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.
(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.
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.