The differential rate law for the reaction

H_{2}+I_{2}\rightarrow 2HI  is

  • Option 1)

    -\frac{d\left [ H_{2} \right ]}{dt}=-\frac{d\left [ I_{2} \right ]}{dt}=-\frac{d\left [ HI \right ]}{dt}

  • Option 2)

    \frac{d\left [ H_{2} \right ]}{dt}=\frac{d\left [ I_{2} \right ]}{dt}=\frac{1}{2}\frac{d\left [ HI \right ]}{dt}

  • Option 3)

    \frac{1}{2}\frac{d\left [ H_{2} \right ]}{dt}=\frac{1}{2}\frac{d\left [ I_{2} \right ]}{dt}=-\frac{d\left [ HI \right ]}{dt}

  • Option 4)

    -2\frac{d\left [ H_{2} \right ]}{dt}=-2\frac{d\left [ I_{2} \right ]}{dt}=\frac{d\left [ HI \right ]}{dt}

 

Answers (1)

As we learnt in 

Instantaneous Rate -

The rate of a reaction calculated at a particular instant of time is called Instantaneous Rate

 

- wherein

r_{av}=\frac{-\Delta R}{\Delta t}=\frac{+\Delta P}{\Delta t}

\lim_{\Delta\rightarrow 0}r_{av} = r_{inst}

Formula = \frac{-d[R]}{dt}= \frac{-d[P]}{dt}

 

 H_{2}+I_{2}\rightleftharpoons 2HI

-\frac{d\left [ H_{2} \right ]}{dt}=-\frac{d\left [ I_{2} \right ]}{dt}=+\frac{1} {2}\frac{d\left [ H{I} \right ]}{dt}

or

-\frac{2d\left [ H_{2} \right ]}{dt}=-\frac{2d\left [ I_{2} \right ]}{dt}=+\frac{d\left [ H{I} \right ]}{dt}


Option 1)

-\frac{d\left [ H_{2} \right ]}{dt}=-\frac{d\left [ I_{2} \right ]}{dt}=-\frac{d\left [ HI \right ]}{dt}

Incorrect option

Option 2)

\frac{d\left [ H_{2} \right ]}{dt}=\frac{d\left [ I_{2} \right ]}{dt}=\frac{1}{2}\frac{d\left [ HI \right ]}{dt}

Incorrect option

Option 3)

\frac{1}{2}\frac{d\left [ H_{2} \right ]}{dt}=\frac{1}{2}\frac{d\left [ I_{2} \right ]}{dt}=-\frac{d\left [ HI \right ]}{dt}

Incorrect option

Option 4)

-2\frac{d\left [ H_{2} \right ]}{dt}=-2\frac{d\left [ I_{2} \right ]}{dt}=\frac{d\left [ HI \right ]}{dt}

Correct option

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