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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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Chemistry

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

The equilibrium constant for a certain reaction decreases from 8.84 to 3.25 × 10−2 when the temperature increases from 25°C to 75°C. Estimate the temperature where K = 1.00 for this reaction. Estimate the value of ∆S° for this reaction. (Hint: Manipulate the equation in Exercise 79.)

Interpretation Introduction

Interpretation: The value of ΔSο for the reaction and the temperature where K=100 is to be estimated.

Concept introduction: Thermodynamics is associated with heat, temperature and its relation with energy and work. It helps us to predict whether a process will take place or not. The terms associated with thermodynamics are system, surrounding, entropy, spontaneity and many more.

Explanation

Explanation

Given

The first equilibrium constant ( K1 ) is 8.84 .

The second equilibrium constant ( K2 ) is 3.25×102 .

The initial temperature ( T1 ) is 25 οC=298 K .

The increased temperature ( T2 ) is 75 οC=348 K .

The equilibrium constant for a certain reaction decreases with increase in temperature. It indicates that equilibrium constant and temperature is inversely related. Therefore,

1T1=1298=3.35×103 .

1T1=1348=2.87×103

lnK1=ln8.84=2.2

lnK2=ln3.25×102=3.42

The graph between 1T and lnK is as follows.

Figure 1

Consider the equation.

lnK=ΔHοRT+ΔSοR (1)

Where,

  • ΔHο is the standard enthalpy change.
  • R is gas constant.
  • T is temperature.
  • K is equilibrium constant.

Form the above equation, the slope is ΔHοR and from the graph the slope is

lnK1lnK2[1T11T2] .

Equate ΔHοR with lnK1lnK2[1T11T2] .

ΔHοR=lnK1lnK2[1T11T2]

The value of gas constant, R=8

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