The Second Order Integrated Rate Law

Karuna Samuel Finch, LaRita Williams
  • Author
    Karuna Samuel Finch

    Karuna has taught Middle and High School level Chemistry and Biology for over 10 years. She have a Masters in Chemistry, a professional degree in Education and, is currently pursuing a PhD in Education.

  • Instructor
    LaRita Williams

    LaRita holds a master's degree and is currently an adjunct professor of Chemistry.

Know the relationship between the integrated and differential rate law. Learn about the second-order integrated rate law with examples and a sample question. Updated: 02/17/2022

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Integrated Rate Law

The term rate relates to the amount of change in one entity in proportion to another. In finance, exchange rates are used to calculate how the value of one currency will convert to another. The rate of change in the position of a vehicle is investigated in physics. Similarly, in applied chemistry, reaction rates are investigated and defined as the change in the quantity of the reactant (or the product) with time. For example, if a drug is being synthesized in a laboratory, one may want to know how long it will take for the reactant amount to reduce to a specific value, so that the product may be extracted.

Calculus may be used to infer the connection between a reactant's concentration and time from a rate law. Details about the rate law are covered in the next section. We can derive integrated rate laws that quantitatively give concentration as a function of time by aggregating or "integrating" the instantaneous rates of a reaction from the start of the reaction until some predetermined time. The order of the reaction determines the form of the integrated rate law.

For a reaction that is of the second order (n=2) in which the reactant A is converted to some products: {eq}aA\longrightarrow Products {/eq}


{eq}\begin{align} Rate & = k \Delta[A]^n &\text{(Equation 1)}\\\\ \end{align} {/eq}


Since n= 2, its integration will yield the following form:


{eq}\begin{align} 1/[A]_t& = kt + 1/[A]_0 &\text{(Equation 2)} \end{align} {/eq}


Equation 2 is known as an integrated rate law equation, where k is used to represent the rate constant, and the value of k may be found by displaying a graph of {eq}1/[A]_t {/eq} versus t in this particular case. Equation 3 is used to determine how the concentration of a material changes over time. Therefore, the integrated rate law can be defined as a function of the initial concentration and the actual concentration of one or more reactants after a specific amount of time. The use of integrated rate laws to analyze experimental data allows for the prediction of the makeup of a reaction system at any point, the validation of the rate law, and the calculation of the rate constant. The concept of rate law is explained next.

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The Differential Rate Law

Chemical reactions are reversible; and once the product starts to accumulate, the backward reaction starts. Consider the decomposition of nitrogen dioxide, {eq}2NO_2 \longrightarrow 2NO + O_2 {/eq}. When the reaction begins and in the very moments of this process, the amounts of nitrogen monoxide and oxygen are quite less, enough to be overlooked. Over time, their amounts will grow. If one wants to know the rate of decomposition of {eq}NO_2 {/eq}, a difference between the forward and reverse reaction rates will be considered. Such calculations can get quite complicated. To avoid them, scientists study reaction rates as soon as the process begins. For our reaction in discussion, the rate of the reaction is proportional to the concentration of {eq}NO_2 {/eq} raised to an exponent n.


{eq}\begin{align} Rate \propto [NO_2]^n && \text{(Equation 3)} \end{align} {/eq}

The proportionality symbol, {eq}\propto {/eq} in Equation 3 can be replaced by an equals sign if we introduce a proportionality constant, k, which is called the rate constant for the reaction. This gives us an expression the same as in Equation 1,

{eq}Rate = k[NO_2]^n {/eq}

So, Equation 1 is actually the differential rate law expression and is commonly referred to as the rate law. It is defined as an expression that indicates how the rate is affected by the concentrations of reactants. The numerical values of the exponent n (i.e. the order of the reaction) must be determined experimentally and is not necessarily related to the coefficient used to balance the equation. The rate law describes what is happening in the chemical reaction at a molecular level, but it does not incorporate the concentration of products. This is simply because the reaction is being studied at a time where it has just begun and products have either not appeared or are negligible.

Second-Order Integrated Rate Law

In chemical kinetics, the term order refers to the fluctuation of the reactant concentration and its influence on the reaction rate. The second-order reaction can be understood better by comparing it with the zero and first order reactions. So, in terms of a reactant, what does it mean to be zero, first, or second order? Zero-order reaction rates are unaltered by changes in the reactant's concentration. Whereas, when the reactant concentration changes, the rate of a first-order reaction grows linearly, producing a sloping straight line graph. For example, if the concentration increases two-fold, the rate will double as well. For second-order reactions, the rate rises quadratically as the reactant concentration changes, leading in a parabola curve. This means that doubling the reactant concentration will result in a fourfold increase in rate; or tripling the reactant will lead to a nine-times increase in the rate. It is important to remember that the order of a reaction has to be known through experimentation and can never be deduced via a balanced equation.

Consider a simple reaction that has a single reactant, {eq}aA \longrightarrow Products {/eq}. Assuming this reaction is of the second-order with respect to A (i.e. n=2), the rate law can be expressed as:

{eq}\begin{align} Rate& = - \Delta [A]/ \Delta t \\ &= k [A]^2 \end{align} {/eq}


This rate law can be put into a different form using a calculus operation known as integration, which yields the expression:


{eq}\begin{align} 1/[A]_t& = kt + 1/[A]_0 &\text{(same as equation 2 mentioned earlier)} \end{align} {/eq}


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Frequently Asked Questions

What are integrated laws?

The integrated rate laws are a function of the initial concentration and the actual concentration of one or more reactants at a certain time interval. Their mathematical forms are specific to the order of the reaction they are associated with.

What is the second order rate law?

The equation, 1/At = kt + 1/A0 represents the second order rate law, as it draws a relationship between the concentration of the reactant and time.

What is an example of a second order reaction?

Dimerization of butadiene; 2C4H6(g) ? C8H12, is a second order reaction with respect to the reactant butadiene. This means if the concentration of butadiene is doubled, the rate is increased four times.

What is a 2nd order reaction?

If a reaction is of the second order with respect to a certain reactant, it means that if the concentration of that reactant is increased, the rate of the reaction will increase quadratically.

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