In the reaction, N 2 + 3 H 2 ⟶ 2 NH 3 , the rate of disappearance of H 2 is 0 ⋅ 02 M / s . The rate of appearance of NH 3 is

N 2 + 3 H 2 ⟶ 2 NH 3 Rate of reaction = − dt d ( N 2 ) = − 3 1 dt d ( H 2 ) = 2 1 dt d ( NH 3 ) From rate expression, dt d ( NH 2 ) = 3 2 dt d ( H 2 ) = 3 2 × 0.02 = 0.0133 M / s

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ChemistryChemical KineticsRate of Reactions/Rate LawEasy2 minPYQ_2020

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In the reaction, $N_{2} + 3 H_{2} ⟶ 2 NH_{3}$ , the rate of disappearance of $H_{2}$ is $0 \cdot 02 M / s$ . The rate of appearance of $NH_{3}$ is

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$N_{2} + 3 H_{2} ⟶ 2 NH_{3}$ Rate of reaction $= - \frac{d ( N _{2} )}{dt} = - \frac{1}{3} \frac{d ( H _{2} )}{dt} = \frac{1}{2} \frac{d ( NH _{3} )}{dt}$ From rate expression, $\frac{d ( NH _{2} )}{dt} = \frac{2}{3} \frac{d ( H _{2} )}{dt} = \frac{2}{3} \times 0.02 = 0.0133 M / s$

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

⏱ 2mℹ️ Source: PYQ_2020

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