Question

He rate of effusion of Ne gas through a porous barrier is observed to be 8.45x10 to the negative 4th mol/h. Under the same conditions, the rate of effusion of NH3gas would be ? mol/h.

Answer 1

The ammonia effuses at a rate of `9.20 ×10^"-4" color(white)(l) "mol/h"`.

Explanation:

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

We usually write the formula as:

`color(blue)(bar(ul(|color(white)(a/a)r_2/r_1 = sqrt(M_1/M_2)color(white)(a/a)|)))" "`

where:

`r_1` and `r_2` are the rates of effusion of two gases
`M_1` and `M_2` are the molar masses of the gases

Let `"Ne"` be Gas 1 and `"NH"_3` be Gas 2. Then

`r_1 = 8.45 × 10^"-4"color(white)(l) "mol/h"; M_1 = "20.18 g/mol"`
`r_2 = ?; color(white)(mmmmmmmm)M_2 = "17.03 g/mol"`

`r_2/r_1 = sqrt((20.18 color(red)(cancel(color(black)("g/mol"))))/(17.03 color(red)(cancel(color(black)("g/mol"))))) = 1.089`

`r_2 = 1.089r_1 = 1.089 × 8.45 × 10^"-4"color(white)(l) "mol/h" = 9.20 ×10^"-4" color(white)(l) "mol/h"`

The ammonia effuses at a rate of `9.20 ×10^"-4" color(white)(l) "mol/h"`.