For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of

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Question: For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of 0.5 M, what is the time taken to reach 0.25 M in a second-order reaction? (2023)

Options:

25 s
50 s
10 s
20 s

Correct Answer: 25 s

Solution:

Using t = 1 / (k[A₀]) * (1/[A] - 1/[A₀]), t = 1 / (0.02 * 0.5) * (1/0.25 - 1/0.5) = 25 s.

For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of

Practice Questions

For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of 0.5 M, what is the time taken to reach 0.25 M in a second-order reaction? (2023)

25 s
50 s
10 s
20 s

Questions & Step-by-Step Solutions

For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of 0.5 M, what is the time taken to reach 0.25 M in a second-order reaction? (2023)

Step 1: Identify the given values. The rate constant (k) is 0.02 M⁻¹s⁻¹ and the initial concentration ([A₀]) is 0.5 M. The final concentration ([A]) we want to reach is 0.25 M.
Step 2: Write down the formula for the time (t) in a second-order reaction: t = 1 / (k[A₀]) * (1/[A] - 1/[A₀]).
Step 3: Substitute the values into the formula. Replace k with 0.02, [A₀] with 0.5, and [A] with 0.25: t = 1 / (0.02 * 0.5) * (1/0.25 - 1/0.5).
Step 4: Calculate the denominator: 0.02 * 0.5 = 0.01.
Step 5: Calculate the first part of the equation: 1 / 0.01 = 100.
Step 6: Calculate 1/[A] and 1/[A₀]: 1/0.25 = 4 and 1/0.5 = 2.
Step 7: Substitute these values back into the equation: t = 100 * (4 - 2).
Step 8: Simplify the equation: t = 100 * 2 = 200.
Step 9: The final answer is t = 200 seconds.

Second-Order Reaction Kinetics – Understanding the rate law and integrated rate equations for second-order reactions, including how to apply the formula to calculate time based on concentration changes.

For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of

What’s inside this PDF?

For a reaction with a rate constant of 0.02 M⁻¹s⁻¹ and initial concentration of

Practice Questions

Questions & Step-by-Step Solutions

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