Kinetics of Consecutive Reactions

What is a Consecutive reaction?

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;

\[\displaystyle A\underset{{}}{\overset{{\mathop{k}_{1}}}{\longrightarrow}}B\underset{{}}{\overset{{\mathop{k}_{2}}}{\longrightarrow}}C\]

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 t_max in terms of rate constant K₁ and k₂ and for [B_max], i.e. the maximum concentration of A, B and C as a function of time.

To write, the concentration at time t;

\[\displaystyle \left[ A \right]=a;\left[ B \right]=b;\left[ C \right]=c---(1)\]

Initial concentrations

\[\displaystyle \left[ A \right]={{a}_{0}};\left[ B \right]={{b}_{0}};\left[ C \right]={{c}_{0}}---(2)\]

\[\displaystyle \frac{{d\left[ A \right]}}{{dt}}=a;\frac{{d\left[ B \right]}}{{dt}}=b;\frac{{d\left[ C \right]}}{{dt}}=c---(3)\]

With this notion, the rate equation is,

\[\displaystyle a=-{{k}_{1}}a;b=-{{k}_{1}}a-{{k}_{2}}b;c={{k}_{2}}b--(4)\]

The condition for mass conversation requires is,

\[\displaystyle {{a}_{0}}=a+b+c---(5)\]

\[\displaystyle a={{a}_{0}}-b-c---(6)\]

\[\displaystyle b={{k}_{1}}\left( {{{a}_{0}}-b-c} \right)-{{k}_{2}}b---(7)\]

Where, a₀ =0,

\[\displaystyle b={{k}_{1}}\left( {-b-c} \right)-{{k}_{2}}b={{k}_{1}}\left( {-b-{{k}_{2}}b} \right)-{{k}_{2}}b---(8)\]

The general solution of this equation is,

\[\displaystyle b={{c}_{1}}\exp \left( {{{m}_{1}}t} \right)+{{c}_{2}}\exp \left( {{{m}_{2}}t} \right)---(9)\]

Here, c₁ and c₂ = constant

m₁, m₂ = two roots of equation

\[\displaystyle {{m}^{2}}+\left( {{{k}_{1}}+{{k}_{2}}} \right)m+{{k}_{1}}{{k}_{2}}=0---(10)\]

This quadratic equation gives,

\[\displaystyle {{m}_{1}}=-{{k}_{1}}\text{ and }{{\text{m}}_{2}}=-{{k}_{2}}---(11)\]

Now, substituting the value of m₁ and m₂ into equation (9), we have

\[\displaystyle b={{c}_{1}}\exp \left( {-{{k}_{1}}t} \right)+{{c}_{2}}\exp \left( {-{{k}_{2}}t} \right)---(12)\]

At t=0, b=0 from equation (9)

\[\displaystyle {{c}_{1}}+{{c}_{2}}=0\text{ or }{{c}_{1}}=-{{c}_{2}}----(13)\]

\[\displaystyle b={{c}_{1}}\left[ {\exp \left( {-{{k}_{1}}t} \right)-\exp \left( {-{{k}_{2}}t} \right)} \right]---(14)\]

Now, evaluate,

\[\displaystyle {{c}_{1}}=b={{c}_{1}}\left[ {-{{k}_{1}}\exp \left( {-{{k}_{1}}t} \right)+{{k}_{2}}\exp \left( {-{{k}_{2}}t} \right)} \right]----(15)\]

\[\displaystyle t=0,b={{c}_{1}}\left( {-{{k}_{1}}+{{k}_{2}}} \right)---(16)\]

From equation (8), at t = 0, b=c=0, so, that

\[\displaystyle b={{k}_{1}}{{a}_{0}}----(17)\]

From equations (16) and (17), we get,

\[\displaystyle {{c}_{1}}=\frac{{{{k}_{1}}{{a}_{0}}}}{{({{k}_{2}}-{{k}_{1}})}}---(18)\]

Substituting for c₁ in equation (14), we get,

\[\displaystyle b=\frac{{{{k}_{1}}{{a}_{0}}}}{{({{k}_{2}}-{{k}_{1}})}}\left[ {\exp \left( {-{{k}_{1}}t} \right)-\exp \left( {-{{k}_{2}}t} \right)} \right]---(19)\]

Since a = -k₁a from equation (4)

\[\displaystyle a={{a}_{0}}\exp \left( {-{{k}_{1}}t} \right)---(20)\]

To calculate the value of c, we use the mass conservation requirement viz, a₀=a+b+c,

\[\displaystyle c={{a}_{0}}-a-b={{a}_{0}}\left[ {1-\frac{{{{k}_{2}}\exp \left( {-{{k}_{1}}t} \right)-{{k}_{1}}\exp \left( {-{{k}_{2}}t} \right)}}{{{{k}_{2}}-k{}_{1}}}} \right]---(21)\]

MAXIMUM VALUE OF [B]

Maximum value of b that is, b_max, b= db/dt=0 and b=d²d/dt²,0 from equation (19), we get,

\[\displaystyle {{k}_{1}}\exp \left( {-{{k}_{1}}{{t}_{{\max }}}} \right)={{k}_{2}}\exp \left( {-{{k}_{2}}{{t}_{{\max }}}} \right)---(22)\]

\[\displaystyle \frac{{{{k}_{1}}}}{{{{k}_{2}}}}=\exp \left[ {\left( {{{k}_{1}}-{{k}_{2}}} \right){{t}_{{\max }}}} \right]---(23)\]

Taking log both sides in the above equation

\[\displaystyle {{t}_{{\max }}}=\frac{{\ln \left( {\frac{{{{k}_{1}}}}{{{{k}_{2}}}}} \right)}}{{{{k}_{1}}-{{k}_{2}}}}---(24)\]

From equation (11) b+(k₁+k₂)b+k₁k₂b=0

\[\displaystyle b=-{{k}_{1}}{{k}_{2}}b=\text{A negative quantity-----(25)}\]

Se that t=t_max, b reaches maximum; it does not have a minimum or a point of inflection.

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

\[\displaystyle {{\text{b}}_{{\max }}}=\frac{{{{k}_{1}}{{a}_{0}}}}{{{{k}_{2}}-{{k}_{1}}}}\left\{ {\exp \left[ {-\frac{{{{k}_{1}}\ln \left( {\frac{{{{k}_{1}}}}{{{{k}_{2}}}}} \right)}}{{{{k}_{1}}-{{k}_{2}}}}} \right]-\exp \left[ {-\frac{{{{k}_{2}}\ln \left( {\frac{{{{k}_{1}}}}{{{{k}_{2}}}}} \right)}}{{{{k}_{1}}{{k}_{2}}}}} \right]} \right\}---(26)\]

On simplification,

\[\displaystyle {{b}_{{\max }}}={{u}_{0}}{{\left( {\frac{{{{k}_{2}}}}{{{{k}_{1}}}}} \right)}^{{\frac{{{{k}_{2}}}}{{\left\{ {{{k}_{1}}-{{k}_{2}}} \right\}}}}}}----(27)\]

The time dependence of a, b and c.

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KINETICS OF CONSECUTIVE REACTIONS

What is a Consecutive reaction?

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Bhoomika Sheladiya

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