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Consider the relationship In ( K ) = − Δ H ∘ R T + Δ S ∘ R The equilibrium constant for some hypothetical process was determined as a function of temperature (Kelvin) with the results plotted below. From the plot, determine the values of ∆H° and ∆S° for this process. What would be the major difference in the In (K) versus 1 /T plot for an endothermic process as compared to an exothermic process?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Chapter
Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 79E
Textbook Problem
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Consider the relationship

In ( K ) = Δ H R T + Δ S R

The equilibrium constant for some hypothetical process was determined as a function of temperature (Kelvin) with the results plotted below.

Chapter 16, Problem 79E, Consider the relationship In(K)=HRT+SR The equilibrium constant for some hypothetical process was

From the plot, determine the values of ∆H° and ∆S° for this process. What would be the major difference in the In (K) versus 1/T plot for an endothermic process as compared to an exothermic process?

Interpretation Introduction

Interpretation: The plot of lnK versus 1/T and a relation of lnK with ΔH°,ΔS° is given. The values of ΔH°,ΔS° is to be calculated from the given plot. Any difference in the plot of lnK versus 1/T for endothermic and exothermic reaction is to be stated.

Concept introduction: The plot of lnK versus 1/T is a straight line. If the slope of this plot is negative, then the reaction will be endothermic, whereas if the slope of this plot is positive, then the reaction will be exothermic.

The expression for free equilibrium constant is,

ln(K)=ΔH°RT+ΔS°R

Explanation of Solution

The stated relationship is,

ln(K)=ΔH°RT+ΔS°R

Where,

  • ΔH° is the standard enthalpy change.
  • ΔS° is the standard entropy change.
  • R is the gas law constant (8.3145J/Kmol) .
  • T is the absolute temperature.
  • K is the equilibrium constant.

Compare the given relation with general straight line equation.

y=mx+c

The slope (m) of graph is,

m=ΔH°R (1)

The intercept (c) of graph is,

c=ΔS°R (2)

And,

y=lnKx=1T

The plot between lnK versus 1/T is,

The slope from the graph is 13.3 .

Substitute the value of m and R in equation (1)

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Chapter 16 Solutions

Chemistry: An Atoms First Approach
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