Order of Chemical Reactions

The order of a chemical reaction is defined as the sum of the powers of the concentration of the reactants in the rate equation of that particular chemical reaction.

Consider a general reaction:

\[aA+bB \Rightarrow cC+dD.\]

Suppose the rate expression for this reaction is:

\[-r_a=kA^xB^y.\]

Here, \(x\) and \(y\) indicate how sensitive the rate is to the change in concentration of A and B. Hence, the order of this reaction is \(x + y\). Note that the \(x\) and \(y\) may not respectively correspond to the stoichiometric coefficients of A and B.

Calculate the overall order of a reaction which has the rate expression

\[ k{[A]}^{\frac{1}{2}}{[B]}^{\frac{3}{2}}.\]

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In the given rate law equation, the powers of concentrations of the reactants \({A}\) and \({B}\) are \(\frac{1}{2}\) and \(\frac{3}{2}\) respectively. Thus the order of the reaction is

\[\frac{1}{2}+\frac{3}{2}=2.\]

Order of a reaction is an experimentally determined quantity. It corresponds to the stoichiometric coefficients only for an elementary reaction (a reaction which occurs in just one step). A complex reaction occurs in a series of elementary reactions (multiple steps). Some steps are very fast as they do not require much energy. These steps do not affect the overall rate of reaction. Hence the rate of reaction is mainly determined by the slowest step of the reaction. Order of a reaction, in case of a complex reaction, corresponds to the stoichiometric coefficients of this rate determining step.

Hence order of a reaction gives details about stoichiometry of the rate determining step of the whole reaction mechanism.

Order of a reaction with respect to a reactant can be negative in some cases. Conversion of ozone to oxygen in excess of oxygen follows a negative order with respect to oxygen as can be seen in following rate law;

\[-r_a=k\frac{[O_3]^2}{[O_2]}.\]

Here is the reaction mechanism:

\[\begin{align} O_3 &\to O_2+O^. &\qquad (\text{Rapid Equilibrium})\\ O_3+O^. &\to 2O_2. &\qquad (\text{Rate Determining}) \end{align}\]

For the first reaction, we have

\[k_1=\frac{[O_2][O^.]}{[O_3]},\]

where \( k_1\) is the equilibrium constant.

Rate law can be obtained from the rate determining step:

\[-r_a=k_2[O_3][O^.].\]

Obtaining \([O^.]\) from the first equation,

\[[O^.]=k\frac{[O_3]}{[O_2]}.\]

Putting this in the rate equation and combining two constants yields

\[-r_a=k\frac{[O_3]^2}{[O_2]}. \ _\square\]

A reaction can also have a reaction order with respect to a reactant if the rate is not simply proportional to some power of the concentration of that reactant.

What is the order of reaction with respect to reactants \(A\) and \(B\) for having the following rate law:

\[-r_a=\frac{k_1C_aC_b}{(1+k_2C_a+k_3C_b)^2}?\]

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The above rate expression is too complicated to relate it with simple rate law. So order of reaction is not defined in this case.

A reaction can have more than one order depending upon different concentration of reactants. For example, a reaction having rate law

\[-r_a=\frac{k_1C}{1+k_2C}\]

is found to have a zero order initially when reactants are in high concentration, while the reaction order shifts to first order at the end of reaction when concentration of reactant is low.

A zero order reaction is the one whose rate is independent of concentration of reactant. Rate law for such reaction is

\[-r_a=kC_a^0=k.\]

For a constant volume system,

\[-r_a=-\frac{dC}{dt}=k.\]

Separating the variables and integrating it, we obtain

\[C=-kt+N,\]

where N is the constant of integration whose value can be found to be \(C=C_0\) at \(t=0\), which gives \(N=C_0\).

Then the final integrated rate expression is

\[C=C_0-kt.\]

Example:
Decomposition of \(N_2O\) on platinum surface gives \(N_2\) and \(O_2\). Here platinum surface acts as a catalyst and also controls the rate of reaction. There is also no effect on rate if we increase concentration of \(N_2O\), unless we keep the surface area of platinum constant:

\[2N_2O\to 2N_2+O_2.\]

.A first order reaction is the one in which the rate is proportional to concentration of a single reactant. Consider a liquid phase first order reaction

\[A+B \to C+D.\]

Rate equation for this reaction is \(-r_a=kC_a\). Since for liquid phase reaction

\[r =\frac{dN}{Vdt}=\frac{d(N/V)}{dt}=\frac{dC}{dt},\]

for the above first order reaction, we have

\[-\frac{dC}{dt}=kC.\]

This equation is called differential rate equation of first order equation. Separating variables and integrating it gives

\[\begin{align} -\frac{dC}{C}&=kdt\\ -\ln C&=kt+N, \end{align}\]

where N is the constant of integration. Its value can be found by noting at \(t=0 \implies C=C_0,\)

which gives \[N=-\ln(C_0).\]

Putting this in the main equation and rearranging gives the integrated first order rate equation

\[\ln(C/C_0)=-kt,\]

or in exponential form

\[C=C_0e^{-kt}.\]

That is, concentration of the reactant drops exponentially with time.

Consider an elementary reaction

\[A+B\to C.\]

Rate expression can be written as

\[-r_a=k^,C_aC_b.\]

If one of the reactants, say \(B\), is in relatively high concentration and its concentration changes only a little during the entire reaction, its value can be assumed to be constant and we effectively have a first order reaction with respect to \(A.\) This type of reaction is called pseudo first order reaction:

\[-r_a=(k^,C_b)C_a=kC_a,\]

where \(k\) is a new constant.

Second order reaction involves determination of concentration of both reactants simultaneously. This method of pseudo first order reaction is useful in the study of chemical kinetics. If one of the reactants is expensive, then the other reactant can be taken in excess quantity and the reaction order with respect to first reactant can be easily determined.

• Rate of Chemical Reactions