Question

For the reaction : $\mathrm{H}{}_2+\mathrm{I}{}_2\to 2\mathrm{HI}$ , the differential rate law is

• A. $-\frac{d\left[H_2\right]}{dt}=-\frac{d\left[I_2\right]}{dt}=2\frac{d\left[HI\right]}{dt}$
• B. $-\frac{d\left[H_2\right]}{dt}=-2\frac{d\left[I_2\right]}{dt}=\frac{d\left[HI\right]}{dt}$
• C. $-\frac{d\left[H_2\right]}{dt}=-\frac{d\left[I_2\right]}{dt}=\frac{d\left[HI\right]}{dt}$
• D. $-\frac{d\left[H_2\right]}{dt}=\frac{d\left[I_2\right]}{dt}=\frac{d\left[HI\right]}{dt}$

Answer 1

Answer: (B)

$H_2+I_2⟶2HI$

Rate of reaction $=\frac{-d\left[H_2\right]}{dt}$

$=\frac{-d\left[I_2\right]}{dt}=\frac{1}{2}\frac{d[HI]}{dt}$

or $\frac{-2d\left[H_2\right]}{dt}=\frac{-2d\left[I_2\right]}{dt}$

$=\frac{d[HI]}{dt}$