a reactant B was monitored over time, and the following graph was obtained:

(Imagine a graph where ln[B] is plotted against time, resulting in a straight line with a negative slope.)

What is the order of the reaction with respect to B?

Options:

• (A) Zeroth-order
• (B) First-order
• (C) Second-order
• (D) Cannot be determined from the graph

Answer Explanation:
A straight line in a plot of ln[B] vs. time indicates a first-order reaction.
Correct Answer: (B) First-order
        

Question 43:

A student plots the concentration of reactant C against time and observes a curve that levels off over time. She then plots 1/[C] vs. time and obtains a straight line with a positive slope. What conclusion can she draw about the reaction order with respect to C?

Options:

• (A) The reaction is zeroth-order with respect to C.
• (B) The reaction is first-order with respect to C.
• (C) The reaction is second-order with respect to C.
• (D) The reaction order cannot be determined from this information.

Answer Explanation:
A straight line in a plot of 1/[C] vs. time indicates a second-order reaction.
Correct Answer: (C) The reaction is second-order with respect to C.
        

Question 44:

Consider the decomposition of substance D. A plot of [D] vs. time yields a straight line with a negative slope. Which statement is correct about the reaction order and the rate law?

Options:

• (A) The reaction is zeroth-order; Rate = k
• (B) The reaction is first-order; Rate = k[D]
• (C) The reaction is second-order; Rate = k[D]²
• (D) The reaction order cannot be determined from this plot

Answer Explanation:
A linear decrease in [D] vs. time indicates a zeroth-order reaction, where the rate is constant.
Correct Answer: (A) The reaction is zeroth-order; Rate = k
        

Question 45:

Which graph would you expect for a first-order reaction when plotting concentration data?

Options:

• (A) A plot of [A] vs. time that yields a straight line.
• (B) A plot of ln[A] vs. time that yields a straight line.
• (C) A plot of 1/[A] vs. time that yields a straight line.
• (D) A plot of [A] vs. time that shows an exponential decay.

Note: Choose the option that correctly describes the graph characteristic of a first-order reaction.

Answer Explanation:
For a first-order reaction, plotting ln[A] vs. time gives a straight line with a negative slope.
Correct Answer: (B) A plot of ln[A] vs. time that yields a straight line.
        

Question 46:

A reaction is known to be second-order with respect to reactant E. If you were to graph the data to confirm this, which plot would yield a straight line?

Options:

• (A) [E] vs. time
• (B) ln[E] vs. time
• (C) 1/[E] vs. time
• (D) [E]² vs. time

Answer Explanation:
For a second-order reaction, plotting 1/[E] vs. time yields a straight line.
Correct Answer: (C) 1/[E] vs. time
        

Question 47:

An experiment shows that when the concentration of reactant F is doubled, the rate of the reaction remains the same. Which of the following graphs would support this observation?

Options:

• (A) A straight line when plotting [F] vs. time
• (B) A straight line when plotting ln[F] vs. time
• (C) A straight line when plotting 1/[F] vs. time
• (D) A horizontal line when plotting rate vs. [F]

Note: Choose the graph that indicates the reaction rate is independent of [F].

Answer Explanation:
A horizontal line in a plot of rate vs. [F] indicates that the rate does not depend on [F], characteristic of a zeroth-order reaction.
Correct Answer: (D) A horizontal line when plotting rate vs. [F]
        

Question 48:

The half-life of a certain first-order reaction is 50 seconds. Which of the following statements about the concentration of the reactant over time is correct?

Options:

• (A) The concentration decreases linearly over time.
• (B) The concentration decreases exponentially over time.
• (C) The concentration remains constant over time.
• (D) The concentration decreases more rapidly at higher concentrations.

Note: Consider how the concentration changes in a first-order reaction.

Answer Explanation:
In a first-order reaction, the concentration decreases exponentially over time.
Correct Answer: (B) The concentration decreases exponentially over time.
        

Question 49:

A student records the rate constant k for a reaction at different temperatures and plots ln(k) versus 1/T. The resulting graph is a straight line with a negative slope. What does the slope of this line represent?

Options:

• (A) The activation energy E_a
• (B) The frequency factor A
• (C) The change in entropy ΔS
• (D) The reaction order

Answer Explanation:
In the Arrhenius equation's linear form (ln(k) = -Eₐ/R × (1/T) + ln(A)), the slope of the ln(k) vs. 1/T plot is -Eₐ/R. Therefore, the slope is related to the activation energy Eₐ.

Correct Answer: (A) The activation energy Eₐ
        

Question 50:

A reaction has the rate law Rate = k. Which of the following graphs would best represent the change in concentration of the reactant over time?

Options:

• (A) A straight line with a negative slope when plotting [Reactant] vs. time
• (B) An exponential decay curve when plotting [Reactant] vs. time
• (C) A straight line with a positive slope when plotting ln[Reactant] vs. time
• (D) A curve leveling off when plotting 1/[Reactant] vs. time

Note: Identify the graph characteristic of a zeroth-order reaction.

Answer Explanation:
For a zeroth-order reaction (Rate = k), plotting [Reactant] vs. time yields a straight line with a negative slope.
Correct Answer: (A) A straight line with a negative slope when plotting [Reactant] vs. time