K-diagram

Figure 21.16 k diagram of the mechanism by which retinoblastoma protein (Rb) regulates transcription factor activity. The rb protein binds to the transcription factor, which forms a complex in which the transcription factor for three genes is inactive. Phosphorylation of Rb by a cell division cycle kinase results in dissociation of transcription factor from the complex and hence activation. 

So what a chemist draws as a band in 5, repeated at left in 8 (and the chemist tires and draws 35 lines or just a block instead of Avogadro s number), the physicist will alternatively draw as an E(k) vs. k diagram at right. Recall that k is quantized, and there is a finite but large number of levels in the diagram at right. The reason it looks continuous is that this is a fine dot matrix printer there are Avogadro s number of points jammed in there, and so it s no wonder we see a line. 

Build the E(k) vs. k diagram of the Bloch functions associated with each of the unit cell MOs (or AOs). The shape of these bands is pseudo-sinusoidal/cosinusoidal. 

This may be understood more easily from Fig. 109(a) [185]. Here, we imagine a perturbation that alters the position of the bands with respect to each other vertically on the E vs. k diagram (i.e. in the Brillouin zone). Clearly, such a perturbation will alter the energy of the optical absorption onset and large differential signal will be seen at threshold. The effect is to... 

Figure 4.9. TPD-MS study of the Hz desorption from two 0.5% Rh/CeOz catalysts prepared by impregnation of a low surface area ceria sample (3.5 m g ) from an aqueous solution of either Rh(NOj>3 (N) or RhCh (Cl). The catalysts were reduced with flowing Hz at 573 K, heated in a flow of He at 873 K, and finally treated with Hz at 298 K 373 K (b), 473 K (c), or 573 K . Diagrams taken from (166). Experimental TPD conditions amount of sample 350 mg He flow rate 40 cm .min Heating rate 30 K.min. ...

FIGURE 28-3 Bkx k diagram showing components of a typical apparatus lor HPLC. (Courtesy of Pefkin-Elmer Corp., Norwalk, CT). 

The graph of two concentrations versus each other is called a concentration-diagram [19], which is abbreviated in the following by the term K-diagram. 

One can calculate the initial slope in a K-diagram by use of the system of differential equations according to... 

Accordingly, the initial slopes of a K-diagram can be calculated for any combination of reactants. 

The final slopes are given in Table 2.7 for a variety of cases. In the first column the concentrations are listed which are graphed versus each other to form the K-diagram The next three columns repre-... 

In the K-diagram some equations can be set up for the areas below the concentration functions. In the case of a simple consecutive reaction... 

These points of inflection ate typical for the K-diagram. They are not found in concentration-time diagrams. 

It is difficult to define general rules. However, one is able to simulate K-diagrams using a desk-top computer for assumed mechanisms and to analyse the curves for points of inflection. The principle approach was discussed in Section 2.3.2. Points of inflection are allowed in K- or X-diagrams only, if more than two linear independent steps of reaction s > 2) take place. 

The K-diagram is usually limited to a triangular area below the straight line... 

By use of the time equations, the X- and K-diagrams of the system are given in parametric form. By an elimination of time one obtains... 

Eq. (2.89) gives X- and K-diagrams in parametric form. The elimination of t using... 

This relationship can be graphically checked. One plots the concentration of one component versus the concentrations of all the other components in pairs in a rectangular coordinate system. If linear concentration diagrams (K-diagrams) result, the reaction is uniform. Additionally one obtains informations on the stoichiometric coefficients, the equilibrium constants, and constants of competitive reactions (if the reaction contains competitive steps). 

Example 5-2 K-diagrams for a single reaction step For the reaction... 

The use of the K-diagrams (see Fig. 5.3) for the dependence of a b) allows the determination of K (a is the ordinate, b is the abscissa). Furthermore, the competitive constant can be found for the K-graph d versus c. The latter relationship can be used for control. In this figure the concentrations are added in relative units at the axes. 

Even if 5 > 1, linear K-diagrams may result for the concentrations of some components. This can happen if the columns of these components depend on each other in the scheme in Section 2.1.1.1. Thus for the reaction... 

K-diagrams do not depend on the intensity of the irradiation source, if the quantum yields of both the partial reactions depend in the same way on the amount of light absorbed. However, if there is a difference in the dependences of the quantum yields of the two partial reactions on the light intensity, or if linear independent photochemical and thermal reactions compete with each other, then the K-diagrams will depend on /q. 

Since /q varies with the wavelength of the irradiation source, the K-diagrams depend on the wavelength in those cases in which they depend on the intensity. Furthermore the K-diagram changes if more than one component absorbs the light and starts the photoreaction by this means. 

The partial quantum yields do not depend on irradiation intensity according to eq. (3.29). For this reason the same situation is valid for K-diagrams. The form of the K-diagrams is determined by eq. (3.40). The related constant x contains the ratio k, which varies with wavelength. For this reason the K-diagrams have to depend on the wavelength of the irradiation source. 

K-diagrams lose linearity, if 5 > 1. If 5 = 2 Oust 2 linear independent steps of the reaction form the mechanism), the concentration of each component A, is a linear function of two suitable chosen components A, and A,-. For this reason... 

If a reaction yields non-linear K-diagrams, then KDQ-diagrams allow the cases s = 2 and s>2 to be distinguished. If KDQ-diagrams are either linear or degenerate for all combinations of components, then the number of partial reactions amounts to = 2. If one finds at least one KDQ-diagram non-linear and non-degenerate, then the number of independent partial reactions is s>2. 

---