Answer:
The rate of appearance of I2(g)=0.225M/s
Explanation:
We are given that
[tex]2HI\rightarrow H_2+I_2[/tex]
[tex]\frac{d[HI]}{dt}=-0.45M/s[/tex]
We have to find the rate of appearance of I2(g).
We know that
Rate of reaction=[tex]-\frac{1}{2}\frac{d[HI]}{dt}=\frac{d[I_2]}{dt}=\frac{d[H_2]}{dt}[/tex]
Therefore,
[tex]-\frac{1}{2}\frac{d[HI]}{dt}=\frac{d[I_2]}{dt}[/tex]
Substitute the values
We get
[tex]-\frac{1}{2}\times (-0.45)=\frac{d[I_2]}{dt}[/tex]
[tex]\frac{d[I_2]}{dt}=0.225M/s[/tex]
Hence, the rate of appearance of I2(g)=0.225M/s
