Chapter 27, Problem 27.26P

(a)

Interpretation Introduction

Interpretation:

The value of $\text{α}$ when all the reactants are $100\%$ deuterated during thermogravimetric analysis has to be calculated.

Concept introduction:

Thermogravimetric analysis:

Thermogravimetric analysis is a kind of thermal analysis in that temperature varies over time to determine mass of sample used for thermogravimetric analysis.  This analysis provides various informations such as phase transition, desorption, absorption, chemisorption, decomposition due to thermal, oxidation and reduction.  It is done on thermogravimetric analyzer.  Thermogravimetric analysis is also known as thermal gravimetric analysis (TGA).

Answer to Problem 27.26P

The value of $\text{α}$ when all the reactants are $100\%$ deuterated during thermogravimetric analysis is $0.854976$.

Explanation of Solution

To calculate: The value of $\text{α}$ when all the reactants are $100\%$ deuterated during thermogravimetric analysis.

The reaction occurs during thermogravimetric analysis is

${\text{K(D}}_{\text{x}}{\text{H}}_{\text{1 -x}}{\text{)}}_{\text{2}}{\text{PO}}_{\text{4}}{}_{\text{(s)}}\stackrel{{\text{450}}^{\text{0}}\text{C}}{\to }{\text{ KPO}}_{\text{3}}{}_{\text{(s) }}{\text{+ (D}}_{\text{x}}{\text{H}}_{\text{1 -x}}{\text{)}}_{\text{2}}{\text{O}}_{\text{(g)}}$

The value of $\text{α}$ is defined as weight of product divided by weight of reactant.

$\begin{array}{l}\text{α}\text{=}\frac{\text{118}\text{.0703}}{\text{136}\text{.0853}\text{+}\text{2}\text{.01255}\text{x}}\\ \\ \text{(136}\text{.0853}\text{)α}\text{+}\text{(2}\text{.01255}\text{)αx}\text{=}\text{118}\text{.0703}\\ \\ \text{(2}\text{.01255}\text{)αx}\text{=}\text{118}\text{.0703}\text{-}\text{(136}\text{.0853}\text{)α}\\ \\ \text{While dividing the equation by (2}\text{.01255}\text{)α}\text{we}\text{get}\\ \\ \text{x}\text{=}\frac{\text{118}\text{.0703}}{\text{(2}\text{.01255)α}}\text{-}\frac{\text{(136}\text{.0853}\text{)α}}{\text{(2}\text{.01255}\text{)α}}\\ \\ \text{x}\text{=}\frac{\text{58}\text{.6670}}{\text{α}}\text{-}\text{67}\text{.6183}\end{array}$

When reactant is fully deuterated x becomes one.

$\text{α}\text{=}\frac{\text{58}\text{.6670}}{x+67.6183}=0.854976$

(b)

Interpretation Introduction

Interpretation:

The value of x for a crystal measured in triplicate with given $\text{α}$ has to be calculated.

Concept introduction:

Thermogravimetric analysis:

Thermogravimetric analysis is a kind of thermal analysis in that temperature varies over time to determine mass of sample used for thermogravimetric analysis.  This analysis provides various informations such as phase transition, desorption, absorption, chemisorption, decomposition due to thermal, oxidation and reduction.  It is done on thermogravimetric analyzer.  Thermogravimetric analysis is also known as thermal gravimetric analysis (TGA).

Answer to Problem 27.26P

The value of x for a crystal measured in triplicate with given $\text{α}$ is $\text{0}\text{.85632}$.

Explanation of Solution

To calculate: The value of x for a crystal measured in triplicate with given $\text{α}$.

Given,

$\text{α}\text{=}\text{0}\text{.85677}$

$\begin{array}{l}\text{x}\text{=}\frac{\text{58}\text{.6670}}{\text{α}}\text{-}\text{67}\text{.6183}\\ \\ \text{x}\text{=}\frac{\text{58}\text{.6670}}{0.85677}\text{-}\text{67}\text{.6183}\text{=}\text{0}\text{.85632}\end{array}$

The x value of this crystal is calculated when it measured in triplicate.

(c)

Interpretation Introduction

Interpretation:

The uncertainty in given equation has to be given.

Explanation of Solution

To give: The uncertainty in given equation.

Given,

$\begin{array}{l}\text{function}\text{x}\text{=}\text{f(α)}\\ \\ {\text{e}}_{\text{x}}=\sqrt[]{{\left(\frac{(\partial F)}{(\partial \alpha )}\right)}^2{\text{e}}_{\alpha }{}^2}\end{array}$

$\begin{array}{l}\text{When the x}\text{=}\frac{\text{58}\text{.6670}}{\text{α}}\text{-}\text{67}\text{.6183}\\ \\ \frac{\text{(}\partial \text{F)}}{\text{(}\partial \text{α)}}\text{=}\text{-}\frac{\text{58}\text{.6670}}{{\text{α}}^{\text{2}}}\end{array}$

While substituting the above value in given equation we get

$\begin{array}{l}\text{function}\text{x}\text{=}\text{f(α)}\\ \\ {\text{e}}_{\text{x}}\text{=}\sqrt[]{{\left(\text{-}\frac{\text{58}\text{.6670}}{{\text{α}}^{\text{2}}}\right)}^{\text{2}}{\text{e}}_{\text{α}}{}^{\text{2}}}\\ \\ \text{=}\frac{\text{58}\text{.6670}{\text{e}}_{\text{α}}}{{\text{α}}^{\text{2}}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

From the given uncertainty of ${\text{e}}_{\text{α}}$ , the uncertainty of ${\text{e}}_{\text{x}}$ has to be calculated.

Answer to Problem 27.26P

From the given uncertainty of ${\text{e}}_{\text{α}}$ , the uncertainty of ${\text{e}}_{\text{x}}$ is $\text{0}\text{.856}\text{±}\text{0}\text{.008}$

Explanation of Solution

To give: From the given uncertainty of ${\text{e}}_{\text{α}}$ , the uncertainty of ${\text{e}}_{\text{x}}$.

Given,

${\text{e}}_{\text{α}}=0.0001$

$\begin{array}{l}{\text{When the e}}_{\text{α}}\text{=}\text{0}\text{.0001}\\ \\ {\text{e}}_{\text{x}}\text{=}\frac{\text{(58}\text{.6670)(0}\text{.0001)}}{{\text{(0}\text{.85677)}}^{\text{2}}}\text{=}\text{0}\text{.008}\\ \\ \text{The}\text{ratio}\text{of}\text{D:H}\text{is}\\ \\ \text{x}\text{±}{\text{e}}_{\text{x}}\text{=}\text{0}\text{.856}\text{±}\text{0}\text{.008}\end{array}$

When ${\text{e}}_{\text{x}}$ is $0.001$

$\begin{array}{l}{\text{e}}_{\text{x}}\text{=}\frac{\text{(58}\text{.6670)(0}\text{.001)}}{{\text{(0}\text{.85677)}}^{\text{2}}}\text{=}\text{0}\text{.08}\\ \\ \text{The}\text{ratio}\text{of}\text{D:H}\text{is}\text{0}\text{.86}\text{±}\text{0}\text{.08}\end{array}$