How much faster does nitrogen monoxide (NO) effuse in comparison to dinitrogen tetraoxide (N2O4)
The effusion rate of a gas is determined by 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.
To compare the effusion rates of nitrogen monoxide (NO) and dinitrogen tetraoxide (N2O4), we need to compare their molar masses.
The molar mass of NO is approximately 30.01 g/mol (14.01 g/mol for nitrogen and 16.00 g/mol for oxygen).
The molar mass of N2O4 is approximately 92.02 g/mol (28.01 g/mol for nitrogen and 16.00 g/mol for oxygen, multiplied by 2).
Now, we can compare the ratio of their effusion rates. Let's assume the effusion rate of NO is x, and the effusion rate of N2O4 is y.
According to Graham's law of effusion:
(x/y) = √(M2/M1)
where M1 and M2 are the molar masses of the gases.
Plugging in the values:
(x/y) = √(92.02 g/mol / 30.01 g/mol)
(x/y) = √3.06
(x/y) ≈ 1.75
Therefore, nitrogen monoxide (NO) effuses at a rate approximately 1.75 times faster than dinitrogen tetraoxide (N2O4).
To determine how much faster one gas effuses compared to another, we need to use Graham's law of effusion, which states that the rate of effusion (or the rate at which a gas escapes through a small hole) is inversely proportional to the square root of its molar mass.
Mathematically, Graham's law of effusion can be expressed as:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)
In this case, we have nitrogen monoxide (NO) and
dinitrogen tetraoxide (N2O4).
To apply Graham's law, we need to find the molar masses of NO and N2O4.
The molar mass of NO can be calculated by adding up the atomic masses of nitrogen (N) and oxygen (O):
Molar mass of NO = (1*14.01) + (1*16.00) = 30.01 g/mol
The molar mass of N2O4 can be calculated by adding up the atomic masses of two nitrogen (N) atoms and four oxygen (O) atoms:
Molar mass of N2O4 = (2*14.01) + (4*16.00) = 92.02 g/mol
Now, let's plug these values into Graham's law:
Rate(NO) / Rate(N2O4) = √(Molar mass(N2O4) / Molar mass(NO))
Rate(NO) / Rate(N2O4) = √(92.02 g/mol / 30.01 g/mol)
Simplifying the equation:
Rate(NO) / Rate(N2O4) = √(3.066)
Squaring both sides:
(Rate(NO) / Rate(N2O4))^2 = 3.066
From here, we can solve for the ratio of the rates:
Rate(NO) / Rate(N2O4) = √(3.066) ≈ 1.75
Therefore, nitrogen monoxide (NO) effuses approximately 1.75 times faster than dinitrogen tetraoxide (N2O4).
rate NO
rate N2O4
mm NO = molar mass NO
mmN2O4 = molar mass N2O4.
(rateNO/rate N2O4) - sqrt (mmN2O4/mmNO)_