rate law

Assuming we have the reaction \[\ce{H2 + I2 -> 2HI}\]. The rate of reaction is then given by \[-\frac{\Delta \left[ \ce{H2} \right]}{1\Delta t}=-\frac{\Delta \left[ \ce{I2} \right]}{1\Delta t}=\frac{\Delta \left[ \ce{HI} \right]}{2\Delta t}\]. Now, the rate law is what relates the rate of reaction to the concentration of the reactants.

So relate it, it is important to note that the rate of a reaction is proportional to the concentration (or pressure) of the reactants raised to an experimentally determined exponent called the reaction order. In our example reaction above, we can say that \[-\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\propto \left[ \ce{H2} \right]^{m}\left[ \ce{I2} \right]^{n}\implies -\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}=k\left[ \ce{H2} \right]^{m}\left[ \ce{I2} \right]^{n}\], where \[m\] and \[n\] are our reaction orders, this constant of proportionality \[k\] is then known as our rate constant. The rate constant for almost every real-world reaction can only be determined via experimental data.

In short, reaction order in the rate law \[r=k \left[ A \right]^{m} \left[ B \right]^{n}\cdots\] is the constant \[m\], \[n\] and so on. These constants can both be zero (which means the concentration of that species does not affect the rate of a reaction), positive/negative or even fractional. It essentially tells us how strongly the rate changes when the concentration of a reactant changes. Assume \[m=2\], we then say that the reaction is second order in \[\ce{A}\]. Another reaction order we need to take note is the overall reaction order, which is just calculated by \[m+n+\cdots\]. This constant tells us the overall dependence of rate on concentration.

There are multiple methods to determine reaction order. For chemical reactions that require only one elementary step, e.g. \[\ce{2NO <=> N2O2}\], the forward rate of reaction would be \[r_{\text{fwd}}=k_{\text{fwd}} \left[ \ce{NO} \right]^{2}\], or simply, the reaction orders will be equal to be stoichiometric coefficients of each reactant. However, for chemical reactions that require more than one elementary step, that is not usually the case.

TBA: Methods of determining reaction order.

In a reversible reaction, say the elementary reaction (single-step reaction) \[\ce{ClNO2(g) + NO(g) <=> NO2(g) + ClNO(g)}\], when the rate of forward reaction, \[r_{\text{fwd}}=k_{f}[\ce{ClNO2}][\ce{NO}]\] is equal to the rate of reverse reaction, \[r_{\text{rev}}=k_{r}[\ce{NO2}][\ce{ClNO}]\] is equal, the system is said to be at equilibrium.

There is a caveat, if the reaction is a multi-step reaction, the above only applies to the rates of the broken down individual elementary reactions, not the overall reaction.

Now, if the reaction is specifically a elementary reaction, when in equilibrium, \[k_{f}[\ce{ClNO2}][\ce{NO}]=k_{r}[\ce{NO2}][\ce{ClNO}]\]. Which can be rearranged into \[\frac{k_{f}}{k_{r}}=\frac{[\ce{NO2}][\ce{ClNO}]}{[\ce{ClNO2}][\ce{NO}]}\]. Now if you look closely, \[\frac{k_{f}}{k_{r}}\] is actually also equal to the equilibrium constant, \[K_{c}\] of this reaction.

Assuming we have a two-step reaction, \[\ce{2NO(g) + O2(g) <=> 2NO2(g)}\], the first step is \[\ce{2NO <=> N2O2}\] and second is \[\ce{N2O2 + O2 <=> 2NO2}\]. The second step is much slower, thus is the rate-limiting step. While there are two reactions happening, we are generally only concerned about the net effect of these reactions, so we can just combine them into one equation as it is difficult to measure the concentrations of intermediates that are simultaneously formed and consumed in the reaction.

As we have stated, the rate of any step in a reaction is directly proportional to the concentrations of the reactants consumed in that step, thus the rate of the second reaction is \[r_{\text{second}}=k \left[ \ce{N2O2} \right] \left[ \ce{O2} \right]\]. Since the first step is relatively faster and being limited by the second step, its overall rate of reaction must also be \[\approx k \left[ \ce{N2O2} \right] \left[ \ce{O2} \right]\].

Taking advantage of the fact that the first step in this reaction is reversible, \[\ce{2NO <=> N2O2}\], the forwards and reverse rates are \[r_{f}=k_{f}[\ce{NO}]^{2}\] and \[r_{r}=k_{r}[\ce{N2O2}]\] respectively. Because the first step in this reaction is very much faster than the second, the first step should come to equilibrium first. When that happens, the rate of the forward and reverse reactions for the first step are the same. \[k_{f}[\ce{NO}]^{2}=k_{r}[\ce{N2O2}]\implies [\ce{N2O2}]=\frac{k_{f}}{k_{r}}[\ce{NO}]^{2}\]. Substituting this into \[r_{\text{second}}=k\frac{k_{f}}{k_{r}}[\ce{NO}]^{2}[\ce{O2}]\]. Since they are all constants, we can simplify this equation into \[r_{\text{second}}=k^{\prime}[\ce{NO}]^{2}[\ce{O2}]\].