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The rate and mechamical reaction are stu...

The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction `aA + bBrarr cC+dD` is given as:
`rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt)` or expression can be written as : rate `=K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d)`. At equilibrium, rate =`0`. The constants `K, K_(1), K_(2)` are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism.
The rate of formation of `SO_(3)` in the following reaction,
`2SO_(2)+O_(2)rarr 2SO_(3)` is `10 g sec^(-1)`
The rate of disappearance of `O_(2)` will be:

A

(a) `5 g sec^(-1)`

B

(b) `100 g sec^(-1)`

C

(c ) `20 g sec^(-1)`

D

(d) `2 g sec^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
d
`1/2(d[SO_(3)])/(dt)=(d[O_(2)])/(dt)`
`:. (d[O_(2)])/(dt)=1/2xx10/80 mol sec^(-1)`
`=1/2xx10/80xx32=2 g sec^(-1)`
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The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bBrarr cC+dD is given as: rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt) or expression can be written as : rate =K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d) . At equilibrium, rate = 0 . The constants K, K_(1), K_(2) are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. For the reaction, aArarr bB , log[(-dA)/(dt)]=log[(dB)/(dt)]+0.6 , then a:b is:

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