Integrated Rate Equations

You must already know that calculating the rate of a reaction is extremely important to understand the reaction. But it also necessary to infer the rate law of a reaction to find out the order of the reaction. But what is the order of a reaction you ask? Let’s find out below and also learn about the different types of reactions.

Rate Law

We know that the rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be equal to the stoichiometric coefficient of the reacting species in a balanced chemical equation.

Consider a general reaction, aA + bBÂ â†’ cC + dD where a, b, c,Â d are theÂ stoichiometric coefficients of the reactants and the products. Therefore, the rate law for the above reaction is,

Rate $$\propto$$ [A]x [B]y

whereÂ xÂ andÂ yÂ may or may not be equal to the stoichiometric coefficients of the reactants. Therefore, the rate of the reaction is equal to kÂ [A]x [B]y,Â where k is the rate constant.

âˆ´ -d[R]/dt =Â kÂ [A]x [B]yÂ

Order of Reaction

The order of a reaction is the sum of the powers of the concentrations of the reactants in the rate law expression. In the above general reaction, x and y are the powers. The sum of them will give the order of the reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero-order reaction means that the rate of the reaction is independent of the concentration of reactants.

DownloadÂ Integrated Rate Equations Cheat SheetÂ by clicking on the button below

Zero Order Reaction

So we already know that in a zero order reaction, the rate is independent of the concentration of the reactants. Thus, it means the sum of the powers of the concentrations is zero. It can only be zero when the all the powers are zero. Consider a reaction, RÂ â†’ P. Therefore, the rate law of this reaction is,

RateÂ $$\propto$$ Â [R]0

âˆ´ Rate = -d[R]/dt = k[R]0Â = kÂ Ã— 1

âˆ´ Rate =Â -d[R]/dt = k

âˆ´ d[R] = -kdt

Integrating both sides, [R] = -kt + I……………….(I)

where I is the constant if integration. At t = 0, the concentration of the reactant R = [R]0. Where [R]0 is the initial concentration of the reactant. Substituting this value in the equation I,

[R]0 = -kÂ Ã— 0 + IÂ = I

Substituting this value of I in the equation (I), we get

[R] = -kt +Â [R]0Â ………………(II)

Comparing equation II with the equation of a straight line y= mx +c,Â if we plot [R] against t, we get a straight line with slope = -k and intercept =Â [R]0Â

Therefore, on simplifying equationÂ II, we get

k ={[R]0 – [R]}/t……………..(III)

Example of Zero Order Reaction

Zero-order reactions are very uncommon but they occur under certain condition. An example of aÂ zero-order reaction is decomposition of ammonia,Â 2NH3Â â†’ N2 + 3H2Â

Rate = k[NH3]0 = k

First Order Reaction

In this type of reaction, the sum of the powers of concentrations of reactants in rate law is equal to 1, that is the rate of the reaction is proportional to the first power of the concentration of the reactant. Consider the reaction RÂ â†’ P again. Therefore, the rate law for this reaction is,

RateÂ $$\propto$$ [R]

We know thatÂ [R] = -kt +Â [R]0Â ( from equation II). Taking log of both sides, we get

ln[R] = -kt + ln[R]0Â ………………………….(IV)

âˆ´ ln[R]/[R]0Â = -ktÂ …………………………….(V)

âˆ´ k = (1/t) lnÂ [R]0Â /[R]Â ………………………(VI)

Now, consider equation II again. At time t1 and time t2, the equation II will be [R]1 = -kt1 +Â [R]0Â and [R]2 = -kt2 +Â [R]0Â respectively, where [R]1Â andÂ  [R]2 are concentrations of the reactants at time t1 and t2Â respectively. Subtracting second equation from first one, we get

ln [R]1– ln[R]2 = -kt1 – (- kt2 )

âˆ´ ln[R]1 /[R]2Â = k (t2 – t1)

âˆ´ k = [1/(t2 – t1)] ln[R]1 /[R]2

Now, taking antilog of both sides of equation V, we get [R] = [R]0e-kt

Comparing this equation with equation of a straight line y = mx + c, if we plot ln [R] against t, we get a straight line with slope = -k and intercept = ln[R]0

(Source:Â nonsibihighschool.org)

On removing natural logarithm from equation VI, the first-order reaction can also be written as,

k = 2.303/t log[R]0Â /[R]Â …………..(VII)

If we plot a graph ofÂ log[R]0Â /[R] against t, we get slope = k/2.303

Example of First Order Reaction

An example of aÂ first-order reaction is the hydrogenation ofÂ ethene.

C2H4 + H2Â â†’ C2H6

Therefore the rate of reaction for the above is k [C2H4]. Hence, equations III and VII are the equations of rate constants of zero and first order reactions respectively. We can find rate constants, initial and final concentrations and the time taken for the reaction to occur using these reactions.

A Solved Question for You

Q: The order of a reaction is:

a) can never be zeroÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â b) can never be a fraction

c) must be a whole number onlyÂ  Â  Â  Â  Â  Â  Â d) can be a whole number or a fraction or zero

Solution: d) can be a whole number or a fraction or zero. It depends on the dependency of the rate of reaction on the reactants. If the rate is independent of the reactants, then the order of reaction is zero. Therefore, the rate law of a zero order reaction would be rateÂ Î± [R]0Â whereÂ [R] is the concentration of the reactant.

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