Atom balance

The mole balances took into account glucose, ethanol in the liquid and the gaseous phase, acetic acid, CO2, biomass and NH3. 

Over the whole experience, the recoveries presented in Table 18 for carbon, nitrogen and available electrons were 1.003, 0.982 and 0.99, respectively. The loss of carbon by ethanol stripping amounted to 1.5% of the carbon supplied by glucose. Some error in the nitrogen recovery may be due to a changing biomass composition with the dilution rate whereas a constant nitrogen content was considered. 

Although the recoveries are close to one, the balances were individually checked with the y test. The test function h listed in Table 18 was calculated for a relative error on the measurement of 3% and must be compared with a threshold value of 2.71 for 1 degree of freedom and a significance level of 90%. 

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There are two forms of reacted H2S in the solution HS and. By recalling that uimi is the total concentration of chemically combined H2S in the aqueous caustic soda and conducting an atomic balance on S over Eqs. (8.15) and (8.16), we get... 

Finally, simplify the appearance of the equation by canceling species that appear on both sides of the arrow and check to make sure that charges as well as numbers of atoms balance. In some cases it is possible to simplify the half-reactions before they are combined. 

A Unbalanced reaction Balance 0 atoms Balance Ca atoms Balance S atoms Balance Hg atoms Self Check ... 

Unbalanced reaction Balance H atoms Balance O atoms Balance N atoms Multiply by 2 (whole ) Self Check ... 

Furthermore, by virtue of atomic balance the total number (q) of H atoms must sum to... 

The total number of atoms on one side of the equation must balance the total number of atoms on the other side. This rule is simply an expression of the well-known chemical fact that atoms are neither created nor destroyed during a chemical reaction. Remember that subscripts and superscripts are labels describing charges and sites and are not counted in evaluating the atom balance. 

Step 3 Look for elements that appear only once on each side of the equation but in unequal numbers of atoms. Balance these elements. 

Make sure that all atoms balance and the coefficients are in the lowest whole number ratio. 

Check to See If All Atoms Balance and That the Total Charge on Each Side Is the Same... 

Turn your attention to the O and H atoms. Balance the half-reactions separately, using H O to add O atoms and using H+ to add H atoms. 

In order to secure a continuous quantitative analysis of the outlet gas mixture, including all the species necessary for the evaluation of the nitrogen atomic balance, namely NH3, NO, N02 (reactants), N2 and N20 (products), the gases exiting the reactor were analyzed both by a Mass Spectrometer (MS) (Balzers QMS200) and by a UV analyzer (ABB Limas 11HW) in a parallel... 

The two methods give the same result. Check that the atoms balance—2 N, 3 S, 6 O, and 8 H on each side—and that the charges balance—0 net charge on each side. 

In a solid or liquid at room pressure and temperature, the attractive forces between atoms balance those of thermal agitation and repulsion. As the external pressure on a... 

Continuing to balance, the equation, how could you make the carbon atoms balance (By placing a 2 in front of the C02.) At this point, the equation looks like this ... 

When the number of independent chemical reactions equals C - p, where p is the rank of the atom matrix (mjk), Gibbs free energy is minimized subject to atom balance constraints ... 

These are described in the next section. Note that when atom balances are used, Dluzniewski and Adler (17) show that "fictitious elements" prevent reaction. Consider a reactor that produces ethylbenzene by reaction of benzene and ethylchloride in the presence of AICI3 catalyst. For calculation of phase equilibrium, downstream of the reactor, fictitious element A replaces a hydrogen atom in benzene (C0H5A) and the moles of each species remain unchanged. 

For these systems, mass balances must account for these restrictions and replace the atom balances (2) as constraints for minimization of Gibbs free energy. These mass balances are given by Schott (37) ... 

Next, we must consider the time-dependence of the oxygen atom balance at the interface xt = L. Because, by hypothesis, there can be no oxygen transport through any individual layer, the oxygen anions consumed at Xj = Li by the growth rate (dL,/df)+ of layer i must be derived entirely from the equivalent anion current (1 /.R<° y)) (dLI + 1/dt)- for i = 2, 3,. . . , JV — 1 obtained from the decomposition of layer t + 1 at xt = Lj. The volume of oxide i2/°xy) associated with each oxygen atom in layer i is given by... 

The oxygen atom balance then leads to the requirement... 

Next we must develop growth equations for the two layers L1 and LN. The layer LN in contact with the oxygen requires decomposition of layer to obtain the necessary metal atoms, so the condition given by eqn. (299) imposed by the metal atom balance still holds. Likewise, eqns. (301)—(303) still hold. However, no decomposition of layer N is required to form a layer N + 1 and, in this respect, layer N differs from the inner layers. Nevertheless, there could possibly be oxide evaporation, in which case... 

From a simulation viewpoint units SO, S6 and S7 may be considered blackboxes. On the contrary, SI to S5 are simulated by rigorous distillation columns, as sieve trays. In the steady state all the reactors can be described by a stoichiometric approach, but kinetic models are useful for Rl, R2 and R4 in dynamic simulation [7, 8]. As shown before, the reaction network should be formulated so as to use a minimum of representative chemical species, but respecting the atomic balance. This approach is necessary because yield reactors can misrepresent the process. 

Step 3 Balance any polyatomic ions, such as sulfate, SCU2-, that occur on both sides of the chemical equation as an ion unit. That is, do not split a sulfate ion into 1 sulfur atom and 4 oxygen atoms. Balance this ion as one unit. 

This net ionic equation is balanced only with regard to the numbers of zinc and silver atoms. Because the charge is not balanced, however, the equation is not balanced. We can balance it by doubling the charge on the left side (with a 2 before the Ag ) and keeping the number of silver atoms balanced (with a 2 before the Ag) ... 

Material balances in the experiments in Type I reactors were excellent, typically 100.0% about 0.3%. There was no evidence of the relatively major carbon formation or the excess hydrogen formation which have characterized small metal-walled laboratory reactors (4,5). With the Type II reactors, Cr,+ yields were calculated to give 100% hydrogen and carbon atom balance. 

The application of NullSpace yields the correct stoichiometric number matrix. Thus the calculation from the standpoint of the further transformed Gibbs energy G yields the expected number of constraints introduced by the enzyme mechanism. This reaction is a dramatic example of the difference between chemical reactions and some enzyme-catalyzed reactions. It is the enzymatic mechanism that introduces the three constraints in addition to atom balances. 

In addition, the carbon atom of the nitrile function of III must proceed from the starting lactone in order to maintain the atomic balance. This premise proves to be the starting point of the entire mechanistic sequence if one realizes that the azido group can accommodate those electrons released by the fragmentation suggested previously. In that case, molecular nitrogen would be libe ited as shown in IV, the consequence of which would be zwitterion V (see Scht-.ie... 

You could solve for lij and hi either directly from the two atomic balances or by using the three molecular species balances in conjunction with the stoichiometric equation for the reaction. 

Referring to the flowchart, we see that a balance on atomic carbon involves only one unknown (nco) and a balance on atomic hydrogen also involves one unknown (hh o). but a balance on atomic oxygen involves three unknowns. We will therefore write the C and H balances first, and then the O balance to determine the remaining unknown variable, no. All atomic balances have the form input = output. We will just determine the component amounts calculation of the mole fractions then follows as in the previous part. 

When analyzing subsystems in which reactions occur (the overall system and the reactor), we will count atomic balances for nonreactive subsystems (the recycle mixing point and the separation unit) we will count molecular species balances. 

Overall system (the outer dashed box on the flowchart). 3 unknown variables (ng, n-], n ) - 2 independent atomic balances (C and H) — 1 additional relation (95 % overall propane conversion) => 0 degrees of freedom. We will therefore be able to determine nj, and ng by analyzing the overall system. Let us count these three variables as known at this point. 

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