Compare the rates of effusion of nh3 and o2
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 + 31.01 = 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
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.
\(R\propto \sqrt{\frac{1}{M}}\)
or,
\(\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}\)
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 :
\(\frac{R_1}{R_2}=\sqrt{\frac{32g/mole}{17g/mole}}\)
\(\frac{R_1}{R_2}=1.37\)
\(R_1=1.37\times R_2\)
Therefore, from this we conclude that, the rate of effusion of ammonia gas is 1.37 times of rate of effusion of oxygen gas.