InterviewSolution
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InterviewSolution
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The value of ` K_(p) " is " 1 xx10^(-3) " atm"^(-1) " at " 25^(@) C. " for the reaction " : 2 NO + Cl_(2) hArr2 NOCl`. A flask contains NO at `0*02 " atm at " 25^(@)(C).` Calculate the mol of `Cl_(2)` that must be added if 1 % of the NO is to be converted to NOCl at equilibrium. The volume of the flask is such that ` 0*2` mol of the gas produce 1 atm pressure at ` 25^(@)C`. ( Ignore the probable association of NO to ` N_(2)O_(2)`) |
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Answer» Suppose intial pressure of `Cl_(2)` added is p atm . `{:("Then ",2NO,+,Cl_(2),hArr,2 NOCl),("Intial",0*02 "atm",,p " atm",,),("A eqm. " ,0*02 - (0*02)/100,,(p-(0*01)/100),,(0*02)/100),(,=2 xx 10^(-2)-2 xx10^(-4),,=p-10^(-4),,-2 xx 10^(-4)"atm"),(,=2 xx 10^(-4)(100-1),,,,),(,=198 xx10^(-4) "atm",,,,):}` ` K_(p) = (p_(NOCl)^(2))/(p_(NO)^(2)xx p_(Cl_(2)) )` ` 10^(-3) = ( 2xx 10^(-4))^(2)/ ((198 xx 10^(-4))^(2)xx(p-10^(-4))) ` or ` (p- 10^(-4)) = 4/(198)^(2) xx 1/ (10^(-3))= 0* 102 or p= 0* 102 + 0* 0001 = 0* 1021 "atm "` Volume of the vessel can be calculated as follows : ` PV = nRT or V = (nRT)/P = (0* 2 xx 0* 082 xx 273 )/1 L = 4* 887 L ` To calculate the number of moles of `Cl_(2)` , again apply ` PV = nRT or n = (PV)/ (RT) = (0*1021 xx 4* 887 )/(0*082 xx 298) = 0* 204 "mol"` |
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