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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)?

Sagot :

Lublanaidn

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

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