Two gases HBr and CH4 effuse through a small opening. HBr effuses through the opening at a rate of 4 cm^3 s^-1. At what will the methane gas effuse through the same opening?
mm = molar mass
(rate HBr/rate CH4) = sqrt (mm CH4/mm HBr)
[(4 cm/s)/(rate CH4)] = sqrt (16/80.9)
Solve for rate CH4. Post your work if you get stuck.
yes
Rate(CH4)= 8.99cm/s
To determine the rate at which methane (CH4) gas will effuse through the same opening as hydrogen bromide (HBr) gas, we can use Graham's Law of effusion. According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Step 1: Find the molar masses of the gases.
The molar mass of HBr (hydrogen bromide) is:
1 atom of hydrogen (H) = 1.0079 g/mol
1 atom of bromine (Br) = 79.904 g/mol (rounded to three decimal places)
So, the molar mass of HBr = 1.0079 g/mol + 79.904 g/mol = 80.9129 g/mol (rounded to four decimal places).
The molar mass of CH4 (methane) is:
1 atom of carbon (C) = 12.011 g/mol
4 atoms of hydrogen (H) = 4.032 g/mol (rounded to three decimal places)
So, the molar mass of CH4 = 12.011 g/mol + 4.032 g/mol = 16.043 g/mol (rounded to three decimal places).
Step 2: Use Graham's law to compare the rates of effusion.
The rate of effusion of HBr is given as 4 cm^3 s^-1.
Let's assume the rate of effusion of CH4 is x cm^3 s^-1.
According to Graham's Law, the ratio of the rates of effusion is equal to the square root of the inverse ratio of the molar masses:
Rate of HBr / Rate of CH4 = √(Molar mass of CH4 / Molar mass of HBr)
Thus, we can set up the proportion as follows:
4 cm^3 s^-1 / x cm^3 s^-1 = √(16.043 g/mol / 80.9129 g/mol)
Step 3: Solve for x.
To isolate x, we can cross-multiply the equation:
4 cm^3 s^-1 * √(80.9129 g/mol / 16.043 g/mol) = x cm^3 s^-1
Simplifying the equation,
x = 4 cm^3 s^-1 * √(80.9129 / 16.043)
x = 4 cm^3 s^-1 * √(5.04457)
x ≈ 9.000 cm^3 s^-1 (rounded to three decimal places)
Therefore, methane gas (CH4) will effuse through the same opening at a rate of approximately 9 cm^3 s^-1.