#sf((a))#
#sf(PCl_(5(g))rightleftharpoonsPCl_(3(g))+Cl_(2(g)))#
For which:
#sf(K_p=(p_(PCl_3)xxp_(Cl_2))/(p_(PCl_5))=11.5" ""Atm")#
No dimensions have been given for #sf(K_p)# or any value for the gas constant R. Subsequent calculations suggest it is in atmospheres so I have gone with that.
To find the pressure if no dissociation has taken place we use the ideal gas equation:
#sf(PV=nRT)#
#:.##sf(P=(nRT)/(V))#
To find #sf(n)# we use:
#sf(n=m/M_(r)=2.010/208.24=0.009652)#
#:.##sf(P=(0.009652xx0.082xx600)/(0.475)" ""Atm")#
#sf(P=1.00color(white)(x)"Atm")#
#sf((b))#
We set up an ICE table based on atmospheres:
#sf(" "PCl_(5(g))rightleftharpoonsPCl_(3(g))+Cl_(2(g)))#
#sf(I" "1.00" "0" "0)#
#sf(C" "-x" "+x" "+x)#
#sf(E" "(1.00-x)" "x" "x)#
#:.##sf(K_p=x^2/(1.00-x)=11.5)#
#:.##sf(x^2=11.5(1.00-x))#
#rArr##sf(x^2+11.5x-11.5=0)#
Applying the quadratic formula and ignoring the -ve root we get:
#sf(x=0.9255color(white)(x)"Atm")#
#:.##sf(p_(PCl_5)=1.00-0.9255=0.0745color(white)(x)"Atm")#
#sf((c))#
The total pressure is given by:
#sf(P_("tot")=(1.00-x)+x+x=(1.00+x)#
#:.##sf(P_("tot")=1.00+0.9255=1.9255color(white)(x)"Atm")#
#sf((d))#
The degree of dissociation #sf(alpha)# is the fraction of moles of #sf(PCl_5)# which have dissociated.
Since partial pressure is proportional to the number of moles we get:
#sf(alpha=0.9255/1.00=0.9255)#
#sf(alpha=0.925)#
This tells us that the #sf(PCl_5)# is 92.5% dissociated.
