Get Answers to all your Questions

header-bg qa

For the reaction \left [ N_{2}O_{5}_{(g)}\rightarrow 2NO_{2}_{(g)} + O_{2}_{(g)}\right ] the value of rate of disappearance of N2O5 is given as 6.25 x 10-3 mol L-1 s-1. The rate of formation of NO2 and O2 is given respectively as:

  • Option 1)

    6.25 x 10-3 mol L-1 s-1 and 6.25 x 10-3 mol L-1 s-1

  • Option 2)

    1.25 x 10-2 mol L-1 s-1 and 3.125 x 10-3 mol L-1 s-1

  • Option 3)

    6.25 x 10-3 mol L-1 s-1 and 3.125 x 10-3 mol L-1 s-1

  • Option 4)

    1.25 x 10-2 mol L-1 s-1 and 6.25 x 10-3 mol L-1 s-1

 

Answers (1)

best_answer

As we learnt in concept

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}]

 

 Fix the reaction coefficients. The reaction should be 

N_{2}O_{5}\rightarrow 2NO_{2}+\frac{1}{2}O_{2}\\ \\ then, \frac{-d\left [ N_{2}O_{5} \right ]}{dt}=\frac{1}{2}\frac{d\left [ NO_{2} \right ]}{dt}=2\frac{d\left [ O_{2} \right ]}{dt}

6.25 * 10^{-3} Ms^{-1}=\frac{1}{2}\frac{d\left [ NO_{2} \right ]}{dt}=2\frac{d\left [ O_{2} \right ]}{dt}

\frac{d\left [ NO_{2} \right ]}{dt}=2\times 6.25\times 10^{-3}Ms^{-1}=1.25\times 10^{-2}Ms^{-1}\\ \frac{d\left [ O_{2} \right ]}{dt}=3.125\times 10^{-3}Ms^{-1}


Option 1)

6.25 x 10-3 mol L-1 s-1 and 6.25 x 10-3 mol L-1 s-1

This solution is incorrect 

Option 2)

1.25 x 10-2 mol L-1 s-1 and 3.125 x 10-3 mol L-1 s-1

This solution is correct 

Option 3)

6.25 x 10-3 mol L-1 s-1 and 3.125 x 10-3 mol L-1 s-1

This solution is incorrect 

Option 4)

1.25 x 10-2 mol L-1 s-1 and 6.25 x 10-3 mol L-1 s-1

This solution is incorrect 

Posted by

divya.saini

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks