For the reaction `: 2NO(g) + O_(2)(g) rarr 2NO_(2)(g)`, the enthalpy and entropy changes are - 113.0 kJ `mol^(-1)` and `-145 JK^(-1) mol^(-1)` respectively. Find the temperature below which this reaction is spontaneous.

Explanation

To determine the temperature below which the reaction \(2NO(g) + O_2(g) \rightarrow 2NO_2(g)\) is spontaneous, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Understand the conditions for spontaneity
A reaction is spontaneous when \(\Delta G < 0\). At equilibrium, \(\Delta G = 0\). Therefore, we can set up the equation:

\[ 0 = \Delta H - T \Delta S \]

### Step 2: Rearrange the equation to find temperature
Rearranging the equation gives:

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 3: Substitute the values for \(\Delta H\) and \(\Delta S\)
Given:
- \(\Delta H = -113.0 \, \text{kJ/mol}\)
- \(\Delta S = -145 \, \text{J/K/mol}\)

First, we need to convert \(\Delta H\) from kJ to J:

\[ \Delta H = -113.0 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -113000 \, \text{J/mol} \]

Now we can substitute the values into the temperature equation:

\[ T = \frac{-113000 \, \text{J/mol}}{-145 \, \text{J/K/mol}} \]

### Step 4: Calculate the temperature
Calculating the temperature gives:

\[ T = \frac{113000}{145} \approx 779.31 \, \text{K} \]

### Step 5: Conclusion
The temperature below which the reaction is spontaneous is approximately \(779.31 \, \text{K}\). Therefore, for temperatures below this value, the reaction will be spontaneous.