Chapter 12, Problem 114IP

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Upon dissolving InCl(s) in HCl, In+(aq) undergoes a disproportionation reaction according to the following unbalanced equation: In + ( a q ) → In ( s ) + In 3+ ( a q ) This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of In+(aq) after 1.25 h if the initial solution of In+(aq) was prepared by dissolving 2.38 g InCl(s) in dilute HCl to make 5.00 × 102 mL of solution? What mass of In(s) is formed after 1.25 h?

Interpretation Introduction

Interpretation: The disproportionation reaction of In+ and its half-life is given. The mass of InCl(s) dissolved in HCl to prepare a given volume of solution is also given. By using these values, the concentration of In+ and the mass of In(s) formed after 1.25h is to be determined.

Concept introduction: The half-life is defined as the time required for the concentration of reactant to be reduced to one-half of its initial value.

The integrated rate law for the first order reaction is given by,

ln[A]=kt+ln[A]0

To determine: The rate constant for the given reaction, the concentration of In+ , mass of In(s) formed after 1.25h .

Explanation

Explanation

Given

Half-life of In+ is 667s .

Formula

The rate constant for the first order reaction is calculated using the formula,

k=0.693t1/2

Where,

• k isthe rate constant.
• t1/2 isthe half-life.

Substitute the values of t1/2 in the above equation.

k=0.693t1/2=0.693667s=0.001s1_

Mass of InCl(s) is 2.38g .

Volume of dilute HCl is 500mL .

The molar mass of InCl(s) is 150.271g/mol .

Formula

The number of moles is calculated using the formula,

Numberofmoles=GivenmassMolarmass

Substitute the values of given mass and molar mass of InCl(s) in the above equation.

Numberofmoles=GivenmassMolarmass=2.38g150.271g/mol=0.0158mol_

Volume of dilute HCl is 500mL .

Number of moles of InCl(s) is 0.0158mol .

The conversion of milliliter (mL) into liter (L) is done as,

1mL=103L

Hence, the conversion of 500mL into liter is,

500mL=(500×103)L=0.5L

Formula

The molarity of In+(aq) is calculated using the formula,

MolarityofIn+=MolesofIn+Volumeofsolution

Substitute the values of volume and moles in the above equation.

MolarityofIn+=MolesofIn+Volumeofsolution=0.01580.5L=0.0316M_

This is the initial concentration of In+(aq) .

The time of reaction is 1.25h .

There are 3600s in one hour. Hence, number of seconds in 1.25h is,

1.25h=1.25×3600s=4500s

The initial concentration is 0

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