Chapter 6, Problem 6.160AP

(a)

Interpretation Introduction

Interpretation: One cotton ball soaked in ammonia and another soaked in hydrochloric acid is given to be placed at opposite ends of a $1\text{m}$ glass tube. The vapors are given to diffuse towards the middle of the tube and form a white ring of ammonium chloride at the point where they met. The stated questions based upon the above statement are to be answered.

Concept introduction: According to the Graham’s law of effusion,

$r\propto \frac{1}{\sqrt{M}}$

To determine: The chemical equation for the reaction between ammonia and hydrochloric acid.

Answer to Problem 6.160AP

Solution

The balance chemical equation for the reaction between ammonia and hydrochloric acid is stated as follows.

Explanation of Solution

Explanation

The chemical equation corresponding to the reaction between ammonia and hydrochloric acid is,

${\text{NH}}_{\text{3}}\left(g\right)+\text{HCl}\left(g\right)\to {\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$

The stated equation is already in its balanced form.

(b)

Interpretation Introduction

To determine: The location of the ammonium chloride ring.

Answer to Problem 6.160AP

Solution

The ammonium chloride ring will be formed near the end of hydrochloric acid gas.

Explanation of Solution

Explanation

The tube contains air. Due to the presence of air, the ammonia and hydrochloric acid molecules undergo many collisions with the components present in the air, that is ${\text{O}}_{\text{2}}\text{and}{\text{N}}_{\text{2}}$.

The molecular weight of $\text{HCl}$ is,

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}\text{HCl}=\text{H}+\text{Cl}\\ =1.0079\text{g/mol}+35.453\text{g/mol}\\ =\text{36}\text{.4608}\text{g/mol}\end{array}$

The molecular weight of ${\text{NH}}_{\text{3}}$ is,

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\text{NH}}_{\text{3}}=\text{N}+\text{3H}\\ =14.00\text{g/mol}+3\times 1.0079\text{g/mol}\\ =17.\text{0304}\text{g/mol}\end{array}$

The molecular weight of ${\text{NH}}_{\text{3}}$ is less than that of $\text{HCl}$. Therefore, ${\text{NH}}_{\text{3}}$ is the lighter gas.

The ammonium gas molecules have high velocities due to a smaller mass than $\text{HCl}$.

Therefore, the ring of ammonium chloride will be formed near the end of $\text{HCl}$.

(c)

Interpretation Introduction

To determine: The distance from the ammonia end to the position of the ammonium chloride ring.

Answer to Problem 6.160AP

Solution

The distance from the ammonia end is $\underset{\_}{0.5928m}$

Explanation of Solution

Explanation

According to Graham’s law of effusion,

$r\propto \frac{1}{\sqrt{M}}$ (1)

Where,

• $r$ is the rate of diffusion.
• $M$ is the molar mass.

The molar masses of two gases are assumed to be $M_{\text{HCl}}and{M}_{{\text{NH}}_{\text{3}}}$.

The time for diffusion of same amount of gas is assumed to be $r_{\text{HCl}}\text{and}{r}_{{\text{NH}}_{\text{3}}}$.

For two different molar masses and temperatures the above expression is written as,

$\frac{r_{{\text{NH}}_{\text{3}}}}{r_{{}_{\text{HCl}}}}=\frac{\sqrt{M_{\text{HCl}}}}{\sqrt{M_{{\text{NH}}_{\text{3}}}}}$ (1)

Where,

• $r_{{\text{NH}}_{\text{3}}}$ is the rate of diffusion of ammonia gas.
• $r_{{}_{\text{HCl}}}$ is the rate of diffusion of hydrochloric acid gas.
• $M_{\text{HCl}}$ is the molar mass of hydrochloric acid gas.
• $M_{{\text{NH}}_{\text{3}}}$ is the molar mass of ammonia gas is.

The molecular weight of $\text{HCl}$ is,

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}\text{HCl}=\text{H}+\text{Cl}\\ =1.0079\text{g/mol}+35.453\text{g/mol}\\ =\text{36}\text{.4608}\text{g/mol}\end{array}$

The molecular weight of ${\text{NH}}_{\text{3}}$ is,

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\text{NH}}_{\text{3}}=\text{N}+\text{3H}\\ =14.00\text{g/mol}+3\times 1.0079\text{g/mol}\\ =17.\text{0304}\text{g/mol}\end{array}$

Substitute the required values in equation (1).

$\begin{array}{c}\frac{r_{{\text{NH}}_{\text{3}}}}{r_{{}_{\text{HCl}}}}=\frac{\sqrt{M_{\text{HCl}}}}{\sqrt{M_{{\text{NH}}_{\text{3}}}}}\\ =\sqrt{\frac{36.4609}{17.0304}}\\ =1.465\end{array}$

The distance of the ring from $\text{HCl}$ end is assumed to be $L\text{cm}$.

The time taken by both gases to meet is assumed to be $T$.

Therefore, the distance from ammonia’s end will be $\left(100-L\right)\text{cm}$.

$\begin{array}{c}\frac{\left(\frac{100-L}{T}\right)}{\frac{L}{T}}=1.465\\ \frac{100-L}{L}=1.465\\ 100-L=1.465L\\ \end{array}$

Simplify the above equation,

$\begin{array}{l}L=\frac{100}{2.465}\\ L=40.72\text{cm}\end{array}$

So, $L$ is the distance travelled from $\text{HCl}$ end and the distance travelled from ammonia end is calculated by the formula,

$\text{Distance}=100\text{cm}-\left(\text{Distance}\text{travelled}\text{from}\text{the}\text{HCl}\text{end}\right)$

Substitute the value of the distance travelled from the $\text{HCl}$ end in the above formula.

$\begin{array}{c}\text{Distance}=100\text{cm}-\left(\text{Distance}\text{travelled}\text{from}\text{the}\text{HCl}\text{end}\right)\\ =100\text{cm}-40.72\text{cm}\\ =\text{59}\text{.28}\text{cm}\\ =\underset{\_}{0.5928m}\end{array}$

Therefore, the distance travelled from ammonia end is $\underset{\_}{0.5928m}$.

Conclusion

1. a) The balanced chemical equation for the reaction between ammonia and hydrochloric acid is

   ${\text{NH}}_{\text{3}}\left(g\right)+\text{HCl}\left(g\right)\to {\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$

2. b) The ammonium chloride ring will be formed near the end of hydrochloric acid gas.
3. c) The distance from the ammonia end is $\underset{\_}{0.5928m}$