NO_{2} required for a reaction is produced by the decomposition of N_{2}O_{5} in CCl_{4} as per the equation, 2N_{2}O_{5}(g)\rightarrow 4NO_{2}(g)+O_{2}(g).

The initial concentration of N_{2}O_{5} is 3.00 mol L^{-1} and it is 2.75 mol L^{-1} after 30 minutes. The rate of formation of NO_{2} is:

 

  • Option 1)

    4.167\times 10^{-3}molL^{-1}min^{-1}

     

     

     

  • Option 2)

    1.667\times 10^{-2}molL^{-1}min^{-1}

  • Option 3)

    8.333\times 10^{-3}molL^{-1}min^{-1}

  • Option 4)

    2.083\times 10^{-3}molL^{-1}min^{-1}

 

Answers (1)

(1.667\times 10^{-2}molL^{-1}min^{-1})

2N_{2}O_{5}(g)\rightarrow 4NO_{2}(g)+O_{2}(g)

t=0        3.0 M

t=30        2.75 M

so, \frac{-\Delta [N_{2}O_{5}]}{\Delta t}=\frac{-[2.75-3.0]}{30-0}=\frac{0.25}{30}

from the reaction 

-\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}

\frac{\Delta [NO_{2}]}{\Delta t}=\frac{0.25}{30}\times 2=1.66\times 10^{-2}M/min

=1.66\times 10^{-2}molL^{-1}min^{-1}

 


Option 1)

4.167\times 10^{-3}molL^{-1}min^{-1}

 

 

 

Option 2)

1.667\times 10^{-2}molL^{-1}min^{-1}

Option 3)

8.333\times 10^{-3}molL^{-1}min^{-1}

Option 4)

2.083\times 10^{-3}molL^{-1}min^{-1}

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