To calculate the change in Gibbs free energy ($\Delta G^\circ_{\text{rxn}}$) and determine whether the reaction is spontaneous, we use the Gibbs free energy equation:

$\Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T \Delta S^\circ_{\text{rxn}}$

Step 1: Convert units if necessary

Given data:

• $\Delta H^\circ_{\text{rxn}} = -95 \, \text{kJ}$
• $\Delta S^\circ_{\text{rxn}} = 157 \, \text{J/K}$ ($\text{J}$ should be converted to $\text{kJ}$, so divide by 1000): $\Delta S^\circ_{\text{rxn}} = 0.157 \, \text{kJ/K}$
• $T = 855 \, \text{K}$

Step 2: Plug values into the Gibbs free energy equation

$\Delta G^\circ_{\text{rxn}} = -95 - (855 \times 0.157)$

First, calculate the term $T \Delta S^\circ_{\text{rxn}}$: $T \Delta S^\circ_{\text{rxn}} = 855 \times 0.157 = 134.235 \, \text{kJ}$

Now, calculate $\Delta G^\circ_{\text{rxn}}$: $\Delta G^\circ_{\text{rxn}} = -95 - 134.235 = -229.235 \, \text{kJ}$

Step 3: Determine spontaneity

• If $\Delta G^\circ_{\text{rxn}} < 0$, the reaction is spontaneous.
• Since $-229.235 \, \text{kJ}$ is negative, the reaction is spontaneous.

Final Answer:

$\text{Correct choice: } \mathbf{2.3 \times 10^2 \, \text{kJ}; \text{spontaneous}}$

Would you like further clarification or details?

Here are some additional questions related to this topic:

1. How does temperature affect the spontaneity of a reaction?
2. Can a reaction with positive $\Delta H$ still be spontaneous?
3. What happens to $\Delta G$ if $\Delta S$ is negative?
4. How is the Gibbs free energy equation related to equilibrium constants?
5. Why is it important to use consistent units in thermodynamic calculations?

Tip: Always check the units of $\Delta H$, $\Delta S$, and $T$ to ensure consistency in your calculations!