[" In the reaction,"xA rarr yB],[log[(-d[A])/(dt)]=log[(+d[B])/(dt)]+0.3],[" then the value of "x/y" is "-]

In the reaction x A rarr yB, log{-(d[A])/(dt)}=log{+(d[B])/(dt)}+0.3 Then, x:y is

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) In the following reaction, xA rarr yB log.[-(d[A])/(dt)] = log.[(d[B])/(dt)] + 0.3 where negative isgn indicates rate of disappearance of the reactant. Thus, x:y is:

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) In the following reaction, xA rarr yB log.[-(d[A])/(dt)] = log.[(d[B])/(dt)] + 0.3 where negative isgn indicates rate of disappearance of the reactant. Thus, x:y is:

For a chemical reaction aArarr bB , log[-(d(A))/(dt)]=log[(d[B])/(dt)]+0.3 Then find the approximately ratio of a and b is.

For a chemical reaction aAtoB,log[-(d[A])/(dt)]=log[(d[B])/(dt)]+0.3 then find the approximate ratio and of a and b

For the reaction : xArarryB , "log"_(10)((-d[A])/(dt))="log"_(10)((+d[B])/(dt))+0.3 If the value of log_(10)5=0.7 , the value of x : y is :

In the following reaction, xA rarryB log_(10)[-(d[A])/(dt)]=log_(10)[(d([B]))/(dt)]+0.3010 ‘A’ and ‘B’ respectively can be:

In the following reaction, xA rarryB log_(10)[-(d[A])/(dt)]=log_(10)[(d([B]))/(dt)]+0.3010 ‘A’ and ‘B’ respectively can be:

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: