Q14 / 150 · MHT CET 2008 · Chemistry

For the reaction : $\mathrm{H}_{2}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI}$, the differential rate law is

  • A
    $-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=2 \frac{d[\mathrm{HI}]}{d t}$
  • Correct answer
    B
    $-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-2 \frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}$
  • C
    $-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}$
  • D
    $-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}$

Solution

$\mathrm{H}_{2}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI}$ Rate of reaction $=\frac{-d\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=\frac{-d\left[\mathrm{I}_{2}\right]}{\mathrm{dt}}=\frac{1}{2} \frac{d[\mathrm{HI}]}{\mathrm{d} t}$ or $\quad \frac{-2 d\left[\mathrm{H}_{2}\right]}{d t}=\frac{-2 d\left[\mathrm{I}_{2}\right]}{d t}=\frac{d[\mathrm{HI}]}{d t}$

Hence, the correct answer is option: 'B'.

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