Question

Question: Calculate the relative rate of diffusion of \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) (molar mass \({\rm{2}}{\rm{.0 g/mol}}\)) compared to that of \(^{\rm{2}}{{\rm{H}}_{\rm{2}}}\) (molar mass \({\rm{4}}{\rm{.0 g/mol}}\)) and the relative rate of diffusion of \({{\rm{O}}_{\rm{2}}}\) (molar mass \({\rm{32 g/mol}}\)) compared to that of \({{\rm{O}}_{\rm{3}}}\) (molar mass \({\rm{48 g/mol}}\)).

Short answer

The relative rates of diffusion of \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) and \(^2{{\rm{H}}_{\rm{2}}}\), \({{\rm{O}}_{\rm{2}}}\) and \({{\rm{O}}_{\rm{3}}}\) are \({\rm{1}}{\rm{.414}}\) and \({\rm{1}}{\rm{.22}}\) respectively.

Step-by-step solution

Concept Introduction

The Graham's law states that a gas's rate of diffusion or effusion is inversely related to the square root of its molecular weight.

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases \(A\) and \(B\) are at the same temperature and pressure, then the relative rates of diffusion for \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) and \(^2{{\rm{H}}_{\rm{2}}}\) will be:

\(\begin{array}{l}\frac{{{\rm{ rate of diffusion of}}{{\rm{ }}^{\rm{1}}}{{\rm{H}}_{\rm{2}}}}}{{{\rm{ rate of diffusion of}}{{\rm{ }}^2}{{\rm{H}}_{\rm{2}}}}}{\rm{ = }}\frac{{\sqrt {{{\rm{M}}_{^2{{\rm{H}}_{\rm{2}}}}}} }}{{\sqrt {{{\rm{M}}_{^1{{\rm{H}}_{\rm{2}}}}}} }},\;Where\;\;M\;is{\rm{ }}the{\rm{ }}molar{\rm{ }}mass.\\{\rm{ = }}\sqrt {\frac{{\rm{4}}}{{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1117}}} \\{\rm{ = 1}}{\rm{.414}}{\rm{.}}\end{array}\)

Therefore, the result obtained is \({\rm{1}}{\rm{.414}}\).

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases \(A\) and \(B\) are at the same temperature and pressure, then the relative rates of diffusion for \({{\rm{O}}_{\rm{2}}}\) and \({{\rm{O}}_{\rm{3}}}\) will be:

\(\begin{array}{l}\frac{{{\rm{ rate of diffusion of }}{{\rm{O}}_3}}}{{{\rm{ rate of diffusion of }}{{\rm{O}}_2}}}{\rm{ = }}\frac{{\sqrt {{{\rm{M}}_{{{\rm{O}}_3}}}} }}{{\sqrt {{{\rm{M}}_{{{\rm{O}}_2}}}} }},\;Where\;\;M\;is{\rm{ }}the{\rm{ }}molar{\rm{ }}mass.\\{\rm{ = }}\sqrt {\frac{{{\rm{48}}}}{{{\rm{32}}}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1117}}} \\{\rm{ = 1}}{\rm{.22}}{\rm{.}}\end{array}\)

Therefore, the result obtained is \({\rm{1}}{\rm{.22}}\).