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The rate of a reaction is expressed in d...

The rate of a reaction is expressed in different ways as follows:
`+(1)/(2)(d[C])/(dt)=-(1)/(3)(d[D])/(dt)=+(1)/(4)(d[A])/(dt)=-(d[B])/(dt)`
the reaction is

A

`4A+B rarr 2C+3D`

B

`B+3Drarr4A+2C`

C

`4A+2Brarr2C+3D`

D

`B+((1)/(2))Drarr4A+3`

Text Solution

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The correct Answer is:
B
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The rate of a reaction is expressed in different ways as follows +1/2 (d[C ])/(dt)=-(1)/(5) -(d[D])/( dt)=+(1)/(3 ) [d[A ])/(dt) =-(d[B])/(dt) the reaction is ________.

The rate of the reaction may be expressed by the following different ways . (1)/(2) (d[A])/(dt)=-(1)/(3)(d[B])/(dt)=(-d[C])/(dt) reaction is

d/(dt) (1/t)=

The rate of reaction of certain reaction is expressed as : (1)/(3)(d[A])/(dt)=-(1)/(2)(d[B])/(dt)=-(d[C])/(dt) The reaction is :

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 a gaseous reaction, the rate is expressed in terms of (dP)/(dt) in place of (dC)/(dt) or (dn)/(dt) where C is concentration, n is number of moles and 'P' is pressure of reactant. The three are related as:

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