For the given reaction
`N_2O_4(g) harr 2NO_2(g)`
`K_p` = 600.1
`K_c` = 20
R= 0.083 Lbar/molK
Find the Temperature

Answer

To find the temperature for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) given \( K_p = 600.1 \), \( K_c = 20 \), and \( R = 0.083 \, \text{L bar/mol K} \), we can use the relationship between \( K_p \), \( K_c \), and temperature.

### Step-by-Step Solution:

1. **Identify the Reaction and Calculate \(\Delta N_g\)**:
- The balanced reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
- Count the number of moles of gaseous products and reactants:
- Moles of products (NO2): 2
- Moles of reactants (N2O4): 1
- Calculate \(\Delta N_g\):
\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 1 = 1 \]

2. **Use the Relationship Between \(K_p\) and \(K_c\)**:
- The relationship is given by:
\[ K_p = K_c \cdot R^T \cdot \Delta N_g \]
- Rearranging the equation to solve for \(T\):
\[ T = \frac{K_p}{K_c \cdot R \cdot \Delta N_g} \]

3. **Substituting the Known Values**:
- Substitute \(K_p = 600.1\), \(K_c = 20\), \(R = 0.083 \, \text{L bar/mol K}\), and \(\Delta N_g = 1\):
\[ T = \frac{600.1}{20 \cdot 0.083 \cdot 1} \]

4. **Calculate the Denominator**:
- Calculate \(20 \cdot 0.083\):
\[ 20 \cdot 0.083 = 1.66 \]

5. **Calculate Temperature \(T\)**:
- Now substitute back to find \(T\):
\[ T = \frac{600.1}{1.66} \approx 361.5 \, \text{K} \]

### Final Answer:
The temperature \(T\) is approximately \(361.5 \, \text{K}\).

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