# Question 0f696

Sep 6, 2017

$K \approx 9.226 \times {10}^{- 5}$

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 $\Delta G$ (at a given temperature and pressure).

We know that:

$\Delta {G}_{\text{tot}} = \Delta {G}_{1} + \Delta {G}_{2} + . . .$

And so, at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, the $\Delta {G}^{\circ}$ for the overall reaction is related to the individual $\Delta {G}^{\circ}$ values.

$\Delta {G}_{1}^{\circ} + \Delta {G}_{2}^{\circ} = \Delta {G}_{\text{rxn"^@ = -"8.80 kJ/mol}}$

And you were given that $\Delta {G}_{2}^{\circ} = - \text{31.6 kJ/mol}$. As a result, for the first reaction we have:

$\Delta {G}_{1}^{\circ} = \Delta {G}_{\text{rxn}}^{\circ} - \Delta {G}_{2}^{\circ}$

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

$=$ $\text{22.8 kJ/mol}$

Now, $\Delta G$ at NONSTANDARD conditions (not at ${25}^{\circ} \text{C}$ but still at $\text{1 atm}$) is related to the STANDARD change in Gibbs' free energy $\Delta {G}^{\circ}$:

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$

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

At chemical equilibrium, $\Delta G = 0$, and $Q = K$, so:

$\underline{\Delta {G}^{\circ} = - R T \ln K}$

Knowing $\Delta {G}_{1}^{\circ}$, the change in the Gibbs' free energy for the first reaction step, we can then get the equilibrium constant for that first reaction step:

$\textcolor{b l u e}{K} = {e}^{- \Delta {G}_{1}^{\circ} / R T}$

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

$= \underline{\textcolor{b l u e}{9.226 \times {10}^{- 5}}}$