To determine whether the reaction is spontaneous at 356 K, we need to calculate the Gibbs free energy change ($\Delta G$) using the following equation:

$\Delta G = \Delta H - T \Delta S$

where:

• $\Delta G$ is the Gibbs free energy change,
• $\Delta H$ is the enthalpy change,
• $T$ is the temperature in Kelvin,
• $\Delta S$ is the entropy change.

Step 1: Convert the given values to consistent units

Given:

• $\Delta H = -93 \, \text{kJ/mol}$
• $\Delta S = -198 \, \text{J/mol K}$

Since $\Delta H$ is in kJ/mol and $\Delta S$ is in J/mol K, we need to convert $\Delta H$ into J/mol to match the units:

$\Delta H = -93 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -93000 \, \text{J/mol}$

Step 2: Substitute the values into the Gibbs free energy equation

Now, substitute the values into the equation:

$\Delta G = -93000 \, \text{J/mol} - (356 \, \text{K} \times -198 \, \text{J/mol K})$

Step 3: Perform the calculation

First, calculate the term $T \Delta S$:

$T \Delta S = 356 \, \text{K} \times -198 \, \text{J/mol K} = -70488 \, \text{J/mol}$

Then, calculate $\Delta G$:

$\Delta G = -93000 \, \text{J/mol} - (-70488 \, \text{J/mol}) = -93000 \, \text{J/mol} + 70488 \, \text{J/mol} = -22512 \, \text{J/mol}$

Step 4: Convert the result back to kJ/mol

Convert $\Delta G$ back to kJ/mol:

$\Delta G = -22.512 \, \text{kJ/mol}$

Conclusion:

Since $\Delta G$ is negative, the reaction is spontaneous. The closest answer is:

B. Yes, the $\Delta G$ is -22.5 kJ/mol.

Would you like further details or have any questions?

Here are some related questions:

1. What does a negative $\Delta G$ indicate about the spontaneity of a reaction?
2. How would the spontaneity of the reaction change if the temperature was increased?
3. What is the significance of the enthalpy change $\Delta H$ being negative?
4. How does entropy $\Delta S$ influence the Gibbs free energy?
5. What is the relationship between $\Delta G$, $\Delta H$, and $\Delta S$ in determining spontaneity?

Tip: Always ensure that the units of $\Delta H$ and $\Delta S$ are consistent before performing calculations for $\Delta G$.