# Using Graham's law of effusion, if #"CO"_2# takes 32 sec to effuse, how long will hydrogen take?

##
using Graham's law of effusion, if co2 takes 32sec to effuse how long will the hydrogen take?

using Graham's law of effusion, if co2 takes 32sec to effuse how long will the hydrogen take?

##### 1 Answer

#### Explanation:

**Graham's Law of Effusion** tells you that the *rate of effusion* of a gas is **inversely proportional** to the square root of the mass of the particles of gas.

This can be written as

#color(blue)("rate of effusion " prop color(white)(a)1/sqrt("molar mass"))#

Essentially, the rate of effusion of a gas will depend on how *massive* its molecules are. The **heavier** the molecules, the *slower* the rate of effusion.

Likewise, the **lighter** the molecules, the *faster* the rate of effusion.

Right from the start, you should look at your two gases, carbon dioxide, **less time** for the hydrogen gas to effuse when compared with the carbon dioxide.

This is a valid prediction because the molecules of hydrogen gas are **smaller**, i.e. **lighter**, than the molecules of carbon dioxide.

Look for the *molar mass** of the two gases

#"For CO"_2: " 44.01 g mol"^(-1)#

#"For H"_2: " 2.016 g mol"^(-1)#

The rates of effusion of the two gases can be written as

#"rate"_(CO_2) prop 1/sqrt("44.01 g mol"^(-1))#

and

#"rate"_(H_2) prop 1/sqrt("2.016 g mol"^(-1))#

This means that you can write

#"rate"_(CO_2)/"rate"_(H_2) = 1/sqrt("44.01 g mol"^(-1)) * sqrt("2.016 g mol"^(-1))/1#

#"rate"_(CO_2)/"rate"_(H_2) = sqrt((2.016 color(red)(cancel(color(black)("g mol"^(-1)))))/(44.01color(red)(cancel(color(black)("g mol"^(-1)))))) = 0.214#

This means that you have

#"rate"_(H_2) = "rate"_(CO_2) * 1/0.241#

#"rate"_(H_2) = 4.15 * "rate"_(CO_2)#

As predicted, hydrogen gas will effuse at a **faster rate** than carbon dioxide.

Therefore, if it takes **times faster**, you can say that you have

#t_(H_2) = "32 s"/4.15 = color(green)("7.7 s")#

The answer is rounded to two **sig figs**.

**SIDE NOTE** *The answer will vary a little depending on which values you take for the molar masses of the two gases.*

*If other values are given to you, for example* *and* *simply redo the calculations and follow the same steps shown here*.