# Review of Elementary Reaction Kinetics

### Review of Elementary Reaction Kinetics

We will first explore the kinetics of non-catalyzed reactions as we did with our study of passive and facilitated diffusion. Before we do that, $$A$$ brief review of the two major types of kinetic equations that you studied in general chemistry is in order, using the first order reaction of reactant $$A$$ forming product $$P$$ with $$A$$ rate constant $$k_1$$:

• initial velocity equation ($$v_0$$ as $$A$$ function of the concentration of reactants): $$v_0 = k_1[A]$$. Initial rate graphs are usually based on measurement of product increase with time ($$\frac{dP}{dt}$$) so $$v_0$$ vs. $$A$$ plots have positive slopes. This equation is true for any velocity v, not just the initial velocity as v is always directly proportional to $$[A]$$ at any concentration of $$A$$. However it is easiest to measure the initial velocity of the reaction in the linear part of the decay curve when the rate is linear over $$A$$ narrow range of $$A$$ concentrations.

• integrated rate equation which gives concentration as $$A$$ function of time. $$A$$ differential equation $$v = \frac{d[A]}{dt} = -k_1[A]$$ that on integration leads to the integrated rate equation: $$A = A_0e^{-k_1t}$$ giving $$A$$ as $$A$$ function of time $$t$$.

#### First Order Reactions

Consider the first order reaction

$$A \xrightarrow{k_1} P$$

where $$k_1$$ is the first order rate constant. For these reactions, the velocity of the reaction, $$v$$, is directly proportional to $$[A]$$, or

$v = \dfrac{-d[A]}{dt} = \dfrac{+d[P]}{dT} = k_1[A] \tag{1a}$

The negative sign in -d[A]/dt indicates that the concentration of $$A$$ decreases. The equation could also be written as:

$v = \dfrac{-d[A]}{dt} = -k_1[A] \tag{1b}$

 For the rest of the reactions shown below, adopted convention here is to write all rates (velocities) as $$\frac{d[x]}{dt}$$ as positive numbers. $$A$$ negative sign for $$A$$ term on the right hand side of the differential equation (as in 1b) will indicate that the concentration dependency of that term will lead to an decrease in $$[x]$$ with time. Likewise $$A$$ positive sign for the term on the right hand side of the equation will indicate that concentration dependency of that term will lead to an increase in $$[x]$$ with time.

Using this nomenclature, the following separable differential equation can be written and solved to find $$[A]$$ as $$A$$ function of $$t$$.

$\dfrac{d[A]}{dt} = -k_1A$

$\int _{[A_o]}^{[A]} \dfrac{d[A]}{dt} = -k_1[A]$

evaluate this integral at the limits assuming $$t_0 = 0$$

$ln\; [A] - ln\; [A_o] = - k_1t - 0$

$ln\; [A] = ln\; [A_o] - k_1t$

or

$[A] = [A_o] e^{- k_1t} \tag{2}$

which is characteristic of first-order kinetics.

Equation 2 is an example of an integrated rate equation. The following graphs show plots of $$A$$ vs. $$t$$ and $$\ln\,A$$ vs. $$t$$ for data from $$A$$ first order process. Note that the derivative ($$\frac{dA}{dt}$$) of the graph of $$A$$ vs. $$t$$ is the velocity of the reaction. The graph of $$\ln\, A$$ vs. $$t$$ is linear with $$A$$ slope of $$-k_1$$.The velocity of the reaction (slope of the $$A$$ vs. $$t$$ curve) decreases with decreasing $$A$$, which is consistent with equation 1. Again, the initial velocity is determined from data taken in the first part of the decay curve when the rate is linear and little $$A$$ has reacted (i.e., $$[A] \approx [A_0]$$).

Figure: First Order Reaction kinetics: $$A \rightarrow P$$

Once again, for complete clarification, another way of analyzing the kinetics of $$A$$ reaction, in addition to following the concentration of $$A$$ reactant or product as $$A$$ function of time and fitting the data to an integrated rate equation, is to plot the initial velocity, $$v_o$$, of the reaction as $$A$$ function of concentration of reactants. The initial velocity is the initial slope of $$A$$ graph of the concentration of reactants or products as $$A$$ function of time, taken over $$A$$ range of times such that only $$A$$ small fraction of $$A$$ has reacted, so $$[A]$$ is approximately constant = $$A_o$$. From the first order graph of $$A$$ vs. $$t$$ above, the slope approaches 0 with increasing time as $$[A]$$ approaches 0, which clearly indicates that the reaction velocity depends on $$A$$. For $$A$$ first order process, two equivalent equations, 1a and 1b, can be written.

Equation 1a above is written showing the disappearance of $$A$$ as

$v = -\dfrac{d[A]}{dt} = k_1[A],$]

while Equation 1b above is written showing the appearance of $$A$$ as

$v = \dfrac{d[A]}{dt} = -k_1[A].$

Both equations shows that v is directly proportional to $$A$$. As $$[A]$$ is doubled, the initial velocity is doubled.

Velocity graphs used by biochemists often show the initial velocity of product formation (not reactant decrease) as $$A$$ function of reactant concentration. Hence, as the concentration of product is increasing, the slopes of initial velocity are positive. $$A$$ graph of $$v \, (= \frac{dP}{dt})$$ vs. $$[A]$$ for $$A$$ first order process would have $$A$$ positive slope and be interpreted as showing that the rate of appearance of $$P$$ depends linearly on $$[A]$$.