   # Use the equation in Exercise 79 to determine ∆H ° and ∆S ° for the autoionization of water: H 2 O ( l ) ⇌ H + ( a q ) + OH - ( a q ) T (°C) K W 0 1.14×10 −15 25 1.00×10 −14 35 2.09×10 −14 40. 2.92×10 −14 50. 5.47×10 −14 ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 92AE
Textbook Problem
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## Use the equation in Exercise 79 to determine ∆H° and ∆S° for the autoionization of water: H 2 O ( l ) ⇌ H + ( a q ) + OH - ( a q ) T(°C) KW 0 1.14×10−15 25 1.00×10−14 35 2.09×10−14 40. 2.92×10−14 50. 5.47×10−14

Interpretation Introduction

Interpretation: The values of ΔHοandΔSο for the auto ionization of water are to be determined.

Concept introduction: The standard enthalpy and entropy change are the thermodynamic functions. The superscript on these functions represents the standard form. The term Δ represents the change. The relation between the enthalpy and entropy change of with the Gibb’s free energy change is given by a mathematical formula.

### Explanation of Solution

Given

The reaction is given as,

H2O(l)H+(aq)+OH(aq)

From the exercise 79 it is clear that there is a straight line plot between the lnKVs1T .

Therefore, from the given values of rate constant and time the values of lnK and inverse of time is tabulated as,

T(K)KwlnKw1T(K)2731.14×101534.4083.66×1032981.00×101432.2363.36×1033082.09×101431.4993.25×1033132.92×101431.1653.19×1033235.47×101430.5373.10×103

The relation between the equilibrium constant with entropy and enthalpy is given as,

lnK=ΔHοR(1T)+ΔSοR

Where,

• K is equilibrium constant.
• R is gas constant.
• ΔHο is enthalpy change.
• ΔSο is enthalpy change.
• T is temperature in Kelvin.

Compare the above equation with the straight line equation

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