# Question #0f696

##### 1 Answer

#K ~~ 9.226 xx 10^(-5)#

This is just a fancy way of making use of the **state function property** of the Gibbs' free energy, i.e.

- ...that it does not matter what path you take, as long as you know what the
**initial**and**final**states are, and... - ...the steps in a path
**add**to give the overall#DeltaG# (at a given temperature and pressure).

We know that:

#DeltaG_"tot" = DeltaG_1 + DeltaG_2 + . . . #

And so, at **overall** reaction is related to the *individual*

#DeltaG_1^@ + DeltaG_2^@ = DeltaG_"rxn"^@ = -"8.80 kJ/mol"#

And you were given that

#DeltaG_1^@ = DeltaG_"rxn"^@ - DeltaG_2^@#

#= -"8.80 kJ/mol" - (-"31.6 kJ/mol")#

#=# #"22.8 kJ/mol"#

Now,

#DeltaG = DeltaG^@ + RTlnQ# where

#Q# is the reaction quotient (for NOT being at equilibrium).

At *chemical equilibrium*,

#ul(DeltaG^@ = -RTlnK)#

Knowing *first* reaction step, we can then get the **equilibrium constant** for that *first* reaction step:

#color(blue)(K) = e^(-DeltaG_1^@ // RT)#

#= e^(-"22.8 kJ/mol" // ("0.008314472 kJ/mol"cdot"K" cdot (22 + "273.15 K"))#

#= ulcolor(blue)(9.226 xx 10^(-5))#