   Chapter 12, Problem 94CWP

Chapter
Section
Textbook Problem

# The thiosulfate ion (S2O32−) is oxidized by iodine as follows: 2S 2 O 3 2 − ( a q ) + I 2 ( a q ) → S 4 O 6 2 − ( a q ) + 2I − ( a q ) In a certain experiment, 7.05 × 10−3 mol/L of S2O32− is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of S2O32−. Calculate the rate of production of iodide ion.

Interpretation Introduction

Interpretation: The amount of thiosulfate ion consumed in a certain time is given. By using these values, the rate of consumption of S2O32 and andthe rate of production of iodine ion in the given reaction is to be calculated.

Concept introduction: The rate of consumption of a reactant in a chemical reaction is defined as the change in concentration of reactant with time, whereasthe rate of production of a product is defined as the change in concentration of product.

Example: Suppose a chemical reaction is as follows:

AB

The rate of consumption of reactant A in above chemical reaction will be calculated as:

RA=(Δ[A]Δt)

The rate of production of product B in above chemical reaction will be calculated as:

RB=(Δ[B]Δt)

To determine: The rate of consumption of  thiosulfate ion.

Explanation

Given

The change in the concentration of S2O32 is 7.05×103mol/L .

The change in time or time interval is 11.0s .

The chemical reaction is as follows:

2S2O32(aq)+I2(aq)S4O62(aq)+2I(aq)

Rate of consumption of thiosulfate ion is calculated by using the formula,

RS2O32=(Δ[S2O32]Δt)

Where,

• Δ[S2O32] is the change in concentration of thiosulfate ion.
• Δt is the time interval.
• RS2O32 is the rate of consumption of thiosulfate ion.

Substitute the values of Δ[S2O32] and Δt in the above equation.

RS2O32=(7.05×103mol/L11.0s)=6.40×10-4mol/L·s_

To determine: The rate of production of  iodine ion.

Given

The change in the concentration of S2O32 is 7.05×103mol/L .

The change in time or time interval is 11

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