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The rate of the reaction 2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2} can be written in three ways:

\frac{-d\left [ N_{2}O_{5} \right ]}{dt}=k\left [ N_{2}O_{5} \right ]

\frac{d\left [ NO_{2} \right ]}{dt}=k{}'\left [ N_{2}O_{5} \right ]

\frac{d\left [O_{2} \right ]}{dt}=k{}''\left [ N_{2}O_{5} \right ]

The relationship between k and k' and between k and k" are:

  • Option 1)

    k'=2k  k'=k

  • Option 2)

    k'=2k ; k"=k/2

  • Option 3)

    k'=2k ; k"=2k

  • Option 4)

    k'=k ; k"=k

 

Answers (1)

best_answer

As we learned in

Rates in presence of stoichiometry of reactants/products -

When stoichiometry coefficients of reactants/ products are not equal to one, the rate of disappearance of  & the rate of appearance of products is divided by their respective stoichiometric coefficients

- wherein

e.g.\:2HI(g)\rightarrow H_{2}(g)+I_{2}(g)

r=\frac{-1}{2}.\frac{d}{dt}[HI]

=\frac{+d}{dt}[H2]=\frac{+ d}{dt}[I_{2}]

 

 R = -\frac{1}{2}\frac{d[N_2O_5]}{dt}=\frac{1}{4}\frac{d[NO_2]}{dt}=\frac{d[O_2]}{dt}

R=+\frac{1}{2}k[N_2O_5]=\frac{1}{4}{k}'[N_2O_5]={k}''[N_2O_5]

k=+\frac{1}{2} \ {k}'=+2{k}''

k'=2k\ and\ k''=\frac{k}{2}


Option 1)

k'=2k  k'=k

incorrect

Option 2)

k'=2k ; k"=k/2

correct

Option 3)

k'=2k ; k"=2k

incorrect

Option 4)

k'=k ; k"=k

incorrect

Posted by

Plabita

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