Moles of I2/I mixture at 1.3 atm can be calculated from gas law equation ,n= PV/

Question

Moles of I2/I mixture at 1.3 atm can be calculated from gas law equation ,n= PV/RT

Given P = 1.3 atm, V= 10L, R=0.0821 L.atm/mole.K and T= 1473K

Hence no of moles, n= 1.3*10/(0.0821*1473)=0.11 moles

Mass of the mixture = 20 gm, let x= mass of I2, 20-x= mass of I

Moles of I2= x/254 and moles of I= (20-x)/126.9

Total moles of mixture = x/254+(20-x)/126.9= 0.11

Hence x/254+20/126.9-x/126.9= 0.11, x/126.9-x/254= 0.048, x=12.2 gm

Mass of I2= 12.2 gm and mass of I= 20-12.2= 7.8 gm

Moles of I2= 12.2/254 =0.048 and I= 7.8/126.9=0.062

Concentrations = moles/Volume, Concentrations : I2= 0.048/10=0.0048 and I= 0.062/10=0.0062

Q= reaction coefficient for the reaction I2(g)ßà2I(g), Q= [I]2/ [I2] =0.0062*0.0062/0.0048=0.008<Kc equilibrium constant so forward reaction takes place.

Let x= drop in concentration of I2 to reach equilibrium, at equilibrium [I2] =0.0048-x and [I]=0.0062+2x

Now KC= (0.0062+2x)2/(0.0048-x)= 0.0258, when solved for x using excel, x=0.00151

Hence at equilibrium [I2] =0.0048-0.00151=0.00329 and [I] =0.0062+2*0.00151=0.00922

Moles at Equilibrium [I]=0.00329*10=0.0329 and [I] =0.00922*10=0.0922

Total moles,n = 0.0329+0.0922=0.1251, Pressure after equilibrium is reached is P= nRT/V= 0.1251*0.0821*1473/10=1.51 atm

Explanation / Answer

ion III) (10 points) Consider the dissociation of iodine V At 1473 K. K pressure of 1.30 atm, what are the mole fractions of the two gases in the 12 /I mixture? (b) After a ce : 12(8)--21(g) 0.0258, 20.0-g of an 12 /1 mixture exerts a pressure l .30 atm in a v c olume of 10.0 L. (a) At the ases in the 12 /I mixture? (b) After a certain period at equilibrium? (Hint: The molar mass of 12 253.8 g/mol)

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