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### Course: Chemistry archive>Unit 15

Lesson 2: Relationship between reaction concentrations and time

# Second-order reaction (with calculus)

Deriving the integrated rate law for second order reactions using calculus. How you can graph second order rate data to see a linear relationship.

## Want to join the conversation?

• What's the highest order a reaction can be theoretically?
• I'm not sure that there is a theoretical limit to it, but there is definitely a real limit. Reactions of order 3 or above are very rare. In reactions it is necessary for the reacting components to meet in a particular orientation in one instant, and the probability of this occurring decreases as the number of molecules increases. Let me know if that helped :)
• How would you figure out which integrated rate equation (first, second, zero) to use based on the units of the rate constant?
• There is a pattern.
If n is the order of the reaction, the units of k are (mol·L⁻¹)^(1-n)s⁻¹.
If n = 0, the units are (mol·L⁻¹)^(1-0)= (mol·L⁻¹)¹s⁻¹ = mol·L⁻¹s⁻¹
If n = 1, the units are (mol·L⁻¹)^(1-1)= (mol·L⁻¹)⁰s⁻¹ = s⁻¹
If n = 2, the units are (mol·L⁻¹)^(1-2)= (mol·L⁻¹)⁻¹s⁻¹ = L·mol⁻¹s⁻¹
If n = 3, the units are (mol·L⁻¹)^(1-3)= (mol·L⁻¹)⁻²s⁻¹ = L²·mol⁻²s⁻¹

To go the other way, you reverse the process.
If the units of k are (mol·L⁻¹)^n, The order of the reaction is 1 - n.
If the units of k are mol·L⁻¹s⁻¹, n = 1. The order is 1 - 1 = 0.
If the units of k are s⁻¹, n = 0. The order is 1 - 0 = 1.
If the units of k_are L·mol⁻¹s⁻¹, n = -1. The order is 1 – (-1) = 2.
If the units of k are L²mol⁻²s⁻¹, n = -2. The order is 1 – (-2) = 3.
Then you use the integrated rate law that corresponds to the order of the reaction.
• What if the reaction is A+B->C, and is second order? (R=k[A][B]) Then how do you derive the integrated rate law?
• What is the law that he used to integrate [A]^-2*d[A] can anyone tell me that?that is what formula he used?
• I thought 2nd order is 2A ---> products. Im confused now about the difference between 1st and 2nd order.
(1 vote)
• The order respect to a reactant does not necessarily correspond to the coefficients of the reaction, this because a reaction can occur in multiple elementary reactions.
For example the reaction 2NO + Br2 ----> 2NOBr has the experimental rate law
Rate = k {NO} {Br2}
This can be explained considering that the reaction occurs in 2 steps:
NO + Br2 ----> NOBr2 ( The first reaction is the slow step, therefore it dictates the rate)
NOBr2 + NO -----> 2NOBr
-----------------------------------------
2NO + Br2 ----> 2NOBr ( The sum of the 2 reactions above gives the original reaction)
• Why did you not move the negative to the other side when integrating like you did in the beginning of the first order derivation video?
• My guess is because the initial concentration is in the denominator and so a larger value indicates a smaller denominator..hence... the concentration is getting lower as time continues
(1 vote)
• why a stright line for second order reaction ?
how could u indicate it ?
(1 vote)
• Any equation of the form y=mx plus b is essentially a straight line. (where m is the 'slope', that is how fast the y value changes with respect to the x value and b is the y intercept, i.e. the value of y when x is zero) why? figure it out because its so much fun doing it yourself...
• When do you know to use the integrated rate law?
(1 vote)
• whenever the question states concentration of a substance with respect to time, it is an integrated rate law.
In other words, what would decreasing the value of `1/[A]` mean about the reaction? And Is this result physically possible?