For A + B rightleftharpoons C, the DeltaG_f^@ are "402.0 kJ/mol", "387.7 kJ/mol", and "500.8 kJ/mol", respectively. What is DeltaG_(rxn)^@? If both entropy and enthalpy changes are positive for this reaction at 25^@ "C", which one drives it?

Jul 13, 2017

See the explanation below...

As with many other thermodynamic functions, they can be added together because they are extensive. This forms the basis for:

$\Delta {G}_{r x n}^{\circ} = {\sum}_{P} {\nu}_{P} {\Delta}_{f} {G}_{P}^{\circ} - {\sum}_{R} {\nu}_{R} {\Delta}_{f} {G}_{R}^{\circ}$,

where:

• ${\Delta}_{f} {G}^{\circ}$ is the change in Gibbs' free energy due to forming the substance from its elements in their standard states. This is the Gibbs' free energy of formation.
• $\nu$ is the stoichiometric coefficient.
• $P$ and $R$ stand for product and reactant, respectively.
• Standard conditions here are defined to be $\text{298.15 K}$ and $\text{1 bar}$. Check your book to see if you still use $\text{1 atm}$.

This gives for

$1 A + 1 B r i g h t \le f t h a r p \infty n s 1 C$,

$\Delta {G}_{r x n}^{\circ}$

$= {\overbrace{\left(1 \cdot 402.0\right)}}^{\text{Products" - overbrace((1cdot387.7 + 1cdot500.8))^"Reactants}}$ $\text{kJ/mol}$

= ???

You should have one decimal place.

Regardless of the magnitude of $\Delta {G}_{r x n}^{\circ}$ at ${25}^{\circ} \text{C}$ and standard pressure, as long as it is negative, your reaction is spontaneous AT ${25}^{\circ} \text{C}$ and standard pressure. It is important to recognize that $\Delta {G}^{\circ} \ne \Delta G$ in general.

But if $\Delta {H}_{r x n}^{\circ}$ and $\Delta {S}_{r x n}^{\circ}$ are both positive, you have

$\Delta {G}_{r x n}^{\circ} = \Delta {H}_{r x n}^{\circ} - T \Delta {S}_{r x n}^{\circ}$

$= \left(+\right) - \left(+\right) \left(+\right) = \left(+\right) - \left(+ '\right)$

That is, if $\Delta {S}_{r x n}^{\circ}$ is large or the temperature is high, the reaction is spontaneous.

The spontaneous reaction is driven by the remainder of the equation that is negative when one term goes to zero, i.e.

$\textcolor{b l u e}{\underline{\Delta {G}_{r x n}^{\circ}}} = \left(0\right) - \left(+\right) \left(+\right) \textcolor{b l u e}{\underline{< 0}}$ when $\Delta {H}_{r x n}^{\circ} = 0$

$\textcolor{red}{\underline{\Delta {G}_{r x n}^{\circ}}} = \left(+\right) - \left(+\right) \left(0\right) \textcolor{red}{\underline{> 0}}$ when $\Delta {S}_{r x n}^{\circ} = 0$

One can see that if $\Delta {S}_{r x n}^{\circ} = 0$, the influence of the entropy is no longer there, and the reaction is no longer spontaneous.

So, the entropy drives the spontaneous (i.e. FORWARD) reaction at ${25}^{\circ} \text{C}$ and standard pressure.