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))
rArrsf(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.
