# Second Order & Pseudo-First Order Reactions

There are two general types of second-order reactions depending on the character of the reacting species. The first type involves two different species ($$A$$ and $$B$$) reacting:

$$A +B \xrightarrow{k_2} P$$

with $$k_2$$ as the second-order rate constant. For this irreversible reactions, the rate (velocity) of the reaction, $$v$$, is directly proportional to $$[A]$$ and $$[B]$$:

$$v = \dfrac{d[A]}{dt} = \dfrac{d[B]}{dt} = -k_2[A][B] \tag{3a}$$

For the second type of second order reaction involves two identical species ($$A$$ and $$A$$) reacting:

$$A +A \xrightarrow{k_2} P$$  or  $$2A \xrightarrow{k_2} P$$

also with $$k_2$$ as the second-order rate constant.

For this irreversible reactions, the rate (also known as velocity) of the reaction, ($$v$$), is directly proportional to $$[A]$$ and $$[B]$$:

$$v = \dfrac{d[A]}{dt} = -k_2[A][A] = -k_2[A]^2 \tag{3b}$$

#### Case 1: Pseudo-First Order Kinetics

The first case involves on of the two reactant having a signifncatly higher concentration than the other, e.g.,

$[B] \gg [A]$

under these conditions, $$[B]$$ essentially never changes during the reaction (i.e., relatively), so equation (3a) becomes

$$v = -k_2[A][B] \approx -(k_2[B]) [A] = -k_1' [A] \tag{4a}$$

where $$k_1'$$ is the pseudo first order rate constant (= $$k_2[B]$$ ) for the reaction. The reaction appears to be first order, depending only on $$[A]$$.

#### Case 2: Second Order Kinetics

The only reactant is $$A$$ which must collide with another $$A$$ to form $$P$$, as illustrated in equation 3b. The derivations below apply to this special case. The following differential equation can be written and solved to find [A] as a function of t.

$$v = \dfrac{d[A]}{dt} = +2\dfrac{d[P]}{dT} = -k_2[A][A] = - k_2[A]^2 \tag{5}$$

$\dfrac{dA}{dt}=-k_2A^2$

$\int_{A_o}^A \dfrac{dA}{A^2} = \int_{A_o}^A A^{-2}\,dA = -k_2 \int_0^t dt$

$\left.\dfrac{A^{n+1}}{n+1} \right|_{A_o}^A = \left.\dfrac{A^{-1}}{-1}\right|_{A_o}^A =\left. -k_2\,t \right|_0^t$

$-\dfrac{1}{A}-\left(-\dfrac{1}{A_0}\right)= -k_2\,t$

$\dfrac{1}{A} = \dfrac{1}{A_o}+k_2t \tag{6}$

The following graphs show plots of $$A$$ vs. $$t$$ and $$1/A$$ vs. $$t$$ for data from a second order process.

Figure: Second Order Reaction: $$A + A \rightarrow P$$

Note that just from a plot of $$A$$ vs. $$t$$, it would be difficult to distinguish a first from second order reaction. If the plots were superimposed, you would observe that at the same concentration of $$A$$ (10 for example as in the linked plots), the $$v_o$$ of a first order reaction would be proportional to 10 but for a second order reaction to 102 or 100. Therefore, the second order reaction is faster (assuming similarity in the relative magnitude of the rate constants) as indicated by the steeper negative slope of the curve. However, at low $$A$$ (0.1 example), the $$v_o$$ of a first order reaction would be proportional to 0.1 but for a second order reaction to 0.12 or 0.01. Therefore, the second order reaction is slower.