To determine the nature of the reaction [tex]\(I_2(s) \rightarrow I_2(g)\)[/tex] at [tex]\(500 \, \text{K}\)[/tex], we will use the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) as our main criterion. The given data includes:
- [tex]\(\Delta H = 62.4 \, \text{kJ/mol}\)[/tex] (enthalpy change)
- [tex]\(\Delta S = 0.145 \, \text{kJ/(mol·K)}\)[/tex] (entropy change)
- [tex]\(T = 500 \, \text{K}\)[/tex] (temperature)
The formula to calculate the Gibbs free energy change is:
[tex]\[
\Delta G = \Delta H - T\Delta S
\][/tex]
Substitute the given values into the formula:
[tex]\[
\Delta G = 62.4 \, \text{kJ/mol} - 500 \, \text{K} \times 0.145 \, \text{kJ/(mol·K)}
\][/tex]
Perform the calculation step-by-step:
1. Calculate the term [tex]\(T \Delta S\)[/tex]:
[tex]\[
500 \, \text{K} \times 0.145 \, \text{kJ/(mol·K)} = 72.5 \, \text{kJ/mol}
\][/tex]
2. Subtract this from [tex]\(\Delta H\)[/tex]:
[tex]\[
\Delta G = 62.4 \, \text{kJ/mol} - 72.5 \, \text{kJ/mol}
\][/tex]
3. Complete the subtraction:
[tex]\[
\Delta G = 62.4 - 72.5 = -10.1 \, \text{kJ/mol}
\][/tex]
Since [tex]\(\Delta G\)[/tex] is negative ([tex]\(\Delta G = -10.1 \, \text{kJ/mol}\)[/tex]), this indicates that the reaction is spontaneous at [tex]\(500 \, \text{K}\)[/tex].
Using the provided choices:
A. Its entropy decreases - This is incorrect. Given [tex]\(\Delta S = 0.145 \, \text{kJ/(mol·K)}\)[/tex], the entropy increases.
B. It is spontaneous - This is correct because [tex]\(\Delta G\)[/tex] is negative.
C. It proceeds rapidly - Gibbs free energy does not give information about the reaction rate.
D. It is exothermic - This is incorrect. Since [tex]\(\Delta H = 62.4 \, \text{kJ/mol}\)[/tex] is positive, the reaction is endothermic.
Thus, the correct answer is:
B. It is spontaneous.
