Consider the reaction $$N_2O_4(g) \rightleftharpoons 2NO_2(g)$$. The temperature at which $$K_C = 20.4$$ and $$K_P = 600.1$$, is ________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K$$^{-1}$$ mol$$^{-1}$$]

For the reaction $$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$$, the change in moles of gas is $$\Delta n = 2 - 1 = 1$$.

The relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C \cdot (RT)^{\Delta n}$$. Substituting $$\Delta n = 1$$:

$$K_P = K_C \cdot RT$$

Solving for $$T$$:

$$T = \frac{K_P}{K_C \cdot R} = \frac{600.1}{20.4 \times 0.0831}$$

Computing the denominator: $$20.4 \times 0.0831 = 1.69524$$

$$T = \frac{600.1}{1.69524} = 354.0 \text{ K}$$

Rounding to the nearest integer, the temperature is 354 K.

The answer is 354.