Question #26e77

1 Answer
Jan 29, 2017

Answer:

#sf((a))# 1.00 Atm

#sf((b))# 0.0745 Atm

#sf((c))# 1.92 Atm

#sf((d))# #sf(alpha=0.925)#

Explanation:

#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.