KINETICS OF CONSECUTIVE REACTIONS

The reaction is complete in two or more steps, one after the other called a consecutive reaction.

These reaction features are illustrated in the following example.

Example: Consider two consecutive first-order reactions;

\[\displaystyle A\underset<<>><\overset<<\mathop_>><\longrightarrow>>B\underset<<>><\overset<<\mathop_>><\longrightarrow>>C\]

Suppose K1 +K2 at time t=0 only A is present and [B]=[C]=0the reaction proceeds,[B] reaches a maximum and falls off after that. Derive an expression for tmax in terms of rate constant K1 and k2 and for [Bmax], i.e. the maximum concentration of A, B and C as a function of time.

To write, the concentration at time t;

\[\displaystyle \left[ A \right]=a;\left[ B \right]=b;\left[ C \right]=c---(1)\] \[\displaystyle \left[ A \right]=>;\left[ B \right]=_>;\left[ C \right]=_>---(2)\]>>=a;\frac<>>=b;\frac<>>=c---(3)\]

With this notion, the rate equation is,

\[\displaystyle a=-<_>a;b=-<_>a-<_>b;c=<_>b--(4)\]The condition for mass conversation requires is,

\[\displaystyle >=a+b+c---(5)\]-b-c---(6)\]\left( >-b-c> \right)-<_>b---(7)\]\left( \right)-<_>b=<_>\left( _>b> \right)-<_>b---(8)\]The general solution of this equation is,

\[\displaystyle b=<_>\exp \left( _>t> \right)+<_>\exp \left( _>t> \right)---(9)\]m1, m2 = two roots of equation

\[\displaystyle <^>+\left( _>+_>> \right)m+_>_>=0---(10)\]This quadratic equation gives,

\[\displaystyle <_>=-_>\text< and >>_>=-_>---(11)\]Now, substituting the value of m1 and m2 into equation (9), we have

\[\displaystyle b=<_>\exp \left( _>t> \right)+<_>\exp \left( _>t> \right)---(12)\]At t=0, b=0 from equation (9)

\[\displaystyle <_>+<_>=0\text< or ><_>=-<_>----(13)\]\left[ _>t> \right)-\exp \left( _>t> \right)> \right]---(14)\]\left( _>+_>> \right)---(16)\]From equation (8), at t = 0, b=c=0, so, that

\[\displaystyle b=<_>>----(17)\]From equations (16) and (17), we get,

\[\displaystyle <_>=\frac_>>>><<(_>-_>)>>---(18)\]Substituting for c1 in equation (14), we get,

\[\displaystyle b=\frac<<_>>>><<(_>-_>)>>\left[ <\exp \left( <-_>t> \right)-\exp \left( <-_>t> \right)> \right]---(19)\]Since a = -k1a from equation (4)

\[\displaystyle a=>\exp \left( _>t> \right)---(20)\]To calculate the value of c, we use the mass conservation requirement viz, a0=a+b+c,

MAXIMUM VALUE OF [B]

Maximum value of b that is, bmax, b= db/dt=0 and b=d 2 d/dt 2 ,0 from equation (19), we get,

\[\displaystyle <_>\exp \left( _>_<<\max >>>> \right)=<_>\exp \left( _>_<<\max >>>> \right)---(22)\]>><<_>>>=\exp \left[ <\left( <_>-_>> \right)_<<\max >>>> \right]---(23)\]Taking log both sides in the above equation

\[\displaystyle <_<<\max >>>=\frac<<\ln \left( <\frac_>>>_>>>> \right)>>_>-_>>>---(24)\]<_>b=\text\]Se that t=tmax, b reaches maximum; it does not have a minimum or a point of inflection.

Substituting equation (24) into equation (19), we get,

\[\displaystyle <_<<\max >>>=_><<\left( <\frac_>>>_>>>> \right)>^<<\frac_>>><<\left\< <_>-_>> \right\>>>>>>----(27)\]The time dependence of a, b and c.

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About the author

Bhoomika Sheladiya

BSc. (CHEMISTRY) 2014- Gujarat University
MSc. (PHYSICAL CHEMISTRY) 2016 - School of Science, Gujarat University

Junior Research Fellow (JRF)- 2019
AD_HOC Assistant Professor-(July 2016 to November 2021)