Which gas diffuses most rapidly at STP? Ar, Kr, N2, or O2? Why?
N2
which is the least dense?
density=molmass/22.4
so the gas with the least molmass is the most rapid diffuser. Nitrogen at 28, Oxygen at 32, Argon at 40, Kr at...
The gas that diffuses most rapidly at STP (Standard Temperature and Pressure) is hydrogen (H2), not Ar, Kr, N2, or O2.
Hydrogen has the smallest molecular mass (2 g/mol) among all the given gases. According to Graham's law of effusion and diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Mathematically, the rate of diffusion is given by:
Rate of gas A / Rate of gas B = square root of (Molar mass of gas B / Molar mass of gas A)
Since hydrogen has the smallest molar mass, it will have the highest rate of diffusion among the given gases.
However, among the options provided (Ar, Kr, N2, and O2), nitrogen (N2) will diffuse most rapidly because it has a smaller molar mass (28 g/mol) compared to the other gases. Oxygen (O2) has a molar mass of 32 g/mol, which is slightly higher than nitrogen, so nitrogen will diffuse slightly faster than oxygen.
To determine which gas diffuses most rapidly at STP (standard temperature and pressure), we need to consider the molecular properties of the gases.
The rate of diffusion depends on several factors, including the molecular mass, temperature, and the size of the gas molecules. According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Let's calculate the molar masses of the gases in question:
- Ar (Argon): 39.95 g/mol
- Kr (Krypton): 83.80 g/mol
- N2 (Nitrogen): 28.02 g/mol
- O2 (Oxygen): 32.00 g/mol
Using Graham's law, we can calculate the ratio of the diffusion rates of each gas compared to nitrogen (N2) because nitrogen has the lowest molar mass among the given gases:
Rate of diffusion (Ar) = Square root of (Molar mass of N2 divided by Molar mass of Ar)
Rate of diffusion (Kr) = Square root of (Molar mass of N2 divided by Molar mass of Kr)
Rate of diffusion (O2) = Square root of (Molar mass of N2 divided by Molar mass of O2)
Calculating these ratios, we get:
- Ar: Square root of (28.02 g/mol ÷ 39.95 g/mol) ≈ 0.84
- Kr: Square root of (28.02 g/mol ÷ 83.80 g/mol) ≈ 0.51
- O2: Square root of (28.02 g/mol ÷ 32.00 g/mol) ≈ 0.74
From the calculations, we can see that the rate of diffusion is highest for Ar (Argon) as it has the lowest molar mass compared to the other gases. Therefore, Ar diffuses most rapidly at STP.
In summary, according to Graham's law of diffusion, Ar (Argon) diffuses most rapidly at STP because it has the lowest molar mass among the given gases.