Chapter 20, Problem 117IP

(a)

Interpretation Introduction

Interpretation: The reaction between ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ and ${\text{PH}}_{\text{3}}$ at $900\text{K}$ is given. The answers are to be stated for the given options.

Concept introduction: The number of moles is calculated using ideal gas law,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

To determine: The mass of $\text{InP}$ when $2.56\text{L}$ ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ at $2.00\text{atm}$ is allowed to react with $1.38\text{L}$ ${\text{PH}}_{\text{3}}$ at $3.00\text{atm}$ (assuming the reaction has $87\%$ yield).

Explanation of Solution

Explanation

Given

Temperature is $900\text{K}$.

Pressure of ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ is $2.00\text{atm}$.

Volume of ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ is $2.56\text{L}$.

Pressure of ${\text{PH}}_{\text{3}}$ is $3.00\text{atm}$.

Volume of ${\text{PH}}_{\text{3}}$ is $1.38\text{L}$.

Formula

Number of moles is calculated as,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$ (1)

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,V,R$ and $T$ for ${\text{PH}}_{\text{3}}$ in the above equation.

$\begin{array}{c}{n}_{{\text{PH}}_{\text{3}}}=\frac{PV}{RT}\\ =\frac{\text{2}\text{.00}\text{atm}\times \text{1}\text{.38}\text{L}}{(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})\left(\text{900}\text{K}\right)}\\ =\underset{\_}{0.056mol}\end{array}$

Substitute the values of $P,V,R$ and $T$ for ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ in equation (1).

$\begin{array}{c}{n}_{{\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}}=\frac{PV}{RT}\\ =\frac{\text{2}\text{.00}\text{atm}\times \text{2}\text{.56}\text{L}}{(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})\left(\text{900}\text{K}\right)}\\ =\underset{\_}{0.0693mol}\end{array}$

To determine the mass of $\text{InP}$is $\underset{\_}{7.1g}$.

Moles of ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ is $\text{0}\text{.0693}\text{mol}$.

Moles of ${\text{PH}}_{\text{3}}$ is $\text{0}\text{.056}\text{mol}$.

The moles of ${\text{PH}}_{\text{3}}$ is less than the moles of ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$. Hence, ${\text{PH}}_{\text{3}}$ will acts as limiting reagent.

Formula

Mass of $\text{InP}$ is calculated using the formula,

$\text{Mass}\text{of}\text{InP}=\text{Moles}\text{of}{\text{PH}}_{\text{3}}\times \frac{\text{1}\text{mol}\text{InP}}{\text{1}\text{mol}{\text{PH}}_{\text{3}}}\times \frac{\text{Molar}\text{mass}\text{of}\text{InP}}{\text{1}\text{mol}\text{InP}}$

Substitute the values of moles of ${\text{PH}}_{\text{3}}$ and the molar mass of $\text{InP}$ in the above equation.

$\begin{array}{c}\text{Mass}\text{of}\text{InP}=\text{Moles}\text{of}{\text{PH}}_{\text{3}}\times \frac{\text{1}\text{mol}\text{InP}}{\text{1}\text{mol}{\text{PH}}_{\text{3}}}\times \frac{\text{Molar}\text{mass}\text{of}\text{InP}}{\text{1}\text{mol}\text{InP}}\\ =0.056\text{mol}{\text{PH}}_{\text{3}}\times \frac{\text{1}\text{mol}\text{InP}}{\text{1}\text{mol}{\text{PH}}_{\text{3}}}\times \frac{\text{145}\text{.79}}{\text{1}\text{mol}\text{InP}}\\ =8.164\text{g}\end{array}$

Since, it is assumed that the reaction has $87\%$ yield. Therefore mass of $\text{InP}$ formed is,

$8.164\text{g}\times \frac{87}{100}=\underset{\_}{7.1g}$

(b)

Interpretation Introduction

Interpretation: The reaction between ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ and ${\text{PH}}_{\text{3}}$ at $900\text{K}$ is given. The answers are to be stated for the given options.

Concept introduction: The number of moles is calculated using ideal gas law,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

To determine: The wavelength of light if $2.03\times {10}^{-19}\text{J}$ of energy is emitted by light; if this wavelength is visible to human eye.

Explanation of Solution

Given

Energy is $2.03\times {10}^{-19}\text{J}$.

Formula

The wavelength of light is calculated using the formula,

$\begin{array}{c}E=\frac{hc}{\lambda }\\ \lambda =\frac{hc}{E}\end{array}$

Where,

• $\lambda$ is the wavelength.
• $c$ is the velocity of light $(3\times {10}^8\text{m/s})$
• $E$ is the energy.
• $h$ is the Plank’s constant $(6.626\times {10}^{-34}\text{J/s})$.

Substitute the values of $E,h$ and $c$ in the above equation.

$\begin{array}{c}\lambda =\frac{hc}{E}\\ =\frac{(6.626\times {10}^{-34}\text{J/s})(3\times {10}^8\text{m/s})}{2.03\times {10}^{-19}\text{J}}\\ =979\times {10}^{-9}\text{m}\end{array}$

The conversion of meter $\left(\text{m}\right)$ into nanometer $\left(\text{nm}\right)$ is done as,

$\text{1}\text{m}={\text{10}}^{\text{9}}\text{nm}$

Hence,

The conversion of $979\times {10}^{-9}\text{m}$ into nanometer is,

$\begin{array}{c}\text{979}\times {\text{10}}^{-\text{9}}\text{m}=\left(\text{979}\times {\text{10}}^{-\text{9}}\times {\text{10}}^{\text{9}}\right)\text{nm}\\ =\underset{\_}{979nm}\end{array}$

This wavelength is not visible to human eye.

The calculated value of wavelength of light is $\text{979}\text{nm}$.

This wavelength of light belongs to infrared region. Hence, this wavelength is not visible to human eye.

(c)

Interpretation Introduction

Interpretation: The reaction between ${\text{In(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ and ${\text{PH}}_{\text{3}}$ at $900\text{K}$ is given. The answers are to be stated for the given options.

Concept introduction: The number of moles is calculated using ideal gas law,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

To determine: If $\text{InP}$ becomes n-type or p-type when small number of phosphorous atoms are replaced by the atoms of electronic configuration $\left[\text{Kr}\right]{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{4}}$.

Explanation of Solution

The material $\text{InP}$ becomes n-type when small number of phosphorous atoms are replaced by the atoms of electronic configuration $\left[\text{Kr}\right]{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{4}}$.

The given electronic configuration $\left[\text{Kr}\right]{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{4}}$ belongs to group ${15}^{\text{th}}$. This is the electronic configuration of arsenic. Arsenic is the pentavalent element. Hence, $\text{InP}$ becomes n-type.