(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
