For reaction, 2 NOCl ( g ) ⇌ 2 NO ( g ) + Cl 2 ( g ) , K c at 42 7 ∘ C is 3 × 1 0 − 6 L mol − 1 . The value of K p is nearly

2 NOCl ( g ) ⇌ 2 NO ( g ) + Cl 2 ( g ) K p = K c ( RT ) Δ n K p = 3 × 1 0 − 6 ( 0.0821 × 700 ) = 172.41 × 1 0 − 6 = 1.72 × 1 0 − 4

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Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumMiscellaneous/MixedMedium2 minPYQ_2005

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For reaction, $2 NOCl_{(g)} ⇌ 2 NO_{(g)} + Cl_{2 (g)}$ , $K_{c}$ at $42 7^{\circ} C$ is $3 \times 1 0^{- 6} L mol^{- 1}$ . The value of $K_{p}$ is nearly

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$2 NOCl_{(g)} ⇌ 2 NO_{(g)} + Cl_{2 (g)}$ $K_{p} = K_{c} (RT)^{Δ n}$ $K_{p} = 3 \times 1 0^{- 6} (0.0821 \times 700) = 172.41 \times 1 0^{- 6} = 1.72 \times 1 0^{- 4}$

Stream:AIIMSSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2005

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