The differential rate law for the reaction H 2 + I 2 ⇌ 2HI is

Chemistry · JEE Main · NTA Exams — Chemical Kinetics

The differential rate law for the reaction
H₂ + I₂ \(\text { ⇌ }\) 2HI is

\(\frac{-\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d}}=\frac{-\mathrm{d}\left[\mathrm{I}_{2}\right]}{\mathrm{d}}=\frac{-\mathrm{d}[\mathrm{H}]}{\mathrm{d}}\)
\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{I}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{H}]}{\mathrm{dt}}\)
\(\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{H}_{2}\right]}{\mathrm{t}}=\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{I}_{2}\right]}{\mathrm{t}}=\frac{-\mathrm{d}[\mathrm{H}]}{\mathrm{t}}\)
\(-2 \frac{\mathrm{~d}\left[\mathrm{H}_{2}\right]}{\mathrm{t}}=-2 \frac{\mathrm{~d}\left[\mathrm{I}_{2}\right]}{\mathrm{t}}=\frac{\mathrm{d}[\mathrm{H}]}{\mathrm{t}}\)

Answer

(D) -2 ~d [ H _ 2 ] t =-2 ~d [ I _ 2 ] t = d [ H ] t

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