Chemical Kinetics

Reaction Rates

Chemical kinetics is the field that studies the rates of chemical reactions. The speed at which a reaction occurs is its reaction rate.

Some chemical reactions occur very quickly. For example, an explosion is a chemical reaction that happens in less than one second. Some chemical reactions occur much more slowly. Iron will rust when exposed to air. This reaction occurs slowly, over years in dry weather. Iron will rust more quickly in the presence of water. The field that studies the rates of chemical reactions is called chemical kinetics. All industrial processes that produce a specific material use chemical kinetics to optimize its production.

The speed at which a reaction occurs is called the reaction rate. During a chemical reaction, reactants are converted into products. Therefore, the amounts of reactants and products change over time. Reaction rates are studied by quantifying the changes in the amount of reactants and products.

The amounts of both reactants and products can be expressed as concentrations in the reaction mixture. The average reaction rate can be defined based on concentrations of reactants and products over a period of time:
average reaction rate=change in concentrationchange in time\text{average reaction rate}=\frac{\text{change in concentration}}{\text{change in time}}
A chemical reaction occurs when there is an interaction between multiple molecules or atoms. The chance of a collision is greater when there are more molecules or atoms present. Concentration is a factor that affects reaction rate. For a given reaction, reaction rates are higher at higher concentrations of reactants. Reaction rates are lower at lower concentrations of reactants. Thus, as a reaction proceeds and the concentration of reactants changes, the reaction rate also changes. Average reaction rate indicates the average rate over a selected period of time. The actual rate is likely to be different than the average reaction. The instantaneous rate is the rate of a chemical reaction at a particular moment. If the change in one of the reactants is graphed over time for a chemical reaction, the graph will form a curve for which the absolute value of the slope decreases over time. A graph of reactant concentration versus time can be used to calculate instantaneous rates.
The instantaneous rate can be found using the slope of the line tangent to the curve at any time. The slope is the change in the reactant concentration divided by the change in time.
The rate of reaction between nitrogen oxide and carbon monoxide can be defined by any of the four compounds involved in it:
NO2(g)+CO(g)NO(g)+CO2(g){\rm{NO_2}}(g)+{\rm{CO}}(g)\rightarrow{\rm{NO}}(g)+{\rm{CO_2}}(g)
For products, the rate is equal to the concentration increase over time:
rate=Δ[NO]Δt=Δ[CO2]Δt{\rm{rate}}=\frac{\Delta\lbrack\rm{NO}\rbrack}{\Delta t}=\frac{\Delta\lbrack{\rm{CO}}_2\rbrack}{\Delta t}
For reactants, the change in concentration is negative. Therefore, the rate is equal to their concentration change over time multiplied by –1.
rate=Δ[NO2]Δt=Δ[CO]Δt=Δ[NO]Δt=Δ[CO2]Δt{\rm{rate}}=-\frac{\Delta\lbrack{\rm{NO}}_2\rbrack}{\Delta t}=-\frac{\Delta\lbrack\rm{CO}\rbrack}{\Delta t}=\frac{\Delta\lbrack\rm{NO}\rbrack}{\Delta t}=\frac{\Delta\lbrack{\rm{CO}}_2\rbrack}{\Delta t}
Note that the unit for concentration is M, which is mol/L. The unit for time is seconds (s). The unit for reaction rate is therefore M/s or mol/L·s. In the decomposition reaction of hydrogen iodide (HI), the rate of consumption of HI is twice the rate of formation of hydrogen (H2) and iodine (I2) gases. This is shown by the coefficients of each species in the reaction equation:
2HI(g)H2(g)+I2(g)2\rm{HI}(\mathit g)\rightarrow{\rm H}_2(\mathit g)+{\rm I}_2(\mathit g)
The rate in terms of products is straightforward.
rate=Δ[H2]Δt=Δ[I2]Δt{\rm{rate}}=\frac{\Delta\lbrack{\rm{H}}_2\rbrack}{\Delta t}=\frac{\Delta\lbrack{\rm{I}}_2\rbrack}{\Delta t}
The rates of formation of hydrogen gas and iodine gas are half the rate that hydrogen iodide (HI) is consumed. The coefficient of two causes the hydrogen iodide (HI) term to have a multiplier of 12\frac{1}{2}. Also note the negative sign, because hydrogen iodide (HI) is a reactant.
rate=12Δ[H2]Δt{\rm{rate}}=-\frac{1}2\frac{\Delta\lbrack{\rm H}_2\rbrack}{\Delta t}
Hence, the rate in terms of hydrogen iodide consumed (HI) can be written in terms of hydrogen gas formation.
Δ[H2]Δt=12Δ[HI]Δt\frac{\Delta[{\rm H}_2]}{\Delta t}=-\frac{1}{2}\frac{\Delta[{\rm {HI}}]}{\Delta t}
In general, the rate for the reaction
aA+bBcC+dDa{\rm A}+b{\rm B}\rightarrow c{\rm C}+d{\rm D}
can be written as
rate=1aΔ[A]Δt=1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt{\rm{rate}}=-\frac1a\frac{\Delta\lbrack\rm A\rbrack}{\Delta t}=-\frac1b\frac{\Delta\lbrack\rm B\rbrack}{\Delta t}=\frac1c\frac{\Delta\lbrack\rm C\rbrack}{\Delta t}=\frac1d\frac{\Delta\lbrack\rm D\rbrack}{\Delta t}
Step-By-Step Example
Reaction Rate Calculation for the Decomposition of Dinitrogen Pentoxide
Dinitrogen pentoxide (N2O5) breaks down into nitrogen dioxide (NO2) and oxygen gas (O2) with the following reaction:
2N2O5(g)4NO2(g)+O2(g)2{\rm N}{_2}{\rm O}{_5}(\mathit g)\rightarrow4{\rm{NO}}{_2}(\mathit g)+{\rm O}{_2}(\mathit g)
The instantaneous rate of formation of oxygen gas (O2) is 2.0×107M/s2.0\times10^{-7}\;\rm M/\rm s . What is the rate of formation of nitrogen dioxide (NO2) gas and rate of consumption of dinitrogen pentoxide (N2O5)?
Step 1
Write the rate in terms of each product and reactant.
rate=12Δ[N2O5]Δt=14Δ[NO2]Δt=11Δ[O2]Δt{\rm{rate}}=-\frac12\frac{\Delta\lbrack{\rm N}{_2}{\rm O}{_5}\rbrack}{\Delta t}=\frac14\frac{\Delta\lbrack{\rm{NO}}_2\rbrack}{\Delta t}=\frac11\frac{\Delta\lbrack{\rm O}{_2}\rbrack}{\Delta t}
Step 2
The rate of formation of oxygen is given. This can be used to solve for the rate of formation of nitrogen dioxide (NO2) gas and the rate of consumption of dinitrogen pentoxide (N2O5).
12Δ[N2O5]Δt=2.0×107M/sΔ[N2O5]Δt=(2)(2.0×107M/s)=4.0×107M/s\begin{aligned}-\frac12\frac{\Delta\lbrack{\rm N}_2{\rm O}_5\rbrack}{\Delta t}&=2.0\times10^{-7}\;\rm M/\rm s\\\frac{\Delta\lbrack{\rm N}_2{\rm O}_5\rbrack}{\Delta t}&=(-2)(2.0\times10^{-7}\;\rm M/\rm s)\\&=-4.0\times10^{-7}\;\rm M/\rm s\end{aligned}
Note the negative sign, indicating dinitrogen pentoxide is consumed.
Solution
Complete the calculation.
14Δ[NO2]Δt=2.0×107M/sΔ[NO2]Δt=(4)(2.0×107M/s)=8.0×107M/s\begin{aligned}\frac14\frac{\Delta[{\rm{NO}}_2]}{\Delta t}&=2.0\times10^{-7}\;\rm M/\rm s\\\frac{\Delta[{\rm{NO}}_2]}{\Delta t}&=(4)(2.0\times10^{-7}\;\rm M/\rm s)\\&=8.0\times10^{-7}\;\rm M/\rm s\end{aligned}