If for the reaction `A to B" rate" =-(d[A])/(dt)=2(d[B])/(dt)` then, rate law 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 the reaction, aArarr bB , log[(-dA)/(dt)]=log[(dB)/(dt)]+0.6 , then a:b is:

For the reaction aA+bBtocC," if "-3(d[A])/(dt)=-(d[B])/(dt)=+1.5(d[C])/(dt), then a, b and c respectively are

ArarrB , Rate of this reaction is expressed by -(d[A])/(dt) " or "+(d[B])/(dt) . What is the significance of (-) sign and (+) sign in this case ?

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) A reaction follows the given concentration-time graph. The rate for this reaction at 20 s will be

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) A reaction follows the given concentration-time graph. The rate for this reaction at 20 s will be