Can you help me out with this one please. Thank you. The following thermodynamics data were obtained for dissolving KNO3 in water: The slope ofthe straight line defined by graph of In Ksp VS. 1/T is -4.0 x 103 K Average AS = +1.4 x 10² J/(K- mol)a) What is the AH? Does the process of dissolving KNO3 require heat or not? Explain briefly.b) What does the entropy change indicate?

Answered: The following thermodynamics data were…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Can you help me out with this one please. Thank you.

The following thermodynamics data were obtained for dissolving KNO3 in water: The slope of
the straight line defined by graph of In Ksp VS. 1/T is -4.0 x 103 K Average AS = +1.4 x 10² J/(K- mol)
a) What is the AH? Does the process of dissolving KNO3 require heat or not? Explain briefly.
b) What does the entropy change indicate?

Expert Solution

Step 1

The equation that relates equilibrium constant and Gibb's free energy is;

$\begin{array}{rcl} ∆ G^{o} & = & - RT \ln K_{eq} \\ ∆ G^{o} & = & ∆ H^{o} - T ∆ S^{o} \\ - RT \ln K_{eq} & = & ∆ H^{o} - T ∆ S^{o} \\ \ln K_{eq} & = & - \frac{∆ H^{o}}{RT} + \frac{∆ S^{o}}{R} \\ on comparing with straight line equation, \\ y & = & mx + c \\ A plot of {lnK}_{eq} versus \frac{1}{T} gives a slope of \frac{- ∆ H^{o}}{R} and intercept of \frac{∆ S^{o}}{R} \end{array}$

Step 2

Given,

$\begin{array}{rcl} slope & = & \frac{- ∆ H}{R} \\ slope & = & - 4.0 \times 10^{3} K \\ R & = & 8.314 J / K mol \\ ∆ H & = & - (Slope \times R) \\ = & - (- 4.0 \times 10^{3}) \times 8.314 \\ ∆ H & = & 3.32 \times 10^{4} J / K mol \end{array}$

The process of dissolving KNO₃requires heat as the value of $∆ H$ is positive.

It means the process is endothermic in nature.

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