Kinetics of Second-Order Reaction

The video tutorial explains how to calculate the time required for the concentration of a reactant in a second-order reaction to decrease from 0.68 M to 0.22 M. It begins with a derivation of the integrated rate law for a second-order reaction, followed by a step-by-step calculation using the derived formula. The tutorial is divided into two parts: the first part focuses on the derivation, while the second part applies the formula to solve the problem.

What is the rate constant for the second-order reaction discussed in the video?

• 0.300 inverse molar inverse second

• 0.680 inverse molar inverse second

• 0.780 inverse molar inverse second

• 0.220 inverse molar inverse second

In the context of the video, what does the term 'separable differential equation' refer to?

• An equation that is linear

• An equation that involves only constants

• An equation that cannot be integrated

• An equation that can be solved by separating variables

What is the significance of the negative sign in the integrated rate law for a second-order reaction?

• It indicates the reaction is exothermic

• It shows the reaction rate is decreasing

• It is a result of the integration process

• It means the reaction is endothermic

What assumption is made about the initial time (t1) in the problem?

• t1 is equal to the final time

• t1 is equal to zero

• t1 is equal to the rate constant

• t1 is equal to the concentration of A

What is the final concentration of A in the problem?

• 0.50 molar

• 0.78 molar

• 0.68 molar

• 0.22 molar

What is the main formula used for a second-order reaction in the video?

• ln[A] = kt + ln[A0]

• 1/[A] = -kt + 1/[A0]

• ln[A] = -kt + ln[A0]

• 1/[A] = kt + 1/[A0]

How long does it take for the concentration of A to decrease from 0.68 molar to 0.22 molar?

• 3.94 seconds

• 5.94 seconds

• 4.94 seconds

• 2.94 seconds