Question #6cee9

2 Answers
Apr 18, 2015

The ratio of rates Cl_2:N_2=0.63Cl2:N2=0.63

Graham's Law states that:

Rate of effusion prop1/sqrt(M_r)1Mr

Where M_rMr is the relative molecular mass.

For two gases:

R_1/R_2=sqrt(M_(r(2))/(M_(r(1)))R1R2=Mr(2)Mr(1)

M_r[Cl_2]=(35.5xx2)=71Mr[Cl2]=(35.5×2)=71

M_r[N_2]=(14xx2)=28Mr[N2]=(14×2)=28

So R_(Cl_2)/(R_(N_2)RCl2RN2=sqrt((28)/(71))=0.63=2871=0.63

Use Grahams's law, which states that the rate of effusion of a gas is inversely proportional to the square root of molar mass.

r_"eff" prop 1/sqrt(M_M)reff1MM

This means that the rate of effusion of chlorine gas will be

r_"chlorine" prop 1/sqrt(M_("M chlorine"))rchlorine1MM chlorine

Likewise, the rate of effusion of nitrogen gas will be

r_"nitrogen" prop 1/sqrt(M_("M nitrogen"))rnitrogen1MM nitrogen

If you divide these two expressions, you'll get the ratio of effusion rates of chlorine and nitrogen

r_"chlorine"/r_"nitrogen" = (1/sqrt(M_("M chlorine")))/(1/sqrt(M_("M nitrogen"))) = sqrt(M_("M nitrogen"))/sqrt(M_("M chlorine"))rchlorinernitrogen=1MM chlorine1MM nitrogen=MM nitrogenMM chlorine

The numerical value of this ratio will be

r_"chlorine"/r_"nitrogen" = (sqrt(28.014cancel("g/mol")))/(sqrt(70.906cancel("g/mol"))) = color(green)(0.629)

Here is a video which shows how to solve a different problem using Graham's law.

Video from: Noel Pauller

Hope this helps!