Question

Compare the rates of effusion of nh3 and o2

Answer 1

Answer : The rate of effusion of ammonia gas is 1.37 times of rate of effusion of oxygen gas.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

`R\propto \sqrt{(1)/(M)}`

or,

`(R_1)/(R_2)=\sqrt{(M_2)/(M_1)}`

where,

`R_1` = rate of effusion of ammonia gas

`R_2` = rate of effusion of oxygen gas

`M_1` = molar mass of ammonia gas = 17 g/mole

`M_2` = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula, we get :

`(R_1)/(R_2)=\sqrt{(32g/mole)/(17g/mole)}`

`(R_1)/(R_2)=1.37`

`R_1=1.37* R_2`

Therefore, from this we conclude that, the rate of effusion of ammonia gas is 1.37 times of rate of effusion of oxygen gas.

Answer 2

According the Graham's Law of effusion, if we let M₁ and R₁ be the mass and effusion rate of NH₃ and M₂ and R₂ be the mass and effusion rate of O₂, then:

R₂/R₁ = √(M₁/M₂).

Since M₁ = 14.01 + 3*1.01 = 17.04 and M₂ = 16 * 2 = 32, we see that:

R₂/R₁ = √(17.04/32) = 0.73
==> R₂ = 0.73 * R₁.

Thus, O₂ effuses at 73% the speed that NH₃ does