# 6.3: Kinetics without Enzymes

## Single Step Reactions

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. You learned two methods to analyze kinetic data:

• initial velocity method: In this method, the initial rate, v0, is measured as a function of the concentration of reactants. The initial velocity, v0, 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. Under these condition, [A] over this short range of time is approximately constant and equal to Ao. Initial rate graphs are often based on measurement of product increase with time (dP/dt) so v0 vs A plots have positive slopes. The velocity at time t along the A vs t curve, dA/dt, is constantly changing as [A] decreases since the velocity depend on the [A]. To reiterate, the initial velocity of the reaction is the slope of the initial linear part of the decay curve when the rate is essentially linear over a narrow range of A concentrations.
• integrated rate method: In this method a differential equation which gives the change of A or P with time (dA/dt or dP/dt) is integrated to give a rate equation giving the concentration of A or P as a function of time. .

#### First Order Reaction:

$\ce{A ->[k_{1}] P}$

where k1 is the first order rate constant. For these reactions, the velocity of the reaction, $$v$$, is directly proportional to [A], or

$v=-\frac{d A}{d t}=+\frac{d P}{d t}=k_{1} A \label{6.3.A.1}$

The negative sign in -d[A]/dt indicates that the concentration of A decreases. The equation could also be written as:$$v=\frac{d A}{d t}=-k_{1} A$$

For the rest of the reactions shown in this chapter, I will follow the convention of writing all velocities expressed as d[x]/dt as positive numbers. A negative sign for a term on the right hand side of the differential equation will indicate that the concentration dependency of that term will lead to a 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.

The differential Equation 6.3.A.1 can solved to find [A] as a function of t.

\begin{aligned} \int_{A_{O}}^{A} \frac{d A}{A}=-k_{1} \int_{0}^{t} d t & \\ \ln A-\ln A_{0}=&-k_{1} t \\ \ln A=\ln A_{0}-k_{1} t \end{aligned}

$A=A_{0} e^{-k_{1} t}$

Equation 6.3.A.2 is an example of an integrated rate equation. The following graphs show plots of A vs t and lnA vs t for data from a first order process. Note that the derivative of the graph of A vs t (dA/dt) is the velocity of the reaction. The graph of lnA vs t is linear with a slope of -k1. 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. That is [A] is approximately equal to [A0].

The figure below shows two ways to plot 1st order reaction data. The left graph shows the exponential decay of A with time and the corresponding rise in P when A0 is 0 and k1=2. The other shows the linear fall of ln[A] vs time.

 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, vo, 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 = Ao. 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 can be written showing the disappearance of A as v = -d[A]/dt = k1[A], while appearance of A as v = d[A]/dt = -k1[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 (= 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].

#### Second Order/Pseudo First Order Reactions

where $$k_2$$ is the second order rate constant. For the first of these irreversible reactions, the velocity of the reaction, v, is directly proportional to [A] and [B], or

$v= \dfrac{dA}{dt} = -k_2[A][B] \label{eq3}$

We will consider two special cases of this reaction type:

1. [B] >> [A]. Under these conditions, [B] never changes, so Equation \ref{3} becomes

4. v = -(k2[B]) [A] = -k1' [A]

where k1' is the pseudo first order rate constant (= k2[B] ) for the reaction. The reaction appears to be first order, depending only on [A].

1. The only reactant is A which must collide with another A to form P, as illustrated in the second reaction above. 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.

$v=\frac{d A}{d t}=2 \frac{d P}{d t}=-k_{2} A^{2}$

Solving the differential equation for A give the following:

$\begin{array}{c}{\frac{d A}{d t}=-k_{2} A^{2}} \\ {\int_{A_{0}}^{A} \frac{d A}{A^{2}}=\int_{A_{0}}^{A} A^{-2} d A=\int_{A_{0}}^{A} A^{-2} d A=-k_{2} \int_{0}^{t} d t} \\ {\left.\left.\left.\frac{A^{n+1}}{n+1}\right]_{A_{0}}^{A}=\frac{A^{-1}}{-1}\right]_{A_{0}}^{A}=-k_{2} t\right]_{A_{0}}^{A}} \\ {-\frac{1}{A}-\left(\frac{1}{A_{0}}\right)=-k_{2} t}\end{array}$ or

$\frac{1}{A}=\frac{1}{A_{0}}+k_{2} t$

The following graphs show plots of A vs t and 1/A vs t for a second order process when A0 is 0 and k2=1. The right graph shows the linear rise of 1/[A] with time.

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), the vo 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 vo of a first order reaction would be proportional to 0.1 but for a second order reaction to 0.12 or 0.01. Therefore, at low A, the second order reaction is slower.

The interactive graphs below show the first and second order conversion of reactant A to product. Change the sliders to see how the curves are different.

## Multi-Step Reactions

#### Reversible First Order Reactions

The following differential equation for this reaction scheme can be written and solved as shown below:

\begin{aligned} \mathrm{v}=& \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=-\mathrm{k}_{\mathrm{f}} \mathrm{A}+\mathrm{k}_{\mathrm{r}} \mathrm{P} \\ & \mathrm{A}=\frac{\mathrm{A}_{0}\left(\mathrm{k}_{\mathrm{r}}+\mathrm{k}_{\mathrm{f}}\left[\mathrm{e}^{-\left(\mathrm{k}_{\mathrm{f}}+\mathrm{k}_{\mathrm{r}}\right) \mathrm{t}}\right]\right.}{\mathrm{k}_{\mathrm{f}}+\mathrm{k}_{\mathrm{r}}} \\ \mathrm{P}=& \mathrm{A}_{0}-\left(\frac{\mathrm{A}_{0}\left(\mathrm{k}_{\mathrm{r}}+\mathrm{k}_{\mathrm{f}}\left[\mathrm{e}^{-\left(\mathrm{k}_{\mathrm{f}}+\mathrm{k}_{\mathrm{r}}\right) \mathrm{t}}\right]\right.}{\mathrm{k}_{\mathrm{f}}+\mathrm{k}_{\mathrm{r}}}\right) \end{aligned}

A graphs of A and P vs t for this reaction at two different sets of values of k1 and k2 are shown below.

Change the sliders on the interactive graph below to see how changing the relative values of the forward and reverse rate constants affects the concentrations at which the concentration plateaus are reached.

We all grew up on mathematical graphs that give you valuable insight into textual descriptions and data tables from which the graphs were made.  These graphs are enhanced when you can use sliders to change constants as in the case for the reversible reaction A ↔ P above.  Even then you might not infer that when the reaction has reached equilibrium, product is still being made from reactant and reactant from product since the equilibrium is dynamic.  To add insight into both simple and complex reactions, animations that show the continual disappearance of reactants and products are valuable.

This book will incorporate many animation to show the continue dynamic reactions. We will use reactants/products concentrations curves as a function of time, often called progress curves, to create the animation.  All of the animations in this book were made by Shraddha Nayak.

It is easy to write the equations (differential) to show how the rate of disappearance of a reactant A (for example), dA/dt, depends on concentration of the immediate participants in the reaction.  It is not so easy to derive the equations (as we did above) for the progress curve which shows how [A] changes with time t (i.e. [A] = f(t)).  Luckily, many free programs have been developed that produce numerical solutions to the differential equations and give progress curve graphs like [A] = f(t).  Two interrelated, freely available programs include Copasi and Vcell, can be used to hundred of cellular reactions simultaneously. They use a format called Systems Biology Markup Language (SBML) for describing and storing computational models. We will use Vcell in this book as it is very simple to use.  All of the coding to describe the reactions is built into this book and hidden from you.  All you will see are the output results, but you will also be given the opportunity to change constants (much as you do with the sliders on the graphs directly incorporated into this book) and see the results.

The table below shows the reaction diagram (left), graphical results  of the progress curve(middle), and animations for the reversible reaction of A ↔ P for the conditions shown below.  The reactant A and product are called species and are shown as green spheres.  The yellow square is a reaction node indicating a reaction connects A to P.  The lines connects the species that participate in the reaction.   The velocities (slope of the concentration vs time curve at any given time) are called fluxes, J, in Vcell and many other similar programs.  When we get to metabolism, we will talk about fluxes of metabolites through pathways.  Also fluxes are used to describe the rate of movement of solute through membranes.

 Rx:      k1 = 0.2,   k2 = 0,4,  At=0 = A0 = 10 This video is a placeholder until Shraddha's animations are avaiable. Case 1:  kf=2, kr=4 Case 2:  kf=4, kr=2 Vcell File: AtoP_Rev_Uncatalyzed

#### Consecutive First Order Reactions

The following differential equations can be written for these reactions:

\begin{aligned} \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}} &=-\mathrm{k}_{1} \mathrm{A} \\ \frac{\mathrm{d} \mathrm{B}}{\mathrm{dt}}=& \mathrm{k}_{1} \mathrm{A}-\mathrm{k}_{2} \mathrm{B} \\ \frac{\mathrm{d} \mathrm{C}}{\mathrm{dt}} &=\mathrm{k}_{2} \mathrm{B} \end{aligned}

Here are the solutions to the differential equations:

\begin{array}{c}{\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{k}_{1} \mathrm{t}}} \\ {\mathrm{B}=\frac{\mathrm{k}_{1} \mathrm{A}_{0}}{\mathrm{k}_{2}-\mathrm{k}_{1}}\left(\mathrm{e}^{-\mathrm{k}_{1} \mathrm{t}}-\mathrm{e}^{-\mathrm{k}_{2} \mathrm{t}}\right)} \\ {\mathrm{C}=\mathrm{A}_{0}-\mathrm{A}-\mathrm{B}=\mathrm{A}_{0}\left[1+\frac{1}{\mathrm{k}_{1}-\mathrm{k}_{2}}\left(\mathrm{k}_{2} \mathrm{e}^{-\mathrm{k}_{1} \mathrm{t}}-\mathrm{k}_{1} \mathrm{e}^{-\mathrm{k}_{2} \mathrm{t}}\right]\right.}\end{array}

Graphs of A, B, and C vs t for these reaction for a fixed value of k1 and k2 are shown below.

Change the sliders on the interactive graph below to see how changing the relative values of the forward and reverse rate constants affects the the curves.

You can imagine that the deriving the equations for the completely reversible reactions of A ↔ B ↔ C would be really difficult.  No problem!  We can simply set up the reactions in Vcell, choice the build in equation based on mass action, and let the program numerically solve the equations to give progress curve graphs.

The table below shows the reaction diagram (left), graphical results  of the progress curves (middle), and animations for both the irreversible  A→B→ C and reversible  A↔ B↔C reactions for the conditions shown.  Note that in the reaction diagram, the arrow goes in only one direction, left to right, simply showing that species are connected.  In the Vcell program, the equations for the reversible reaction were used for both cases, but for the irreversible reaction, the rate constants for the backwards reaction were set to 0.

 Rx:  Case 1: A → B → C     k1 = 0.2,   k2 = 0.6.  A0 = 1.   Case 2:  A ↔ B ↔ C    k1f = 0.2, k1r = 0.1,  k2f = 0.6, k2r = 0.3 A0 = 1 This video is a placeholder until Shraddha's animations are avaiable. Vcell File: Case 1:  AtoBtoC_NoEnz Case 2: AtoBtoC_RevNoEnz

VCell Activity

Run the following simulation for the fully reversible reaction of A ↔ B  ↔  C using these parameters:  k1f = 0.2, k1r = 0.1,  k2f = 0.6, k2r = 0.3 A0 = 1