Consider the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) .$ The temperature (in kelvin) at which $K_{C} = 20.4$ and $K_{P} = 600.1$ is (Round off to the Nearest Integer) Assume all gases are ideal and $R = 0.0831 L b a r K^{- 1} m o l^{- 1}$ ]

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Solution

Here change in number of gaseous moles :
$Δ n_{g} = 2 - 1 = 1$
$K_{P} = K_{C} (R T)^{Δ n_{g}}$

$600.1 = 20.4 (0.0831 \times T)^{1}$

$T = 353.99 K \approx 354 K$
Correct answer is 354

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Consider the reaction N2O4(g)⇌2NO2(g). The temperature (in kelvin) at which KC=20.4and KP=600.1 is (Round off to the Nearest Integer) Assume all gases are ideal and R=0.0831LbarK−1mol−1]

Here change in number of gaseous moles : Δng=2−1=1 KP=KC(RT)Δng 600.1=20.4(0.0831×T)1 T=353.99K≈354K Correct answer is 354

Consider the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) .$ The temperature (in kelvin) at which $K_{C} = 20.4$ and $K_{P} = 600.1$ is (Round off to the Nearest Integer) Assume all gases are ideal and $R = 0.0831 L b a r K^{- 1} m o l^{- 1}$ ]

Here change in number of gaseous moles :
$Δ n_{g} = 2 - 1 = 1$
$K_{P} = K_{C} (R T)^{Δ n_{g}}$

$600.1 = 20.4 (0.0831 \times T)^{1}$

$T = 353.99 K \approx 354 K$
Correct answer is 354