A gaseous sample of a certain compound follows the postulates of the kinetic mol

Question

A gaseous sample of a certain compound follows the postulates of the kinetic molecular theory and has a most probable speed pf 264.13 m/s at a certain temperature.

A) The molecules in this sample are a binary compound of sulfur and nitrogen that is 30.4 mass percent nitrogen. The molar mass of this compound is between one hundred fifty and two hundred g/mole. Determine the molecular formula of this compound and the temperature of this sample in degrees Celsius.

B) This gaseous sample, at the above temperature, is in a 5.00 L container at a pressure of 280 torr. Determine the collision frequency of these molecules with a 2.00 mm2 area of the container wall. Express your answer in moles/second.

Explanation / Answer

A) Most probable speed=cmp=1.4(RT/M)^1/2 where R=gas const   M=molar mass T= temperature

264.13 m/s=1.4(8.314 kg m2s-2 k-1)(T)/M)^1/2

Or, T/M=4281.04 k mol/kg………………..(1)

Now , let the molecular formula be SxNy

So we can write , 0.30 *M=y*14

And 0.70*M=x*32

Or ,0.30M/0.70M=14y/32x

Or,y/x=0.99=1(approx)

Empirical formula(SN)n

But given that molar mass of this compound is between one hundred fifty and two hundred g/mole

(SN)n=150g/mol     to 200g/mol

(32 +14)n=150g/mol     to 200g/mol

Or, (46)n=150g/mol       or , (46)n=200g/mol   

Or,n=3.26=3             to n=4.34=4

With n=3

(SN)3=138g/mol            

With n=4

(SN)4=184g/mol         

Molecular mass can be taken as the average of these two,M=138g/mol +184 g/mol/2=161g/mol

Using eqn( 1)

, T/M=4281.04 k mol/kg

T=4281.04 k mol/kg *(161g/mol *10^-3 kg/g)=689K

T=689K  

M=161g/mol

B) T=689 K

P=280 torr=280*0.0013157 atm/torr=0.37 atm

Collision frequency=Z1=(2)^1/2 * (d)^2* Cav*n’

Where (d)=collision diameter=(sum of radius of atoms N and S)=65pm+100pm=165pm=165*10^-12 m

Cav=average speed=(8RT/M)^1/2

=(8*8.314 kgm2s-2 k-1 mol-1 *689K/3.14* 161 *10^-3 kg/mol)^1/2

=301.08 m/s

N’=no of molecules per unit volume=N/V=NP/RT

=(0.37 atm) *(6.022*10^23 molecules/mol )(10^3 Lm-3)/0.0821 atm L k-1 mol-1*(689K)

=3.9218 *10^24 molecules/m3

Z1=(2)^1/2 *3.14 * (165*10^-12 m)^2 *(301.08 m/s)*( 3.9218 *10^24 molecules/m3)

=1.43*10^8 collisions/s

Z11=collisions per unit volume=1/2 Z1 n’=0.5*(1.43*10^8 collisions/s)*( 3.9218 *10^24 molecules/m3)

=2.80*10^32 collisions m-3 s-1

In volume=5L=5*10^-3 m3

Collision frequency=(2.80*10^32 collisions m-3 s-1)*( 5*10^-3 m3)=14*10^29 collisions/s

In terms of moles/S

Collision frequency=14*10^29 collisions/s (1 collision per pair of molecules)

                                     =(14*10^29 collisions/s)*2 molecules/collision

                                   =28*10^29 molecules/s

                                       =28*10^29 molecules/s/ (6.022*10^23 molecules/mol)

                                     =4.65*10^6 mol/s

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