For the reaction, H_2+I_2→2HI, the differential rate law is

Correct answer: (A) -dH_2/dt=-(d[I_2])/dt=1/2(d[HI])/dt. H_2+I_2→2HI Rate of reaction, =(-dH_2)/dt=(-d[I_2])/dt=1/2(d[HI])/dt Or =(-2dH_2)/dt=(-2d[I_2])/dt=(d[HI])/dt

Q34 / 196 · NTA Abhyas (NEET) 2020 · Chemistry · Chemical Kinetics

For the reaction, $H_{2} + I_{2} \to 2 H I$ , the differential rate law is

Correct answer
A
$- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}$
B
$- \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}$
C
$- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}$
D
$- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = - \frac{d [H I]}{d t}$

Solution

H_{2} + I_{2} \to 2 H I

Rate of reaction,

= \frac{- d [H_{2}]}{d t} = \frac{- d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}

O r = \frac{- 2 d [H_{2}]}{d t} = \frac{- 2 d [I_{2}]}{d t} = \frac{d [H I]}{d t}

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

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

Solution

More Chemical Kinetics Questions

For the reaction, $H_{2} + I_{2} \to 2 H I$ , the differential rate law is