For the reaction H 2 + I 2 → 2HI, the differential rate law is:

Chemistry · NEET · NTA Exams — Chemical Kinetics

For the reaction H₂ + I₂ → 2HI, the differential rate law is:

\(-\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[_{2}\right]}{\mathrm{dt}}=2 \frac{\mathrm{~d}[\mathrm{HI}]}{\mathrm{dt}}\)
\(-2 \frac{d\left[H_{2}\right]}{d t}=-2 \frac{d\left[l_{2}\right]}{d t}=\frac{d[H]}{d t}\)
\(-\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{l}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}\)
\(-\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{l}_{2}\right]}{\mathrm{Zdt}}=\frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}\)

Answer

(C) - d [ H _ 2 ] dt =- d [ l _ 2 ] dt = d [ HI ] dt

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.