Chapter 3, Problem 3D.6E

(a)

Interpretation Introduction

Interpretation:

Time period required for $\text{CO}$ to effuse has to be determined.

Concept Introduction:

Effusion is explained as movement of the gas molecule through small hole.

Diffusion can be explained as when one gas molecule mix with another gas molecule by random motion.

Graham’s law of effusion gives relation of rate of effusion with molar mass. Rate of effusion and square root of molar mass are inversely related. The mathematical expression is as follows:

  $\frac{\text{Effusion rate of A}}{\text{Effusion rate of B}}=\sqrt{\frac{M_{\text{B}}}{M_{\text{A}}}}$

Here,

$M_{\text{B}}$ is molar mass of gas $\text{B}$.

$M_{\text{A}}$ is molar mass of gas $\text{A}$.

Answer to Problem 3D.6E

Time period required for $\text{CO}$ to effuse is $73.2\text{ s}$.

Explanation of Solution

The expression for molar masses of $\text{CO}$ and $\text{Ne}$ gases and their time for effusion is as follows:

  $\frac{\text{Time for CO to effuse}}{\text{Time for Ne to effuse}}=\sqrt{\frac{M_{\text{CO}}}{M_{\text{Ne}}}}$        (1)

Rearrange equation (1) to calculate value of time for $\text{CO}$ to effuse.

  $\text{Time for CO to effuse}=\left(\text{Time for Ne to effuse}\right)\left(\sqrt{\frac{M_{\text{CO}}}{M_{\text{Ne}}}}\right)$        (2)

Substitute $28.01\text{ g/mol}$ for $M_{\text{CO}}$, $20.18\text{ g/mol}$ for $M_{\text{Ne}}$, and $62.1\text{ s}$ for time for $\text{Ne}$ to effuse in equation (2).

  $\begin{array}{c}\text{Time for CO to effuse}=\left(62.1\text{ s}\right)\left(\sqrt{\frac{28.01\text{ g/mol}}{20.18\text{ g/mol}}}\right)\\ \approx 73.2\text{ s}\end{array}$

Hence, time period required for $\text{CO}$ to effuse is $73.2\text{ s}$.

(b)

Interpretation Introduction

Interpretation:

Time period required for ${\text{NH}}_3$ to effuse has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.6E

Time period required for ${\text{NH}}_3$ to effuse is $57.0\text{ s}$.

Explanation of Solution

The expression for molar masses of ${\text{NH}}_3$ and $\text{Ne}$ gases and their time for effusion is as follows:

  $\frac{{\text{Time for NH}}_3\text{ to effuse}}{\text{Time for Ne to effuse}}=\sqrt{\frac{M_{{\text{NH}}_3}}{M_{\text{Ne}}}}$        (3)

Rearrange equation (1) to calculate value of time for ${\text{NH}}_3$ to effuse.

  ${\text{Time for NH}}_3\text{ to effuse}=\left(\text{Time for Ne to effuse}\right)\left(\sqrt{\frac{M_{{\text{NH}}_3}}{M_{\text{Ne}}}}\right)$        (4)

Substitute $17.03\text{ g/mol}$ for $M_{{\text{NH}}_3}$, $20.18\text{ g/mol}$ for $M_{\text{Ne}}$, and $62.1\text{ s}$ for time for $\text{Ne}$ to effuse in equation (2).

  $\begin{array}{c}{\text{Time for NH}}_3\text{ to effuse}=\left(62.1\text{ s}\right)\left(\sqrt{\frac{17.03\text{ g/mol}}{20.18\text{ g/mol}}}\right)\\ \approx 57.0\text{ s}\end{array}$

Hence time period required for ${\text{NH}}_3$ to effuse is $57.0\text{ s}$.

(c)

Interpretation Introduction

Interpretation:

Time period required for $\text{Kr}$ to effuse has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.6E

Time period required for $\text{Kr}$ to effuse is $127\text{ s}$.

Explanation of Solution

The expression for molar masses of $\text{Kr}$ and $\text{Ne}$ gases and their time for effusion is as follows:

  $\frac{\text{Time for Kr to effuse}}{\text{Time for Ne to effuse}}=\sqrt{\frac{M_{\text{Kr}}}{M_{\text{Ne}}}}$        (5)

Rearrange equation (1) to calculate value of time for $\text{Kr}$ to effuse.

  $\text{Time for Kr to effuse}=\left(\text{Time for Ne to effuse}\right)\left(\sqrt{\frac{M_{\text{Kr}}}{M_{\text{Ne}}}}\right)$        (6)

Substitute $83.80\text{ g/mol}$ for $M_{\text{Kr}}$, $20.18\text{ g/mol}$ for $M_{\text{Ne}}$, and $62.1\text{ s}$ for time for $\text{Ne}$ to effuse in equation (2).

  $\begin{array}{c}\text{Time for Kr to effuse}=\left(62.1\text{ s}\right)\left(\sqrt{\frac{83.80\text{ g/mol}}{20.18\text{ g/mol}}}\right)\\ \approx 127\text{ s}\end{array}$

Hence time period required for $\text{Kr}$ to effuse is $127\text{ s}$.

(d)

Interpretation Introduction

Interpretation:

Time period required for $\text{NO}$ to effuse has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.6E

Time period required for $\text{NO}$ to effuse is $75.7\text{ s}$.

Explanation of Solution

The expression for molar masses of $\text{NO}$ and $\text{Ne}$ gases and their time for effusion is as follows:

  $\frac{\text{Time for NO to effuse}}{\text{Time for Ne to effuse}}=\sqrt{\frac{M_{\text{NO}}}{M_{\text{Ne}}}}$        (7)

Rearrange equation (1) to calculate value of time for $\text{NO}$ to effuse.

  $\text{Time for NO to effuse}=\left(\text{Time for Ne to effuse}\right)\left(\sqrt{\frac{M_{\text{NO}}}{M_{\text{Ne}}}}\right)$        (8)

Substitute $30.01\text{ g/mol}$ for $M_{\text{NO}}$, $20.18\text{ g/mol}$ for $M_{\text{Ne}}$, and $62.1\text{ s}$ for time for $\text{Ne}$ to effuse in equation (2).

  $\begin{array}{c}\text{Time for NO to effuse}=\left(62.1\text{ s}\right)\left(\sqrt{\frac{30.01\text{ g/mol}}{20.18\text{ g/mol}}}\right)\\ \approx 75.7\text{ s}\end{array}$

Hence time period required for $\text{NO}$ to effuse is $75.7\text{ s}$.