Question

determine the value of $\delta s_{sur}$ at 298 k, and predict whether or not the reaction is spontaneous at this temperature.\
$\ce{a(g) \
ightarrow b(g) + c(g)}$\
$\delta h^\circ_{rxn} = 21.16\\ \text{kj/mol}$\
$\delta s^\circ_{rxn} = 15.6\\ \text{j/k·mol}$\
\bigcirc $\delta s_{sur} = 71.0\\ \text{j/k·mol}$, spontaneous\
\bigcirc it is not possible to calculate $\delta s_{sur}$ or predict the spontaneity of this reaction without more information.\
\bigcirc $\delta s_{sur} = 0.071\\ \text{j/k·mol}$, spontaneous\
\bigcirc $\delta s_{sur} = -0.071\\ \text{j/k·mol}$, nonspontaneous\
\bigcirc $\delta s_{sur} = -71.0\\ \text{j/k·mol}$, nonspontaneous

Explanation:

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

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

Step2: Convert units of $\Delta H_{rxn}^{\circ}$

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

Step3: Calculate $\Delta S_{sur}$

Using the formula $\Delta S_{sur} = -\frac{\Delta H_{rxn}^{\circ}}{T}$ with $T = 298\ \text{K}$:
$\Delta S_{sur} = -\frac{21160\ \text{J/mol}}{298\ \text{K}} \approx -71.0\ \text{J/K·mol}$ (note: the negative sign indicates the surroundings lose entropy as the reaction is endothermic, absorbing heat from the surroundings).

Step4: Determine spontaneity

To determine spontaneity, we consider the total entropy change $\Delta S_{total} = \Delta S_{rxn}^{\circ} + \Delta S_{sur}$.
We have $\Delta S_{rxn}^{\circ} = 15.6\ \text{J/K·mol}$ and $\Delta S_{sur} \approx -71.0\ \text{J/K·mol}$.
So $\Delta S_{total} = 15.6 - 71.0 = -55.4\ \text{J/K·mol}$ (negative). For a reaction to be spontaneous, $\Delta S_{total}$ should be positive (at constant $T$ and $P$). Since $\Delta S_{total}$ is negative here, the reaction is nonspontaneous.

Answer:

$\Delta S_{sur} = -71.0\ \text{J/K·mol}$, nonspontaneous (the option corresponding to this is the last one with $\Delta S_{sur} = -71.0\ \text{J/K·mol}$, nonspontaneous)