What is the ratio predicted from Graham’s law for rates of diffusion for \[N{H_3}/HCl\]?
Answer
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Hint: Graham's Law states that a gas's rate of diffusion or effusion is inversely proportional to its molecular weight squared. To put it another way, at constant pressure and temperature, molecules or atoms with lower molecular masses can effuse faster than molecules or atoms with higher molecular masses.
Complete answer:
We use the following equation,
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}\]
Where,
\[Rat{e_1} = \] The rate of effusion for the first gas
\[Rat{e_2} = \] The rate of effusion for the second gas
\[{M_1} = \]Molar mass of gas 1
\[{M_2} = \]Molar mass of gas 2
We may infer from Graham's law that the rate of diffusion of a gas is inversely proportional to the square root of the gas's molecular mass.
\[rat{e_1} \propto \dfrac{1}{{\sqrt {{m_1}} }}\] and \[rat{e_2} \propto \dfrac{1}{{\sqrt {{m_2}} }}\]
That is,
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\sqrt {{m_2}} }}{{\sqrt {{m_1}} }}\]
We know that the molar mass of \[HCl\] is \[36.458g{(mol)^{ - 1}}\] and the molar mass of \[N{H_3}\] is \[17.031g{(mol)^{ - 1}}\].
By substituting the values to the equation, we get
\[\dfrac{{{r_{N{H_3}}}}}{{{r_{HCl}}}} = \dfrac{{\sqrt {36.458} }}{{\sqrt {17.031} }}\]
\[ = 1.46\]
This shows that ammonia gas diffuses at a much slower rate of 1.46 than hydrogen chloride gas.
Additional Information:
Remember that from the previous statement we know that a gas's rate of diffusion or effusion is inversely proportional to its molecular weight squared. If one gas has four times the molecular weight of another, it can diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate. Gases that are heavier disperse more slowly.
Note:
The molar mass is proportional to the mass density at the same temperature and pressure. As a result, the square roots of the mass densities of various gases are inversely proportional to their diffusion rates.
Complete answer:
We use the following equation,
\[\dfrac{{Rat{e_1}}}{{Rat{e_2}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}\]
Where,
\[Rat{e_1} = \] The rate of effusion for the first gas
\[Rat{e_2} = \] The rate of effusion for the second gas
\[{M_1} = \]Molar mass of gas 1
\[{M_2} = \]Molar mass of gas 2
We may infer from Graham's law that the rate of diffusion of a gas is inversely proportional to the square root of the gas's molecular mass.
\[rat{e_1} \propto \dfrac{1}{{\sqrt {{m_1}} }}\] and \[rat{e_2} \propto \dfrac{1}{{\sqrt {{m_2}} }}\]
That is,
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\sqrt {{m_2}} }}{{\sqrt {{m_1}} }}\]
We know that the molar mass of \[HCl\] is \[36.458g{(mol)^{ - 1}}\] and the molar mass of \[N{H_3}\] is \[17.031g{(mol)^{ - 1}}\].
By substituting the values to the equation, we get
\[\dfrac{{{r_{N{H_3}}}}}{{{r_{HCl}}}} = \dfrac{{\sqrt {36.458} }}{{\sqrt {17.031} }}\]
\[ = 1.46\]
This shows that ammonia gas diffuses at a much slower rate of 1.46 than hydrogen chloride gas.
Additional Information:
Remember that from the previous statement we know that a gas's rate of diffusion or effusion is inversely proportional to its molecular weight squared. If one gas has four times the molecular weight of another, it can diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate. Gases that are heavier disperse more slowly.
Note:
The molar mass is proportional to the mass density at the same temperature and pressure. As a result, the square roots of the mass densities of various gases are inversely proportional to their diffusion rates.
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