Chapter 5, Problem 5.63P

(a)

Interpretation Introduction

Interpretation: The increasing order of effusion rate for the mixture of disulfur difluoride, dinitrogen tetrafluoride and sulfur tetrafluoride gases placed in an effusion apparatus needs to be determined.

Concept Introduction:

Graham's Law of effusion of gases states that at same conditions, the effusion rate of two gases vary inversely with the square roots of molar masses of the gases. The mathematical expression for the Graham’s law for two gases ‘a’ and ‘b’ can be written as:

  $\frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}=\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}$

Answer to Problem 5.63P

The rate of effusion of gases must be:

  ${\text{SF}}_{\text{4 }}{\text{< N}}_{\text{2}}{\text{F}}_{\text{4}}{\text{ < S}}_{\text{2}}{\text{F}}_{\text{2}}$

Explanation of Solution

As the molar mass increases, the rate of effusion decreases. The molar mass of gases are:

• Disulfur difluoride = 102.13 g/mol
• Dinitrogen tetrafluoride =104.00 g/mol
• Sulfur tetrafluoride = 108.07 g/mol

Hence, the rate of effusion of gases must be:

  ${\text{SF}}_{\text{4 }}{\text{< N}}_{\text{2}}{\text{F}}_{\text{4}}{\text{ < S}}_{\text{2}}{\text{F}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation: The ratio of effusion rate of gases disulfur difluoride and dinitrogen tetrafluoride is placed in an effusion apparatus needs to be determined.

Concept Introduction:

Graham's Law of effusion of gases states that at same conditions, the effusion rate of two gases vary inversely with the square roots of molar masses of the gases. The mathematical expression for the Graham’s law for two gases ‘a’ and ‘b’ can be written as:

  $\frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}=\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}$

Answer to Problem 5.63P

  $\frac{{\text{Rate}}_{\text{(disulfur difluoride )}}}{{\text{Rate}}_{\text{(dinitrogen tetrafluoride}}{}_{\text{)}}}\text{=}1.0$

Explanation of Solution

Graham's Law of effusion of gases states that at the same conditions the rate of effusion of two different gases are inversely proportional to the square roots of their molar masses. The mathematical expression for the Graham’s law for two gases ‘a’ and ‘b’ can be written as:

  $\begin{array}{l}\frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}=\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}\\ \frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}=\frac{{\text{Time}}_{\text{(b}}{}_{\text{)}}}{{\text{Time}}_{\text{(a)}}}=\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}\end{array}$

As the molar mass increases, the rate of effusion decreases. The molar mass of gases are:

• Disulfur difluoride = 102.13 g/mol
• Dinitrogen tetrafluoride = 104.00 g/mol

Substitute the values of molar mass to calculate the rate of effusion:

  $\begin{array}{l}\frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}\text{=}\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}\\ \frac{{\text{Rate}}_{\text{(disulfur difluoride )}}}{{\text{Rate}}_{\text{(dinitrogen tetrafluoride}}{}_{\text{)}}}\text{=}\sqrt{\frac{{\text{Molar mass }}_{\text{(dinitrogen tetrafluoride)}}}{{\text{Molar mass}}_{\text{(disulfur difluoride}}{}_{\text{)}}}}\\ \frac{{\text{Rate}}_{\text{(disulfur difluoride )}}}{{\text{Rate}}_{\text{(dinitrogen tetrafluoride}}{}_{\text{)}}}\text{=}\sqrt{\frac{\text{104}\text{.00}}{\text{102}\text{.13}}}\\ \frac{{\text{Rate}}_{\text{(disulfur difluoride )}}}{{\text{Rate}}_{\text{(dinitrogen tetrafluoride}}{}_{\text{)}}}\text{=}1.0\end{array}$

(c)

Interpretation Introduction

Interpretation: The molar mass of X gas that is added to gas mixture and effuses at 0.935 times at the rate of ${\text{SF}}_{\text{4}}$ needs to be determined

Concept Introduction:

Graham's Law of effusion of gases states that at same conditions, the effusion rate of two gases vary inversely with the square roots of molar masses of the gases. The mathematical expression for the Graham’s law for two gases ‘a’ and ‘b’ can be written as:

  $\frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}=\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}$

Answer to Problem 5.63P

  ${\text{Molar mass}}_{\text{(X}}{}_{\text{)}}\text{=123}\text{.6g/mol}$

Explanation of Solution

As the molar mass increases, the rate of effusion decreases. The molar mass of gas:

• Sulfur tetrafluoride = 108.07 g/mol

Substitute the values to calculate the molar mass:

  $\begin{array}{l}\frac{{\text{Rate}}_{\text{(a)}}}{{\text{Rate}}_{\text{(b}}{}_{\text{)}}}\text{=}\sqrt{\frac{{\text{Molar mass }}_{\text{(b)}}}{{\text{Molar mass}}_{\text{(a}}{}_{\text{)}}}}\\ \frac{{\text{Rate}}_{\text{(X )}}}{{\text{Rate}}_{\text{(SF4}}{}_{\text{)}}}\text{=}\frac{{\text{Rate}}_{\text{(SF4}}{}_{\text{)}}\text{×0}\text{.935}}{{\text{Rate}}_{\text{(SF4}}{}_{\text{)}}}\text{=}\sqrt{\frac{{\text{Molar mass }}_{\text{(SF4)}}}{{\text{Molar mass}}_{\text{(X}}{}_{\text{)}}}}\\ \text{0}\text{.935=}\sqrt{\frac{\text{108}\text{.07 g/mol }}{{\text{Molar mass}}_{\text{(X}}{}_{\text{)}}}}\\ \frac{\text{108}\text{.07 g/mol }}{{\text{Molar mass}}_{\text{(X}}{}_{\text{)}}}\text{=0}\text{.874}\\ {\text{Molar mass}}_{\text{(X}}{}_{\text{)}}\text{=}\frac{\text{108}\text{.07 g/mol }}{\text{0}\text{.874}}\\ {\text{Molar mass}}_{\text{(X}}{}_{\text{)}}\text{=123}\text{.6g/mol}\end{array}$

Thus, molar mass of unknown gas is 123.6 g/mol.