The rate of a gas phase reaction involving the conversion of reactant ‘i’ to product is expressed as ____(a) (-ri)=-\(\frac{1}{V^2}\frac{dN_i}{dt} \)(b) (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt} \)(c) (-ri)=-\(\frac{dN_i}{dt} \)(d) (-ri)=-V\(\frac{dN_i}{dt} \)I had been asked this question by my college professor while I was bunking the class.I want to ask this question from Liquid and Gas Phase Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering

The rate of a gas phase reaction involving the conversion of reactant

1.	The rate of a gas phase reaction involving the conversion of reactant ‘i’ to product is expressed as ____(a) (-ri)=-\(\frac{1}{V^2}\frac{dN_i}{dt} \)(b) (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt} \)(c) (-ri)=-\(\frac{dN_i}{dt} \)(d) (-ri)=-V\(\frac{dN_i}{dt} \)I had been asked this question by my college professor while I was bunking the class.I want to ask this question from Liquid and Gas Phase Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» CORRECT answer is (b) (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt} \) The best explanation: The RATE of change of the gaseous component ‘i’ in DUE COURSE of the reaction is the change in number of moles of ‘i’ per unit time per unit volume of the reaction mixture. HENCE, (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt}. \)

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