KINETICS OF OPPOSING OR REVERSIBLE REACTIONS

What is Reversible reaction?

The reaction at which reactant is converted into product and product is converted into reactant process takes place simultaneously is known as Reversible reaction.

The kinetics of such reactions are usually studied at the initial stages of the process when the products are too low a concentration to set up the opposing reaction rate.

Consider a simple case for an opposing reaction; that reaction order is first for reverse and forward reactions.

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

Here,

K1 = rate constant for the forward reactions

K2 = rate constant for the reverse reactions

a is the initial concentration of the reactant A. (supposed that initial concentration of B=0). After time t, initial concentration A will be (a-x), and the product will be x.

The rate of forwarding reaction A will be (a-x), and the reverse reaction k1-x. The net rate of the formation of B is,

\[\displaystyle \frac{{dx}}{{dt}}={{k}_{1}}\left( {a-x} \right)-{{k}_{{-1}}}x---(1)\]

And  xe is the concentration of B at equilibrium; when the net rate is zero, then,

\[\displaystyle {{k}_{1}}\left( {a-x{}_{e}} \right)-{{k}_{{-1}}}{{x}_{e}}=0\]

or

\[\displaystyle {{k}_{1}}\left( {a-x{}_{e}} \right)={{k}_{{-1}}}{{x}_{e}}---(2)\]

or

\[\displaystyle {{k}_{{-1}}}={{k}_{1}}\frac{{\left( {a-x{}_{e}} \right)}}{{x{}_{e}}}---(3)\]

Now, substituting the value of k1 in equation (2), we get,

\[\displaystyle \frac{{dx}}{{dt}}={{k}_{1}}\left( {a-x} \right)-{{k}_{1}}\left( {\frac{{a-x{}_{e}}}{{x{}_{e}}}} \right)x={{k}_{1}}\frac{{\left( {x{}_{e}-x} \right)a}}{{x{}_{e}}}---(4)\]

Now, separating the variables, we get,

\[\displaystyle \frac{{dx}}{{x{}_{e}-x}}=\frac{a}{{x{}_{e}}}{{k}_{1}}at---(5)\]

Integrating the above equation & we get,

\[\displaystyle -\ln \left( {x{}_{e}-x} \right)=\frac{a}{{x{}_{e}}}\left( {{{k}_{1}}t+I} \right)---(6)\]

Here, I = integration constant

or

\[\displaystyle -\frac{{x{}_{e}}}{a}\ln \left( {x{}_{e}-x} \right)={{k}_{1}}t+I---(7)\]

If t = 0 and x =0, so we have,

\[\displaystyle I=-\frac{{x{}_{e}}}{a}\ln x{}_{e}---(8)\]

Now, substitute the value of I in equation (7), and we get,

\[\displaystyle -\frac{{x{}_{e}}}{a}\ln \left( {x{}_{e}-x} \right)={{k}_{1}}t--\frac{{x{}_{e}}}{a}\]

or

\[\displaystyle {{k}_{1}}t=\frac{{x{}_{e}}}{a}\ln \left( {\frac{{x{}_{e}}}{{x{}_{e}-x}}} \right)\]

or

\[\displaystyle {{k}_{1}}=\frac{{x{}_{e}}}{{at}}\ln \left( {\frac{{x{}_{e}}}{{x{}_{e}-x}}} \right)---(9)\]

Here, equation (9) k1 in terms of easily measurable quantities.

From equation (3), we get,

\[\displaystyle (1+{{k}_{{-1}}})={{k}_{1}}\frac{a}{{x{}_{e}}}=\frac{1}{t}\ln \left( {\frac{{x{}_{e}}}{{x{}_{e}-x}}} \right)---(10)\]

K1, a and xe, the rate constant k-1 for the reverse reaction can be calculated.

Assume that initially a small amount of B, say b moles, is present; the net rate of formation of B is,

\[\displaystyle \frac{{dx}}{{dt}}={{k}_{1}}\left( {a-x} \right)-{{k}_{{-1}}}\left( {a-x} \right)---(11)\]

At equilibrium dx/dt =0,

\[\displaystyle {{k}_{1}}\left( {a-x{}_{e}} \right)={{k}_{{-1}}}\left( {b+{{x}_{e}}} \right)---(12)\]

or

\[\displaystyle {{k}_{{-1}}}={{k}_{1}}\frac{{\left( {a-x{}_{e}} \right)}}{{\left( {b+{{x}_{e}}} \right)}}---(13)\]

Now, substituting the value of k-1 in equation (11), we have,

\[\displaystyle \frac{{dx}}{{dt}}={{k}_{1}}\left( {a-x} \right)-{{k}_{1}}\frac{{\left( {a-x{}_{e}} \right)\left( {b+x} \right)}}{{\left( {b+{{x}_{e}}} \right)}}\]
\[\displaystyle ={{k}_{1}}\left( {\frac{{a+b}}{{b+{{x}_{e}}}}} \right)\left( {x{}_{e}-x} \right)---(14)\]

Separating the variables,

\[\displaystyle \left( {\frac{{b+{{x}_{e}}}}{{a+b}}} \right)\left( {\frac{{dx}}{{x{}_{e}-x}}} \right)={{k}_{1}}dt---(15)\]

Integration yields,

\[\displaystyle \frac{{\left( {b+{{x}_{e}}} \right)}}{{\left( {a+b} \right)}}\left[ {-\ln \left( {x{}_{e}-x} \right)} \right]={{k}_{1}}dt+I---(16)\]

At t=0, x=0

\[\displaystyle I=\frac{{\left( {b+{{x}_{e}}} \right)}}{{\left( {a+b} \right)}}\left( {-\ln {{x}_{e}}} \right)---(17)\]

Now, substituting the value of I in equation (16), we get,

\[\displaystyle {{k}_{1}}=\frac{{b+{{x}_{e}}}}{{t(a+b)}}\ln \left( {\frac{{{{x}_{e}}}}{{{{x}_{e}}-x}}} \right)---(18)\]

From equation (13),

\[\displaystyle {{x}_{e}}=\frac{{{{k}_{1}}a-{{k}_{{-1}}}b}}{{{{k}_{1}}+{{k}_{{-1}}}}}---(19)\]

Now, substitute the value of xe into equation (18); we have,

\[\displaystyle \left( {{{k}_{1}}+{{k}_{{-1}}}} \right)=\frac{1}{t}\ln \left( {\frac{{{{x}_{e}}}}{{{{x}_{e}}-x}}} \right)---(20)\]

Equation (20) is similar to equation (10), obtained when B is not present initially. 

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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)

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