# For the reaction 2"C"(gr) + "O"_2(g) -> 2"CO"(g), what are the changes in entropy of the system, surroundings, and universe at "298 K"? DeltaH_(rxn)^@ = -"221.0 kJ" at this temperature and "1 atm".

## In a table somewhere, it was found that: ${S}^{\circ} \left(C \left(g r\right)\right) = \text{5.7 J/mol"cdot"K}$ ${S}^{\circ} \left({O}_{2} \left(g\right)\right) = \text{205.2 J/mol"cdot"K}$ ${S}^{\circ} \left(C O \left(g\right)\right) = \text{197.7 J/mol"cdot"K}$

Jul 15, 2017

$\Delta {S}_{s y s}^{\circ} = \text{178.8 J/K}$

$\Delta {S}_{s u r r}^{\circ} = \text{741.6 J/K}$

DeltaS_(univ)^@ = DeltaS_(t ot)^@ = ?

(Read further if you don't quite get why you should be able to determine $\Delta {S}_{u n i v}^{\circ}$ with the above results.)

Well, once you find what you are given in the table... (I would think you know how to recognize atomic symbols and phases)

${S}^{\circ} \left(C \left(g r\right)\right) = \text{5.7 J/mol"cdot"K}$

${S}^{\circ} \left({O}_{2} \left(g\right)\right) = \text{205.2 J/mol"cdot"K}$

${S}^{\circ} \left(C O \left(g\right)\right) = \text{197.7 J/mol"cdot"K}$

The process is the same as it would be for any other thermodynamic function. We conventionally define the reaction to BE the system:

$\Delta {S}_{r x n}^{\circ} = \Delta {S}_{s y s}^{\circ} = {\sum}_{P} {\nu}_{P} {S}_{P}^{\circ} - {\sum}_{R} {\nu}_{R} {S}_{R}^{\circ}$,

where $\nu$ is the stoichiometric coefficient, ${S}^{\circ}$ is the standard molar entropy (relative to $\text{0 K}$), and $P$ and $R$ mean product and reactant, respectively.

=> color(blue)(DeltaS_(sys)^@) = ["2 mols" cdot overbrace(197.7)^(CO(g))] "J/mol"cdot"K" - ["2 mols" cdot overbrace(5.7)^(C(gr)) + "1 mol" cdot overbrace(205.2)^(O_2(g))] "J/mol"cdot"K"

$=$ $\textcolor{b l u e}{\text{178.8 J/K}}$,

which should make physical sense because you formed more mols of gas than you started with, increasing the entropy of the system.

You evidently have to use $\Delta {H}_{r x n}^{\circ}$ as well, which is a big hint... though we have to assume it was for the SYSTEM.

If we assume the reaction was done at constant temperature and pressure (it was done at $\text{298 K}$ and $\text{1 atm}$ here), then we can say that:

$\Delta {G}_{s y s}^{\circ} = \Delta {H}_{s y s}^{\circ} - T \Delta {S}_{s y s}^{\circ} = 0$

$\Delta {G}_{s u r r}^{\circ} = \Delta {H}_{s u r r}^{\circ} - T \Delta {S}_{s u r r}^{\circ} = 0$

because the Gibbs' free energy is a function of temperature and pressure in a mechanically-closed system (no mass in or out). This means:

$\Delta {S}_{s u r r}^{\circ} = \frac{\Delta {H}_{s u r r}^{\circ}}{T}$

Since heat is flowing out of the system ($\Delta {H}_{s y s}^{\circ} < 0$), heat is flowing into the surroundings, i.e. $\Delta {H}_{s u r r}^{\circ} = - \Delta {H}_{s y s}^{\circ}$.

So, you should see a familiar equation...

$\textcolor{b l u e}{\Delta {S}_{s u r r}^{\circ}} = - \frac{\Delta {H}_{s y s}^{\circ}}{T}$

= -(-221.0 cancel"kJ")/("298 K") xx "1000 J"/cancel"1 kJ"

$=$ $\textcolor{b l u e}{\text{741.6 J/K}}$

So the surroundings become much more entropic; this is a fairly large positive value, at least when it comes to entropy.

Lastly, the universe gains entropy equal to the change in entropy of everything that has just occurred, so just like with Hess's Law:

$\Delta {S}_{s y s}^{\circ} + \Delta {S}_{s u r r}^{\circ} = \textcolor{b l u e}{\Delta {S}_{u n i v}^{\circ}}$

= "178.8 J/K" + "741.6 J/K" = color(blue)("920.4 J/K")