The following mechanism has been proposed for the reaction of NO with Br_{2}  to form \dpi{100} NOBr.

NO_{(g)}+Br_{2(g)}\rightleftharpoons NOBr_{2(g)}

NOBr_{2(g)}+NO_{(g)}\rightarrow 2 NOBr_{(g)}

If the second step is the rate determining step, the order of the reaction with respect to   NO_{(g)}  is 

  • Option 1)

    1

  • Option 2)

    0

  • Option 3)

    3

  • Option 4)

    2

 

Answers (1)

As we learnt in 

Rate Determining Step -

The overall rate of reaction is controlled by the slowest step in a reaction called the rate determining step.

- wherein

e.g.

2H_{2}O_{2}\xrightarrow[alkaline\ medium]{I^{-}}2H_{2}O+O_{2}

r=\frac{-d[H_{2}O_{2}]}{dt}=K[H_{2}O_{2}][I^{-}]

H_{2}O_{2}+I^{-}\underset{slow}{\rightarrow}H_{2}O+IO

H_{2}O_{2}+IO^{-}\underset{fast}{\rightarrow}H_{2}O+I+O_{2}^{-}

 

 

 NO\left ( {g} \right ) + Br_{2}\left ( g \right ) \rightleftharpoons NOBr_{2}\left ( g \right )......................(1)

NOBr_{2}\left ( g \right ) + NO\left ( {g} \right ) \rightleftharpoons 2NOBr_{2}\left ( g \right ) (slow).....................(2)

Rate =K \left [NOBr_{2}\left ( g \right ) \right ] \left [ NO \right ]

From equation (1)

K_{c} = \frac{\left [ NOBr_{2} \right ]}{\left [ NO \right ]\left [ Br_{2} \right ]}

\left [NOBr_{2} \right ] = K_{c} \times \left [ NO \right ] \left [ Br_{2} \right ]

Putting the value of  \left [NOBr_{2} \right ] in rate equation

rate = K\left [ NO \right ]^{2} \left [ Br_{2} \right ] \times K_c

rate = KK_{c} \left [ NO^{2} \right ] \left [ Br_{2} \right ]

order with respect to NO is 2 


Option 1)

1

This option is incorrect

Option 2)

0

This option is incorrect

Option 3)

3

This option is incorrect

Option 4)

2

This option is correct

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