Consider the reaction N₂O₄(g) excess 2NO₂(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 = 0.0831L bar K–1 mol–1)

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answer is 354.

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Detailed Solution

$N_{2} {O_{4}}_{(g)} \to 2 N O_{2 (g)} Δ n_{g} = 2 - 1 = 1 K p = K c . {(R T)}^{∆ n g} 600.1 = 20.4 \times {(0.0831 \times T)}^{1} ∴ T = 353.99 K = 354 K$

Hence, the correct answer is 354.

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