(a)

#S_((s))+O_(2(g))rarrSO_(2(g))#

The expression to use which relates free energy change to the concentrations of all the reacting species is:

#DeltaG_(r)=DeltaG^(@)+RTlnQ " "color(red)((1))#

#Q# is termed the reaction quotient which, in this case, is given by:

#Q=([SO_2])/([O_2])#

As the reaction proceeds to equilibrium the value of #DeltaG_r# increases to zero. When this point is reached we can say from #color(red)((1))# that:

#0=DeltaG^(@)+RTlnQ " "color(red)((2))#

Because we are at equilibrium we can say that #Q=K_c#.

This means that #color(red)((2))# becomes #rArr#

#DeltaG^(@)=-RTlnK_c " "color(red)((3))#

This is an important expression in chemistry as you can predict, given the relevant thermodynamic data, the equilibrium constant of a reaction.

This is what we can now do with this question:

From #color(red)((3))# we get:

#lnK_c=-(DeltaG^(@))/(RT)#

#:.lnK_c=-((-300.4xx10^(3))/(8.31xx298))#

#lnK_c=121.3#

#K_c=4.814#

(b)

Now we just need to insert the values we have been given into eqn #color(red)((1))rArr#

#DeltaG_r=(-300.4xx10^3)+8.31xx298xxln(0.03/0.001)#

#DeltaG_r=(-300.4xx10^3)+8422.66#

#DeltaG_r=-291,977"J"#

#DeltaG_r=-292"kJ"#

Just a side note about the reaction quotient #Q#. Here you can see that it is larger than #K_c# (30 v 4.8). This means that there is a predominance of products so the equilibrium will shift to the left.

The reverse happens if #Q# <#K_c#