Solution Manual for Quantitative Chemical Analysis
Solution Manual for Quantitative Chemical Analysis
9th Edition
ISBN: 9781464175633
Author: Daniel Harris
Publisher: Palgrave Macmillan Higher Ed
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Chapter 6, Problem 6.10P

(a)

Interpretation Introduction

Interpretation:

Equilibrium constant has to be written for given reaction and the vapour pressure of gaseous water ( PH2O ) at 298K has to be calculated.

Concept introduction:

Equilibrium constant (K): In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.  If value of K is small than 1 the reaction should be move to the left and k is more than 1 the reaction should be move to right.

Relating ΔG0 to K:

ΔG0= -RTlnKK  =  e(-ΔG0/RT)

ΔG0-Change in free energyK-Equilibrium constantR-Gas constant (8.314472)T-Temperature

(a)

Expert Solution
Check Mark

Answer to Problem 6.10P

K=PH2O= e-ΔG0/RT=e-(ΔH0-TΔS)/RT

The vapour pressure for given reaction is 4.7×10-4 bar .

Explanation of Solution

Given

Given reaction is,

BaCl2.H2O(s)BaCl2(s)+H2O(g)ΔH0= -63.11kJ/mol at 25°CΔH0= +148J(K.mol) at 25°C  

To calculate: vapour pressure of given equation

K=PBaCl2.PH2OPBaCl2.H2O

For pure solid, the activity is equal to one.

K=PH2O= e-ΔG0/RT=e-(ΔH0-TΔS)/RT

Where,

ΔGO= -(ΔGO-TΔS)

K= e{(63.11×103J/mol-(298.15K)(148JK-1mol-1)/(8.314472J/(K.mol))(298.15K)}= 4.7×10-4 bar

(b)

Interpretation Introduction

Interpretation:

The temperature at which gaseous water ( PH2O ) over BaCl2.H2O(s) will be 1 bar when

ΔH0 and ΔS0 are temperature independent has to be calculated.

(b)

Expert Solution
Check Mark

Answer to Problem 6.10P

The temperature at which gaseous water ( PH2O ) over BaCl2.H2O(s) will be 1 bar is 153°C

Explanation of Solution

Given

Given reaction is,

BaCl2.H2O(s)BaCl2(s)+H2O(g)ΔH0= -63.11kJ/mol at 25°CΔH0= +148J(K.mol) at 25°C  

K=PBaCl2.PH2OPBaCl2.H2O

For pure solid, the activity is equal to one.

K = PH2O

PH2O=1=e-(ΔH0-TΔS)/RTΔH0-TΔSOmust be zero

ΔHO-TΔSO= 0T=ΔHOΔSO= 426K=153°C

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