For the reaction `N_(2)(g) + 3H_(2)(g) rarr 2NH_(2)(g), DeltaH = - 95.4 Kj ` and `Delta S=- 198 .300 JK^(-1)` . Calculate the temperature in centrigrade at which it attains equilibrium.

Answer

To find the temperature at which the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \) attains equilibrium, we can use the relationship between Gibbs free energy (\( \Delta G \)), enthalpy change (\( \Delta H \)), and entropy change (\( \Delta S \)). At equilibrium, \( \Delta G = 0 \).

### Step-by-Step Solution:

1. **Write the equation for Gibbs free energy:**
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, \( \Delta G = 0 \), so we can set up the equation:
\[ 0 = \Delta H - T \Delta S \]

2. **Rearrange the equation to solve for temperature (T):**
\[ T \Delta S = \Delta H \]
\[ T = \frac{\Delta H}{\Delta S} \]

3. **Substitute the given values:**
- \( \Delta H = -95.4 \, \text{kJ} = -95.4 \times 10^3 \, \text{J} \) (since 1 kJ = 1000 J)
- \( \Delta S = -198.3 \, \text{J/K} \)

Now substitute these values into the equation:
\[ T = \frac{-95.4 \times 10^3 \, \text{J}}{-198.3 \, \text{J/K}} \]

4. **Calculate T:**
\[ T = \frac{95.4 \times 10^3}{198.3} \approx 481.0 \, \text{K} \]

5. **Convert temperature from Kelvin to Celsius:**
To convert Kelvin to Celsius, use the formula:
\[ T(°C) = T(K) - 273 \]
\[ T(°C) = 481.0 - 273 = 208.0 \, °C \]

### Final Answer:
The temperature at which the reaction attains equilibrium is \( 208.0 \, °C \).