Question

Question #0f696

Answer 1

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

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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 `25^@ "C"` and `"1 atm"`, the `DeltaG^@` for the overall reaction is related to the individual `DeltaG^@` values.

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

And you were given that `DeltaG_2^@ = -"31.6 kJ/mol"`. As a result, for the first reaction we have:

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

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

`=` `"22.8 kJ/mol"`

Now, `DeltaG` at NONSTANDARD conditions (not at `25^@ "C"` but still at `"1 atm"`) is related to the STANDARD change in Gibbs' free energy `DeltaG^@`:

`DeltaG = DeltaG^@ + RTlnQ`

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

At chemical equilibrium, `DeltaG = 0`, and `Q = K`, so:

`ul(DeltaG^@ = -RTlnK)`

Knowing `DeltaG_1^@`, the change in the Gibbs' free energy for the 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))`