Solveeit Logo
Login

Question

Chemistry Question on Rate of a Chemical Reaction

For the reaction : \ceH2+I2−>2HI\ce{H_2 +I_2 -> 2HI} , the differential rate law is

A

−d[H2]dt=−d[I2]dt=2d[HI]dt-\frac{d\left[H_{2}\right]}{dt}=-\frac{d\left[I_{2}\right]}{dt}=2 \frac{d\left[HI\right]}{dt}

B

−d[H2]dt=−2d[I2]dt=d[HI]dt-\frac{d\left[H_{2}\right]}{dt}=-2\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt}

C

−d[H2]dt=−d[I2]dt=d[HI]dt-\frac{d\left[H_{2}\right]}{dt}=-\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt}

D

−d[H2]dt=d[I2]dt=d[HI]dt-\frac{d\left[H_{2}\right]}{dt}=\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt}

Answer

−d[H2]dt=−2d[I2]dt=d[HI]dt-\frac{d\left[H_{2}\right]}{dt}=-2\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt}

Explanation

Solution

H2+I2⟶2HIH _{2}+ I _{2} \longrightarrow 2 HI

Rate of reaction =−d[H2]dt=\frac{-d\left[ H _{2}\right]}{d t}
=−d[I2]dt=12d[HI]dt=\frac{-d\left[ I _{2}\right]}{d t}= \frac{1}{2} \frac{d[ HI ]}{d t}

or −2d[H2]dt=−2d[I2]dt\frac{-2 d\left[ H _{2}\right]}{d t}=\frac{-2 d\left[ I _{2}\right]}{d t}
=d[HI]dt=\frac{d[ HI ]}{d t}