`P(g)+2Q(g)toPQ_(2)(g),DeltaH=18kJmol^(-1)`
The entropy change of the above reaction `(DeltaS_("system")` is 60 `JK^(-1)mol^(-1))`. At what temperature the reaction becomes spontaneous?

Options

A. Below 200 K
B. Above 300 K
C. Below 300 K
D. Above 200 K and below 300 K.

Explanation

To determine the temperature at which the reaction becomes spontaneous, we need to analyze the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS). The reaction is spontaneous when ΔG is less than 0.

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation:**
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
For the reaction to be spontaneous, we require:
\[ \Delta G < 0 \]

2. **Rearranging the Equation:**
From the equation, we can rearrange it to find the condition for spontaneity:
\[ \Delta H - T \Delta S < 0 \]
This can be rewritten as:
\[ \Delta H < T \Delta S \]
or
\[ T > \frac{\Delta H}{\Delta S} \]

3. **Substituting the Given Values:**
We know:
- ΔH = 18 kJ/mol = 18000 J/mol (since 1 kJ = 1000 J)
- ΔS = 60 J/K·mol

Now, substituting these values into the equation:
\[ T > \frac{18000 \, \text{J/mol}}{60 \, \text{J/K·mol}} \]

4. **Calculating the Temperature:**
Performing the calculation:
\[ T > \frac{18000}{60} = 300 \, \text{K} \]

5. **Conclusion:**
The reaction becomes spontaneous at temperatures greater than 300 K.

### Final Answer:
The reaction becomes spontaneous at temperatures greater than 300 K.