Chapter 5, Problem 5.147QP

Silicon nitride, Si₃N₄, is a material that is used in computer chips as an insulator. Silicon nitride can be prepared according to the chemical equation

$3{\text{SiH}}_4(g)+4{\text{NH}}_3(g)\to {\text{Si}}_3{\text{N}}_4(g)+12{\text{H}}_2(g)$

If you wanted to prepare a surface film of Si₃N₄(s) that had an area of 9.0 mm² and was 4.0 × 10⁵ mm thick, what volume of SiH₄ gas would you need to use at 1.0 × 10⁵ torr and 775 K? The density of Si₃N₂ is 3.29 g/cm³.

Interpretation Introduction

Interpretation:

The volume of ${\text{SiH}}_{\text{4}}$ needed to prepare a surface film of ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ with an area $9.0{\text{ mm}}^{\text{2}}$ and thickness of $4.0\times {10}^5\text{ mm}$ has to be calculated

Concept Introduction:

Ideal Gas Law:

The ideal gas equation is:

$\text{PV = nRT}$

Where,

P is the pressure

V is the volume

T is the temperature

R is molar gas constant

n is the mole

Answer to Problem 5.147QP

The volume of ${\text{SiH}}_{\text{4}}$ needed to prepare a surface film of ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ with an area $9.0{\text{ mm}}^{\text{2}}$ and thickness of $4.0\times {10}^5\text{ mm}$ is $1.2\times {10}^2\text{ L}$

Explanation of Solution

Given data:

In computer chips, Silicon nitride, ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ is used as an insulator.

It can be prepared as per the following chemical equation.

${\text{3SiH}}_{\text{4}}{\text{(g) + 4NH}}_{\text{3}}\text{(g) }\stackrel{}{\to }{\text{ Si}}_{\text{3}}{\text{N}}_{\text{4}}{\text{(g) + 12H}}_{\text{2}}\text{(g)}$

${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ of an area of $9.0{\text{ mm}}^{\text{2}}$ is prepared at $1.0\times {10}^5\text{ torr}$ and 775 K

The density of ${\text{Si}}_{\text{3}}{\text{N}}_2$ is $3.29{\text{ g/cm}}^3$

Calculation of volume of ${\text{SiH}}_{\text{4}}$:

The volume of film to be deposited is obtained by the product of its area and thickness.

Using the density of ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ and the volume calculated as above, the mass of ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ is found as follows,

$\begin{array}{l}{\text{mass}}_{{\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}}=9.0{\text{ mm}}^2\times 4.0\times {10}^{-5}\text{ mm}\times {\left(\frac{\text{1 cm}}{10\text{ mm}}\right)}^3\times \left(\frac{3.29\text{ g}}{{\text{1 cm}}^3}\right)\\ =1.18\times {10}^{-6}{\text{ g Si}}_{\text{3}}{\text{N}}_{\text{4}}\end{array}$

Convert this mass to moles, then convert the moles of ${\text{SiH}}_{\text{4}}$ needed using the reaction stoichiometry as follows,

$\begin{array}{l}{\text{mol}}_{{\text{SiH}}_{\text{4}}}=1.18\times {10}^{-6}{\text{ g Si}}_{\text{3}}{\text{N}}_{\text{4}}\times \left(\frac{{\text{1 mol Si}}_{\text{3}}{\text{N}}_{\text{4}}}{140.3{\text{ g Si}}_{\text{3}}{\text{N}}_{\text{4}}}\right)\times \left(\frac{3{\text{ mol SiH}}_{\text{4}}}{{\text{1 mol Si}}_{\text{3}}{\text{N}}_{\text{4}}}\right)\\ =2.53\times {10}^{-8}{\text{ mol SiH}}_{\text{4}}\end{array}$

Using ideal gas law, the volume of ${\text{SiH}}_{\text{4}}$ needed is calculated as below,

$\begin{array}{l}{\text{V}}_{{\text{SiH}}_{\text{4}}}=\frac{n\text{RT}}{\text{P}}\\ =\frac{\text{(2}\text{.53}\times {\text{10}}^{-8}\text{ mol)(0}\text{.08206 L}·\text{atm/K}·\text{mol)(775 K)}}{1.0\times {10}^{-5}/760\text{ atm}}\\ =1.2\times {10}^2\text{ L}\end{array}$

Therefore, the volume of ${\text{SiH}}_{\text{4}}$ needed is $1.2\times {10}^2\text{ L}$

Conclusion

The volume of ${\text{SiH}}_{\text{4}}$ needed to prepare a surface film of ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ with an area $9.0{\text{ mm}}^{\text{2}}$ and thickness of $4.0\times {10}^5\text{ mm}$ is calculated as $1.2\times {10}^2\text{ L}$