Question

HI dissociates to form I2 and H2: 2HI(g) → H2(g) + I2(g) If the concentration of HI changes at a rate of –0.45 M/s, what is the rate of appearance of I2(g)?

Answer 1

Answer:

The rate of appearance of I2(g)=0.225M/s

Explanation:

We are given that

$2HI\rightarrow H_2+I_2$

$\frac{d[HI]}{dt}=-0.45M/s$

We have to find the rate of appearance of I2(g).

We know that

Rate of reaction=$-\frac{1}{2}\frac{d[HI]}{dt}=\frac{d[I_2]}{dt}=\frac{d[H_2]}{dt}$

Therefore,

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

Substitute the values

We get

$-\frac{1}{2}\times (-0.45)=\frac{d[I_2]}{dt}$

$\frac{d[I_2]}{dt}=0.225M/s$

Hence, the rate of appearance of I2(g)=0.225M/s