Ejercicios FyQPortada del sitio > Chemistry exercises > Matter and gas laws > Application of Graham’s law: hydrogen diffusion rate (8308)
Application of Graham’s law: hydrogen diffusion rate (8308)
Jueves 12 de septiembre de 2024, por
Determine the diffusion rate of hydrogen, knowing that the diffusion rate of oxygen is 2 minutes.
Graham’s law relates the diffusion rates of two gases to their molecular masses. If the gases are A and B, the relationship is:
![\color[RGB]{2,112,20}{\bm{\frac{v_A}{v_B} = \sqrt{\frac{M_B}{M_A}}}}](local/cache-vignettes/L126xH65/4cf3fe04d22fc72e2d726e95ac2bdc33-76a3d.png?1733044089 "\color[RGB]{2,112,20}{\bm{\frac{v_A}{v_B} = \sqrt{\frac{M_B}{M_A}}}}")
For hydrogen and oxygen, both being diatomic, the molecular masses are:
![\left \ce{H2}: 2\cdot 1 = {\color[RGB]{0,112,192}{\bf 2\ u}} \atop \ce{O2}: 2\cdot 16 = {\color[RGB]{0,112,192}{\bf 32\ u}} \right \}](local/cache-vignettes/L179xH52/fa228ad48fcef891303029aae15bb4aa-70373.png?1733044089 "\left \ce{H2}: 2\cdot 1 = {\color[RGB]{0,112,192}{\bf 2\ u}} \atop \ce{O2}: 2\cdot 16 = {\color[RGB]{0,112,192}{\bf 32\ u}} \right \}")
The relationship between their diffusion rates will be:
![\frac{v_{\ce{H_2}}}{v_{\ce{O_2}}}= \sqrt{\frac{32\ \cancel{u}}{2\ \cancel{u}}} = \sqrt{16} = \color[RGB]{0,112,192}{\bf 4}](local/cache-vignettes/L236xH55/d7383bc419efafca3c82287879aa12d2-6ae53.png?1733051299 "\frac{v_{\ce{H_2}}}{v_{\ce{O_2}}}= \sqrt{\frac{32\ \cancel{u}}{2\ \cancel{u}}} = \sqrt{16} = \color[RGB]{0,112,192}{\bf 4}")
This means hydrogen diffuses four times faster than oxygen, so the diffusion rate of hydrogen will be 0.5 minutes, or 30 seconds.