객관식
According to Graham's law, how does the rate of effusion of NH_3 compare to that of He under identical conditions?
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7. Gases
Effusion
객관식
According to Graham's law of effusion, how much faster does H2 effuse compared to O2 under identical conditions?
A
H2 effuses approximately 2 times faster than O2.
B
H2 effuses at the same rate as O2.
C
H2 effuses approximately 4 times faster than O2.
D
H2 effuses approximately 8 times faster than O2.
검증된 단계별 안내1
Recall Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\), where \(r\) is the rate of effusion and \(M\) is the molar mass.
Identify the gases involved: \(H_2\) (hydrogen gas) and \(O_2\) (oxygen gas). Find their molar masses: \(M_{H_2} = 2.02\, \text{g/mol}\) and \(M_{O_2} = 32.00\, \text{g/mol}\).
Set \(r_1\) as the rate of effusion of \(H_2\) and \(r_2\) as the rate of effusion of \(O_2\). Substitute the molar masses into Graham's law formula: \(\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}\).
Calculate the ratio inside the square root: \(\frac{32.00}{2.02}\), which will give a number greater than 1, indicating \(H_2\) effuses faster.
Take the square root of the ratio to find how many times faster \(H_2\) effuses compared to \(O_2\). This final value represents the relative effusion rate.
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