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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

Posted by

Sabhrant Ambastha

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