Question

determine the value of $\delta s_{sur}$ at 213 k, and predict whether or not the reaction is spontaneous at this temperature.\
a(l) $\
ightarrow$ 2b(g) + c(g)\
$\delta h^\circ_{rxn}$ = 21.16 kj/mol\
$\delta s^\circ_{rxn}$ = 25.8 j/k·mol\
\bigcirc $\delta s_{sur}$ = -0.0993 j/k·mol, spontaneous.\
\bigcirc $\delta s_{sur}$ = -99.3 j/k·mol, nonspontaneous.\
\bigcirc $\delta s_{sur}$ = 0.0993 j/k·mol, nonspontaneous.\
\bigcirc it is not possible to calculate $\delta s_{sur}$ or predict the spontaneity of this reaction without more information.\
\bigcirc $\delta s_{sur}$ = 99.3 j/k·mol, spontaneous.

Explanation:

Step1: Recall the formula for $\Delta S_{surr}$

The formula for the entropy change of the surroundings ($\Delta S_{surr}$) is $\Delta S_{surr} = \frac{-\Delta H^{\circ}_{rxn}}{T}$, where $\Delta H^{\circ}_{rxn}$ is the enthalpy change of the reaction and $T$ is the temperature in Kelvin.

Step2: Convert $\Delta H^{\circ}_{rxn}$ to J/mol

Given $\Delta H^{\circ}_{rxn} = 21.16\ \text{kJ/mol}$, we convert it to J/mol: $21.16\ \text{kJ/mol} = 21160\ \text{J/mol}$.

Step3: Calculate $\Delta S_{surr}$

Substitute $\Delta H^{\circ}_{rxn} = 21160\ \text{J/mol}$ and $T = 213\ \text{K}$ into the formula:
$\Delta S_{surr} = \frac{-21160\ \text{J/mol}}{213\ \text{K}} \approx -99.3\ \text{J/K·mol}$.

Step4: Determine spontaneity using $\Delta G^{\circ}$

The Gibbs free energy change is given by $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
First, calculate $T\Delta S^{\circ}$: $T = 213\ \text{K}$, $\Delta S^{\circ} = 25.8\ \text{J/K·mol} = 0.0258\ \text{kJ/K·mol}$.
$T\Delta S^{\circ} = 213\ \text{K} \times 0.0258\ \text{kJ/K·mol} \approx 5.5\ \text{kJ/mol}$.
Now, $\Delta G^{\circ} = 21.16\ \text{kJ/mol} - 5.5\ \text{kJ/mol} \approx 15.66\ \text{kJ/mol} > 0$.
Since $\Delta G^{\circ} > 0$, the reaction is nonspontaneous.

Answer:

$\Delta S_{surr} = -99.3\ \text{J/K·mol}$, nonspontaneous. (Corresponding to the option: $\boldsymbol{\Delta S_{surr} = -99.3\ \text{J/K·mol}, nonspontaneous}$)