At 300 K, the value of K P for the reaction N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) is 2 × 1 0 − 6 atm − 2 . The value of K C in terms of R would be

Δ n = 2 − 4 = − 2 . K P = K C ( RT ) − 2 , so K C = K P ( RT ) 2 = 2 × 1 0 − 6 × ( 300 R ) 2 = 2 × 1 0 − 6 × 9 × 1 0 4 R 2 = 0.18 R 2 . Answer: (A).

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At 300 K, the value of $K_{P}$ for the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ is $2 \times 1 0^{- 6} atm^{- 2}$ . The value of $K_{C}$ in terms of R would be

Options:

Answer:

Solution:

$Δ n = 2 - 4 = - 2$ .

$K_{P} = K_{C} (RT)^{- 2}$ , so $K_{C} = K_{P} (RT)^{2} = 2 \times 1 0^{- 6} \times (300 R)^{2} = 2 \times 1 0^{- 6} \times 9 \times 1 0^{4} R^{2} = 0.18 R^{2}$ .

Answer: (A).

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp

⏱ 2mℹ️ Source: QB

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