   Chapter 12, Problem 57E

Chapter
Section
Textbook Problem

# You and a coworker have developed a molecule that·has shown potential as cobra antivenin (AV). This antivenin works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law Rate   =   k [ AV ] 1 [ V ] 1 You have been given the following data from your coworker: [ V ] 0   =   0.20   M [ AV ] 0   =   1.0   ×   10 − 4   M A plot of ln[AV] versus t(s) gives a straight line with a slope of −0.32 s−1. What is the value of the rate constant (k) for this reaction?

Interpretation Introduction

Interpretation: The rate law of the reaction between cobra venom and antivenom, the initial concentration of venom and antivenom and the slope of a straight line graph are given. The value of rate constant for this reaction is to be calculated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: The value of the rate constant for the given rate law.

Explanation

Explanation

Initial concentration of venom ([V]0) is 0.20M .

Initial concentration of antivenom ([AV]0) is 1.0×104M .

The slope of straight line graph is 0.32s1 .

The rate law is represented as,

Rate=k[AV]1[V]1

It is clear from the above equation that, it is a second order reaction.

The integral rate law equation of second order reaction is,

ln[A]0[B][A][B]0=K([B]0[A]0)t

Where,

• k is the rate constant.
• [A]0 is the initial concentration of reactant (AV)0 .
• [B]0 is the initial concentration of reactant (V)0 .
• [A] is the final concentration of reactant (AV) .
• [B] is the final concentration of reactant (V) .
• t is the time.

The graph between ln[A]0[B][A][B]0 versus time is,

Figure 1

The slope of above graph is K([B]0[A]0)

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