## 5.55 (b)

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

### 5.55 (b)

According to the solutions manual, why do we have to calculate for the limiting reactant? Wouldn't graphite be automatically excluded from the equilibrium constant because it's a solid?

If graphite was found to be the limiting reactant, how would that change the outcome of the solution?

romina_4C
Posts: 100
Joined: Thu Jul 11, 2019 12:17 am

### Re: 5.55 (b)

I believe you have to calculate for the limiting reactant because you don't actually have the equilibrium value of H2O. Once you find out that graphite is the limiting reactant, you can calculate the concentration of H2O consumed and thus have a value of H2O to use (when making the ice table, otherwise, without knowing the limiting reactant, you would not be able to tell the initial amount of H2O gas, and thus would not be able to apply an -x change to it).