For the reaction N 2 + 3 H 2 → 2 NH 3 if Δ t Δ [ NH 3 ] = 2 × 1 0 − 4 mol l − 1 s − 1 , then value of Δ t − Δ [ H 2 ] would be

N 2 + 3 H 2 Δ t − Δ [ N 2 ] ∴ Δ t − Δ [ H 2 ] → 2 NH 3 = − 3 1 Δ t Δ [ H 2 ] = 2 1 Δ t Δ [ NH 3 ] = 2 3 × Δ t Δ [ NH 3 ] = 2 3 × 2 × 1 0 − 4 = 3 × 1 0 − 4 mol L − 1 S − 1

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ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minPYQ_2019

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For the reaction $N_{2} + 3 H_{2} \to 2 NH_{3}$ if $\frac{Δ [ NH _{3} ]}{Δ t} = 2 \times 1 0^{- 4} mol l^{- 1} s^{- 1}$ , then value of $\frac{- Δ [ H _{2} ]}{Δ t}$ would be

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Solution:

$N_{2} + 3 H_{2} \frac{- Δ [ N _{2} ]}{Δ t} ∴ \frac{- Δ [ H _{2} ]}{Δ t} \to 2 NH_{3} = - \frac{1}{3} \frac{Δ [ H _{2} ]}{Δ t} = \frac{1}{2} \frac{Δ [ NH _{3} ]}{Δ t} = \frac{3}{2} \times \frac{Δ [ NH _{3} ]}{Δ t} = \frac{3}{2} \times 2 \times 1 0^{- 4} = 3 \times 1 0^{- 4} mol L^{- 1} S^{- 1}$

Stream:VITEEESubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

⏱ 2mℹ️ Source: PYQ_2019

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