   Chapter 12, Problem 30E

Chapter
Section
Textbook Problem

# The reaction 2I - ( a q ) +S 2 O 8 2- ( a q ) → I 2 ( a q ) +2SO 4 2- ( a q ) was studied at 25°C. The following results were obtained where Rate   =   − Δ [ S 2 O 8 2- ] Δ t [I−]0(mol/L) [S2O82−]0(mol/L) Initial Rate (mol/L · s) 0.080 0.040 12.5 × 10−6 0.040 0.040 6.25 × 10−6 0.080 0.020 6.25 × 10−6 0.032 0.040 5.00 × 10−6 0.060 0.030 7.00 × 10−6 a. Determine the rate law.b. Calculate a value for the rate constant for each experiment and an average value for the rate constant

Interpretation Introduction

Interpretation: A reaction between I and S2O82 and its rate expression is given. The rate law and the value of the rate constant are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

Explanation

Given

The stated reaction is,

2I(aq)+S2O82(aq)I2(aq)+2SO42(aq)

The rate law for the given reaction is calculated by the expression,

Rate=k[I]m[S2O82]n

Where,

• [I] and [S2O82] are the concentrations of I and S2O82 .
• k is the rate constant.

The values of and n are calculated by the comparison of the different rate values from the given table.

The value of m is calculated using the first and second result as only the value of [I] changes in these results. Substitute the values of the concentration of [I] and [S2O82] for the first two experiments in the above expression.

Rate1=k[0.080]m[0.040]n

Rate2=k[0.040]m[0.040]n

According to the given rate values in the table,

Rate2Rate1=6.25×106mol/Ls12.5×106mol/Ls

Therefore,

k[0.040]m[0.040]nk[0.080]m[0.040]n=6.25×106mol/Ls12.5×106mol/Ls(0

(b)

Interpretation Introduction

Interpretation: A reaction between I and S2O82 and its rate expression is given. The rate law and the value of the rate constant are to be calculated.

Concept introduction: The relation between the reaction rate and the concentration of the reactants is stated by the rate law.

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