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 or bond energies.
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 dependence on [B].
- Doubling [A] while keeping [B] constant (Experiments 2 and 3) does not change the rate, indicating zero-order dependence on [A].
The rate law is therefore Rate = k[B].

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 dependence on [B].
- Doubling [A] while keeping [B] constant (Experiments 2 and 3) quadruples the rate, indicating a first-order dependence on [A].
The rate law is therefore Rate = k[A][B].

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, which is characteristic of a first-order reaction. 
The rate law is therefore Rate = k[A].

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

Question 14:

The following data was recorded for the decomposition of a substance:

What is the order of the reaction?

Options:

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

Answer Explanation:
The concentration of B halves at equal intervals, which indicates a first-order reaction. 
The half-life is constant throughout the data, confirming the first-order behavior.

Correct Answer: (B) First-order
        

Question 15:

For the reaction:

H2(g) + I2(g) → 2 HI(g)

In an experiment, the rate law was found to be:

Rate = k[H2][I2]

If the concentration of H₂ is doubled while the concentration of I₂ is halved, what happens to the rate of the reaction?

Options:

• (A) The rate remains the same.
• (B) The rate is doubled.
• (C) The rate is halved.
• (D) The rate is quadrupled.

Answer Explanation:
The rate law is Rate = k[H2][I2]. If [H2] is doubled, the rate increases by a factor of 2. 
If [I2] is halved, the rate decreases by a factor of 2. The net effect is that the rate remains unchanged.

Correct Answer: (A) The rate remains the same.
        

Question 16:

The following reaction was studied:

2 A + B → Products

The rate law for the reaction was found to be:

Rate = k[A]2[B]

If the concentration of A is increased by a factor of 3 while B is kept constant, how does the rate of the reaction change?

Options:

• (A) The rate increases by a factor of 3.
• (B) The rate increases by a factor of 9.
• (C) The rate increases by a factor of 27.
• (D) The rate remains the same.

Answer Explanation:
The rate law shows a second-order dependence on [A], so the rate changes by a factor of [A]2. 
If [A] is increased by a factor of 3, the rate increases by 32 = 9.

Correct Answer: (B) The rate increases by a factor of 9.
        

Question 17:

The decomposition of hydrogen peroxide (H₂O₂) was studied, and the reaction was found to follow first-order kinetics:

2 H2O2(aq) → 2 H2O(l) + O2(g)

Which of the following graphs would produce a straight line?

Options:

• (A) [H₂O₂] vs. time
• (B) ln[H₂O₂] vs. time
• (C) 1/[H₂O₂] vs. time
• (D) [H₂O₂]² vs. time

Answer Explanation:
For a first-order reaction, a plot of ln[H2O2] vs. time yields a straight line with a negative slope.

Correct Answer: (B) ln[H2O2] vs. time
        

Question 18:

In an experiment, the decomposition of a substance X was monitored, and the following data were collected:

What is the order of the reaction, and what is the half-life of X?

Options:

• (A) Zeroth-order; 10 s
• (B) First-order; 10 s
• (C) Second-order; 20 s
• (D) First-order; 20 s

Answer Explanation:
The concentration of X decreases by half every 10 seconds, which is characteristic of a first-order reaction. 
For a first-order reaction, the half-life is constant and does not depend on the initial concentration.

Correct Answer: (B) First-order; 10 s
        

Question 19:

A scientist is studying the reaction between two reactants, A and B, and collects the following data:

What is 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] doubles the rate, indicating first-order dependence on [A].
From experiments 2 and 3: Doubling [B] doubles the rate, indicating first-order dependence on [B].
The rate law is therefore Rate = k[A][B].

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

Question 20:

The decomposition of dinitrogen tetroxide (N₂O₄) is represented by the following equation:

N2O4(g) ⇌ 2 NO2(g)

The reaction is studied at equilibrium, and the following observations are made:

• When the temperature is increased, the concentration of NO₂ increases.
• When the pressure is increased, the concentration of N₂O₄ increases.

What conclusions can be drawn about the thermodynamics of the reaction?

Options:

• (A) The reaction is exothermic and involves a decrease in the number of gas molecules.
• (B) The reaction is endothermic and involves an increase in the number of gas molecules.
• (C) The reaction is exothermic and involves an increase in the number of gas molecules.
• (D) The reaction is endothermic and involves a decrease in the number of gas molecules.

Answer Explanation:
- Increasing temperature shifts the equilibrium toward the products, suggesting the reaction is endothermic.
- Increasing pressure shifts the equilibrium toward the reactants, where fewer gas molecules are present.

The reaction is therefore endothermic and involves an increase in the number of gas molecules.

Correct Answer: (B) The reaction is endothermic and involves an increase in the number of gas molecules.
        

Question 21:

The reaction 2 NO₂(g) ⇌ N₂O₄(g) is studied at equilibrium. The following changes are made:

• The temperature is increased.
• The pressure is decreased.

What will happen to the equilibrium position?

Options:

• (A) The equilibrium will shift to the left, favoring NO₂.
• (B) The equilibrium will shift to the right, favoring N₂O₄.
• (C) The equilibrium will not shift.
• (D) The equilibrium will favor both sides equally.

Answer Explanation:
- Increasing the temperature favors the endothermic direction. Since the decomposition of N2O4 into NO2 is endothermic, the equilibrium shifts to the left.
- Decreasing the pressure favors the side with more gas molecules. NO2 has more molecules than N2O4, so the equilibrium also shifts to the left.

Correct Answer: (A) The equilibrium will shift to the left, favoring NO2.
        

Question 22:

For the reaction:

A + B → C

The rate law is Rate = k[A][B]. What happens to the rate of the reaction if the concentration of A is tripled and the concentration of B is doubled?

Options:

• (A) The rate increases by a factor of 2.
• (B) The rate increases by a factor of 3.
• (C) The rate increases by a factor of 6.
• (D) The rate increases by a factor of 9.

Answer Explanation:
The rate law is Rate = k[A][B]. If [A] is tripled and [B] is doubled:
- Rate ∝ (3[A])(2[B]) = 6 times the original rate.

Correct Answer: (C) The rate increases by a factor of 6.
        

Question 23:

The decomposition of ozone, O₃, occurs in the presence of chlorine atoms (Cl) as a catalyst:

Step 1: O3 + Cl → O2 + ClO (slow)
Step 2: ClO + O → Cl + O2 (fast)

What is the role of chlorine (Cl) in this reaction?

Options:

• (A) Chlorine is a reactant.
• (B) Chlorine is a product.
• (C) Chlorine is an intermediate.
• (D) Chlorine is a catalyst.

Answer Explanation:
Chlorine is consumed in the first step and regenerated in the second step, meaning it is not used up in the overall reaction. This makes it a catalyst.

Correct Answer: (D) Chlorine is a catalyst.
        

Question 24:

A reaction has a rate constant k = 2.5 × 10^-2 s^-1. How long will it take for the concentration of the reactant to decrease to 25% of its initial value in a first-order reaction?

Options:

• (A) 13.9 s
• (B) 27.7 s
• (C) 34.7 s
• (D) 46.2 s

Answer Explanation:
For a first-order reaction: ln([A]0/[A]) = kt
ln(1/0.25) = (2.5 × 10-2)t
ln(4) = 2.5 × 10-2t
t = ln(4) / (2.5 × 10-2)
t ≈ 1.386 / 0.025 ≈ 55.4 s

Correct Answer: (A) 13.9 s
        

Question 25:

For the reaction A → Products, a plot of ln[A] vs. time is linear with a slope of -0.1. What is the rate constant (k) for this reaction?

Options:

• (A) 0.01 s^-1
• (B) 0.05 s^-1
• (C) 0.10 s^-1
• (D) 1.00 s^-1

Answer Explanation:
For a first-order reaction: ln[A] = -kt + ln[A]0
The slope of the line is equal to -k. Given the slope is -0.1, the rate constant k = 0.1 s-1.

Correct Answer: (C) 0.10 s-1
        

Question 26:

The reaction:

2 NO(g) + O2(g) → 2 NO2(g)

was studied, and the rate law was determined to be:

Rate = k[NO]2[O2]

What will happen to the rate of reaction if the concentration of NO is doubled and the concentration of O₂ is tripled?

Options:

• (A) The rate will double.
• (B) The rate will increase by a factor of 12.
• (C) The rate will increase by a factor of 24.
• (D) The rate will remain constant.

Answer Explanation:
The rate law is Rate = k[NO]2[O2].
- Doubling [NO] increases the rate by a factor of 22 = 4.
- Tripling [O2] increases the rate by a factor of 3.

The total increase in rate is 4 × 3 = 12.

Correct Answer: (B) The rate will increase by a factor of 12.
        

Question 27:

For a reaction with the rate law Rate = k[A][B]², what is the overall order of the reaction?

Options:

• (A) 1
• (B) 2
• (C) 3
• (D) 4

Answer Explanation:
The overall order of the reaction is the sum of the exponents in the rate law. For Rate = k[A][B]2, the order is 1 (from [A]) + 2 (from [B]) = 3.

Correct Answer: (C) 3
        

Question 28:

A reaction has the following mechanism:

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

What is the rate law for the reaction based on this mechanism?

Options:

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

Answer Explanation:
The slow step is the rate-determining step, and the rate law is based on the reactants involved in that step. Step 1 involves A and B, so the rate law is Rate = k[A][B].

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

Question 29:

The decomposition of hydrogen peroxide (H₂O₂) in the presence of iodide ions (I^–) as a catalyst occurs in two steps:

Step 1: H2O2 + I- → H2O + IO- (slow)
Step 2: H2O2 + IO- → H2O + O2 + I- (fast)

What is the rate law for this reaction?

Options:

• (A) Rate = k[H₂O₂]
• (B) Rate = k[H₂O₂][I^–]
• (C) Rate = k[H₂O₂][IO^–]
• (D) Rate = k[H₂O₂]²

Answer Explanation:
The rate-determining step is the slow step: H2O2 + I- → H2O + IO-. The rate law is based on the reactants in this step, giving Rate = k[H2O2][I-].

Correct Answer: (B) Rate = k[H2O2][I-]
        

Question 30:

The following reaction mechanism is proposed for a reaction:

Step 1: NO + Cl2 → NOCl2 (slow)
Step 2: NOCl2 + NO → 2 NOCl (fast)

What is the predicted rate law based on this mechanism?

Options:

• (A) Rate = k[NO][Cl₂]
• (B) Rate = k[NO]²[Cl₂]
• (C) Rate = k[NOCl₂][Cl₂]
• (D) Rate = k[Cl₂]

Answer Explanation:
The rate law is determined by the slow step, which involves NO and Cl2. Therefore, the rate law is Rate = k[NO][Cl2].

Correct Answer: (A) Rate = k[NO][Cl2]
        

Question 31:

Consider a first-order reaction with a half-life of 10 minutes. If the initial concentration of the reactant is 1.00 M, what will the concentration be after 30 minutes?

Options:

• (A) 0.50 M
• (B) 0.25 M
• (C) 0.125 M
• (D) 0.0625 M

Answer Explanation:
For a first-order reaction, the concentration halves every half-life. After 30 minutes (3 half-lives):
- Initial concentration = 1.00 M
- After 10 minutes: 1.00 M → 0.50 M
- After 20 minutes: 0.50 M → 0.25 M
- After 30 minutes: 0.25 M → 0.125 M

Correct Answer: (C) 0.125 M
        

Question 32:

For the reaction A + B → Products, the following experimental data were collected:

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][B]³

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].
The rate law is Rate = k[A][B]2.

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

Question 33:

A reaction follows the second-order rate law: Rate = k[A]². If the initial concentration of A is 0.80 M and the rate constant is 0.10 M^-1s^-1, what is the rate of the reaction when [A] = 0.40 M?

Options:

• (A) 0.016 M/s
• (B) 0.020 M/s
• (C) 0.032 M/s
• (D) 0.040 M/s

Answer Explanation:
The rate is calculated using the rate law:
Rate = k[A]2
Substitute the given values:
Rate = (0.10 M-1s-1)(0.40 M)2
Rate = (0.10)(0.16) = 0.016 M/s

Correct Answer: (A) 0.016 M/s
        

Question 34:

The reaction 2NO + O₂ → 2NO₂ is studied, and the rate law is found to be Rate = k[NO]²[O₂]. If the concentration of NO is doubled while keeping [O₂] constant, how does the rate change?

Options:

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

Answer Explanation:
Since the rate law is Rate = k[NO]2[O2], the reaction is second-order in NO. Doubling [NO] increases the rate by a factor of (2)2 = 4.

Correct Answer: (C) The rate quadruples.
        

Question 35:

Consider the reaction A + 2B → Products. The rate law is Rate = k[A][B]². If [A] is increased by a factor of 3 and [B] remains constant, by what factor does the rate increase?

Options:

• (A) 1
• (B) 2
• (C) 3
• (D) 9

Answer Explanation:
The rate law indicates the reaction is first-order in A. Increasing [A] by a factor of 3 increases the rate by a factor of 3.

Correct Answer: (C) 3
        

Question 36:

The following data were collected for a reaction:

What is the rate law for this reaction?

Options:

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

Answer Explanation:
From experiments 1 and 2: Doubling [X] doubles the rate, so the reaction is first-order in X.
From experiments 2 and 3: Doubling [Y] doubles the rate, so the reaction is first-order in Y.
The overall rate law is Rate = k[X][Y].

Correct Answer: (A) Rate = k[X][Y]
        

Question 37:

For the reaction A → B, the concentration of A over time was monitored, and the following data were obtained:

What is the order of the reaction?

Options:

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

Answer Explanation:
The concentration of A decreases by half every 10 seconds, which is characteristic of a first-order reaction.

Correct Answer: (B) First-order
        

Question 38:

The reaction 2A + B → C has the rate law Rate = k[A]²[B]. If [A] is doubled and [B] is halved, what is the effect on the rate?

Options:

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

Answer Explanation:
The rate law is Rate = k[A]2[B]. 
If [A] is doubled, the rate increases by (2)2 = 4. 
If [B] is halved, the rate decreases by 1/2. 
Net effect: Rate = 4 × 1/2 = 2.

Correct Answer: (B) The rate doubles.
        

Question 39:

The reaction 3A → Products is studied, and the rate law is determined to be Rate = k[A]³. If the initial concentration of A is 0.40 M and the rate constant is 0.05 M^-2s^-1, what is the rate of the reaction?

Options:

• (A) 0.008 M/s
• (B) 0.032 M/s
• (C) 0.064 M/s
• (D) 0.080 M/s

Answer Explanation:
The rate law is Rate = k[A]3.
Substitute the given values:
Rate = (0.05 M-2s-1)(0.40 M)3
Rate = (0.05)(0.064) = 0.0032 M/s

Correct Answer: (B) 0.032 M/s
        

Question 40:

The reaction 2A + B → Products has a rate law of Rate = k[A][B]. If the initial concentrations of A and B are 0.50 M and 0.25 M, respectively, and the rate constant is 0.10 M^-1s^-1, what is the initial rate of the reaction?

Options:

• (A) 0.00125 M/s
• (B) 0.0025 M/s
• (C) 0.0050 M/s
• (D) 0.0100 M/s

Answer Explanation:
The rate law is Rate = k[A][B].
Substitute the given values:
Rate = (0.10 M-1s-1)(0.50 M)(0.25 M)
Rate = 0.0125 M/s

Correct Answer: (B) 0.0025 M/s
        

Question 41:

A zeroth-order reaction has a rate constant of 0.20 M/s. If the initial concentration of the reactant is 1.00 M, how long will it take for the concentration to decrease to 0.40 M?

Options:

• (A) 2 seconds
• (B) 3 seconds
• (C) 4 seconds
• (D) 5 seconds

Answer Explanation:
For a zeroth-order reaction:
[A] = [A]₀ - kt
Substitute the given values:
0.40 = 1.00 - (0.20)t
t = (1.00 - 0.40) / 0.20 = 3 seconds

Correct Answer: (B) 3 seconds
        

Question 42:

For a first-order reaction, the rate constant is 0.693 s^-1. What is the half-life of the reaction?

Options:

• (A) 1 second
• (B) 2 seconds
• (C) 0.5 seconds
• (D) 1.5 seconds

Answer Explanation:
The half-life of a first-order reaction is given by:
t₁/₂ = 0.693 / k
Substitute the given rate constant:
t₁/₂ = 0.693 / 0.693 = 1 second

Correct Answer: (A) 1 second
        

Question 43:

The reaction A + 2B → Products has a rate law Rate = k[A][B]². What is the overall reaction order?

Options:

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

Answer Explanation:
The reaction is first-order with respect to A and second-order with respect to B. 
The overall reaction order is the sum of the exponents: 1 + 2 = 3.

Correct Answer: (C) Third-order
        

Question 44:

A second-order reaction has a rate constant of 0.50 M^-1s^-1. If the initial concentration of the reactant is 1.00 M, what is the rate of the reaction?

Options:

• (A) 0.25 M/s
• (B) 0.50 M/s
• (C) 1.00 M/s
• (D) 0.75 M/s

Answer Explanation:
For a second-order reaction, the rate law is:
Rate = k[A]2
Substitute the given values:
Rate = (0.50 M-1s-1)(1.00 M)2
Rate = 0.50 M/s

Correct Answer: (B) 0.50 M/s
        

Question 45:

The decomposition of hydrogen iodide, HI, is found to follow the reaction:

2 HI(g) → H2(g) + I2(g)

The rate law is determined to be Rate = k[HI]². If the concentration of HI is 0.30 M and the rate constant is 0.10 M^-1s^-1, what is the reaction rate?

Options:

• (A) 0.009 M/s
• (B) 0.010 M/s
• (C) 0.027 M/s
• (D) 0.090 M/s

Answer Explanation:
The rate law is Rate = k[HI]2.
Substitute the given values:
Rate = (0.10 M-1s-1)(0.30 M)2
Rate = 0.10 × 0.09 = 0.009 M/s

Correct Answer: (A) 0.009 M/s
        

Question 46:

A first-order reaction has a half-life of 20 seconds. If the initial concentration of the reactant is 0.80 M, what will the concentration be after 60 seconds?

Options:

• (A) 0.10 M
• (B) 0.20 M
• (C) 0.40 M
• (D) 0.60 M

Answer Explanation:
For a first-order reaction:
t₁/₂ = 20 s, and the concentration halves every 20 seconds.
Initial concentration: 0.80 M
After 20 seconds: 0.80 / 2 = 0.40 M
After 40 seconds: 0.40 / 2 = 0.20 M
After 60 seconds: 0.20 / 2 = 0.10 M

Correct Answer: (A) 0.10 M
        

Question 47:

A reaction is found to have a rate law Rate = k[A]^1/2[B]. What is the order of the reaction with respect to A, B, and overall?

Options:

• (A) A: 1, B: 1, Overall: 2
• (B) A: 1/2, B: 1, Overall: 1.5
• (C) A: 1, B: 1/2, Overall: 1.5
• (D) A: 1/2, B: 1/2, Overall: 1

Answer Explanation:
From the rate law, the reaction is:
- Half-order with respect to A (exponent = 1/2)
- First-order with respect to B (exponent = 1)
The overall order is the sum of the exponents: 1/2 + 1 = 1.5.

Correct Answer: (B) A: 1/2, B: 1, Overall: 1.5
        

Question 48:

The reaction A + 2B → Products has a rate law of Rate = k[A][B]. Which of the following changes will result in the greatest increase in the reaction rate?

Options:

• (A) Doubling the concentration of A
• (B) Doubling the concentration of B
• (C) Tripling the concentration of A
• (D) Tripling the concentration of B

Answer Explanation:
The rate law is Rate = k[A][B].
- Doubling A increases the rate by a factor of 2.
- Doubling B increases the rate by a factor of 2.
- Tripling A increases the rate by a factor of 3.
- Tripling B increases the rate by a factor of 3.
Both tripling A and tripling B result in the same rate increase.

Correct Answer: (C) or (D) Both result in the same maximum increase.
        

Question 49:

A student proposes the following mechanism for a reaction:

Step 1: A + B → C (fast equilibrium)
Step 2: C → D + E (slow)
Step 3: D → F (fast)

Which step determines the rate law for the reaction?

Options:

• (A) Step 1
• (B) Step 2
• (C) Step 3
• (D) All steps contribute equally

Answer Explanation:
The slowest step in the mechanism is the rate-determining step, which determines the rate law for the reaction. In this mechanism, Step 2 is the slow step.

Correct Answer: (B) Step 2
        

Question 50:

For a reaction with a rate law Rate = k[A]², which of the following changes will cause the reaction rate to increase by a factor of 16?

Options:

• (A) Doubling [A]
• (B) Tripling [A]
• (C) Quadrupling [A]
• (D) Halving [A]

Answer Explanation:
The rate law is Rate = k[A]2.
- Doubling [A] increases the rate by 2² = 4.
- Tripling [A] increases the rate by 3² = 9.
- Quadrupling [A] increases the rate by 4² = 16.
Therefore, quadrupling [A] increases the rate by a factor of 16.

Correct Answer: (C) Quadrupling [A]