Ammonia equilibrium constant

Essential for synthesis considerations is the abiUty to determine the amount of ammonia present ia an equiUbrium mixture at various temperatures and pressures. ReHable data on equiUbrium mixtures for pressures ranging from 1,000 to 101,000 kPa (10 —1000 atm) were developed early on (6—8) and resulted ia the determination of the reaction equiUbrium constant (9). Experimental data iadicates that is dependent not only on temperature and pressure, but also upon the ratio of hydrogen and nitrogen present. Table 3 fists values for the ammonia equilibrium concentration calculated for a feed usiag a 3 1 hydrogen to nitrogen ratio and either 0 or 10% iaerts (10). 

The production of ammonia is of historical interest because it represents the first important application of thermodynamics to an industrial process. Considering the synthesis reaction of ammonia from its elements, the calculated reaction heat (AH) and free energy change (AG) at room temperature are approximately -46 and -16.5 KJ/mol, respectively. Although the calculated equilibrium constant = 3.6 X 108 at room temperature is substantially high, no reaction occurs under these conditions, and the rate is practically zero. The ammonia synthesis reaction could be represented as follows ... 

The values of the equilibrium constant K listed in Table A are those obtained from data at low pressures, where the gases behave ideally. At higher pressures the mole percent of ammonia observed is generally larger than the calculated value. For example, at 400°C and 300 atm, the observed mole percent of NH3 is 47 the calculated value is only 41. 

Consider the statement The equilibrium constant for a mixture of hydrogen, nitrogen, and ammonia is 3.41. What information is missing from this statement ... 

The equilibrium constant for the decomposition at a certain temperature is 25. Calculate the partial pressures of all the gases at equilibrium if ammonia with a pressure of 1.00 atm is sealed in a 3.0-L flask. 

The equilibrium constant for the solubility reaction is readily calculated. Consider, for example, the reaction by which zinc hydroxide dissolves in ammonia. Again, imagine that the reaction occurs in two steps ... 

Write the equilibrium constant for the ammonia synthesis reaction, reaction C. 

Self-Test 9.7A The equilibrium constant for the ammonia synthesis (reaction C) is K = 41 at 127°C. What is the value of Kc at that temperature ... 

FIGURE 9.10 These graphs show the changes in composition that can be expected when additional hydrogen and then ammonia are added to an equilibrium mixture of nitrogen, hydrogen, and ammonia. Note that the addition of hydrogen results in the formation of ammonia, whereas the addition of ammonia results in the decomposition of some of the added ammonia as reactants are formed. In each case, the mixture settles into a composition in accord with the equilibrium constant of the reaction. 

STRATEGY The synthesis of ammonia is exothermic, and so we expect the equilibrium constant to be smaller at the higher temperature. To use the van t Hoff equation, we need... 

We can also write an equilibrium constant for the proton transfer equilibrium of a base in water. For aqueous ammonia, for instance,... 

As mentioned after Equation (10), the equilibrium constant may be expressed when the reactants are in several phases. As an example, the equilibrium between ammonia in a large cloud droplet and in the gas phase, NH3(aq) and NH3(g), is described by the equilibrium constant expression... 

Here cnh, is the concentration of undissociated ammonia in water. The equilibrium constant for this class of equilibria is often defined in terms of a Henry s Law constant, Xh-... 

For a reaction at equilibrium, the rate of the forward reaction is balanced exactly by the rate of the reverse reaction. For this reason, any equilibrium reaction can be written in either direction. The equilibrium constant for the Flaber synthesis of ammonia, for example, can be expressed in two ways ... 

The direction chosen for the equilibrium reaction Is determined by convenience. A scientist interested in producing ammonia from N2 and H2 would use f. On the other hand, someone studying the decomposition of ammonia on a metal surface would use eq,r Either choice works as long as the products of the net reaction appear in the numerator of the equilibrium constant expression and the reactants appear in the denominator. Example applies this reasoning to the iodine-triiodide reaction. 

Ammonia is an example of a weak base. A weak base generates hydroxide ions by accepting protons from water but reaches equilibrium when only a fraction of its molecules have done so. The equilibrium constant for this type... 

Salts that contain cations of weak bases are acidic. For example, the ammonium cation Is the conjugate acid of ammonia. When ammonium salts dissolve in water, NH4 ions transfer protons to H2 O molecules, generating H3 O and making the solution slightly acidic NH4" ((2 q) + H2 0(/) NH3(c2 q) + H3 O (a q) The equilibrium constant for this reaction can be calculated from Equation and for ammonia (Example ) ... 

In principle, the calculation of concentrations of species of a complexation equilibrium is no different from any other calculation involving equilibrium constant expressions. In practice, we have to consider multiple equilibria whenever a complex is present. This is because each ligand associates with the complex in a separate process with its own equilibrium expression. For instance, the silver-ammonia equilibrium is composed of two steps ... 

Because the two equilibrium constant expressions have similar magnitudes, a solution of the silver-ammonia complex generally has a significant concentration of each of the species that participate in the equilibria. The details of such calculations are beyond the scope of general chemistry. When the solution contains a large excess of ligand, however, each step in the complexation process proceeds nearly to completion. Under these conditions we can apply the standard seven-step approach to a single expression that describes the formation reaction of the complete complex. 

Although still less than 1, this equilibrium constant is large enough to allow solid AgCl to dissolve in strong aqueous ammonia, hi fact, if you open a bottle of commercial polish, you will surely notice the sharp odor of ammonia. These polishes contain a variety of ligands, including ammonia. 

Please note that the correction corresponds to introducing a corrected equilibrium constant K(T)/X . The correction has hardly any influence on the mole fraction of ammonia in the mixture at low pressures, but for ptot = 100 bar and higher the correction becomes significant. The results are presented in the last column of Tab. 2.4 and in Fig. 2.1. It should be noted that correction procedures exist for cases where the mixture does not behave ideally, but this goes beyond the scope of the present treatment. 

We will list the elementary steps and decide which is rate-limiting and which are in quasi-equilibrium. For ammonia synthesis a consensus exists that the dissociation of N2 is the rate-limiting step, and we shall make this assumption here. With quasi-equilibrium steps the differential equation, together with equilibrium condition, leads to an expression for the coverage of species involved in terms of the partial pressures of reactants, equilibrium constants and the coverage of other intermediates. 

The rate of the ammonia production can now be predicted if we can estimate all of the participating equilibrium constants and k. Where possible, one should take experimental values for the different constants. For instance, it is possible to measure the uptake of atomic nitrogen on the Fe or Ru surface and thereby determine... 

For ammonia synthesis, we still need to determine the coverages of the intermediates and the fraction of unoccupied sites. This requires a detailed knowledge of the individual equilibrium constants. Again, some of these may be accessible via experiments, while the others will have to be determined from their respective partition functions. In doing so, several partition functions will again cancel in the expressions for the coverage of intermediates. 

Acetylene is sufficiently acidic to allow application of the gas-phase proton transfer equilibrium method described in equation l7. For ethylene, the equilibrium constant was determined from the kinetics of reaction in both directions with NH2-8. Since the acidity of ammonia is known accurately, that of ethylene can be determined. This method actually gives A f/ acid at the temperature of the measurement. Use of known entropies allows the calculation of A//ac d from AG = AH — TAS. The value of A//acij found for ethylene is 409.4 0.6 kcal mol 1. But hydrocarbons in general, and ethylene in particular, are so weakly acidic that such equilibria are generally not observable. From net proton transfers that are observed it is possible sometimes to put limits on the acidity range. Thus, ethylene is not deprotonated by hydroxide ion whereas allene and propene are9 consequently, ethylene is less acidic than water and allene and propene (undoubtedly the allylic proton) are more acidic. Unfortunately, the acidity of no other alkene is known as precisely as that of ethylene. 

Few inorganic ligands form stable complexes with the beryllium ion in aqueous solution. This is a reflection of the fact that on the one hand Be2+ shows a strong preference for oxygen donor ligands such as water and the hydroxide ion, and on the other hand reacts with the more basic ligands such as ammonia to give the insoluble hydroxide. Reported equilibrium constants are in Table V. 

The problem gives us [N2] = 2.00 M and [H2] = 3.00 M. The next step is to enter the given values into the equilibrium constant expression and rearrange the expression to isolate the ammonia. Some people find it easier to reverse these steps. 

Other types of equilibria can be treated in much the same way as the ones discussed above. For example, there is an equilibrium constant associated with the formation of complex ions. This equilibrium constant is called the formation constant, Kf. Zn(H20)4+ reacts with ammonia to form the Zn(NH3)2+ complex ion according to the following equation ... 

A. System NH3 H S-H20. The dissociation of water (re-action 9) and the second dissociation of H2S (reaction 6) are neglected at given temperature and total molalities of NHo and H2S there remain four unknown molalities in the liquid phase (e.g. NH3, NH4+, H2S and HS ), the composition of the vapor phase and the total pressure, which are calculated from 8 equations The dissociation constants of ammonia and hydrogen sulfide (eqs.I and III) together with the phase equilibrium for hydrogen sulfide (eq. XII) are combined resulting in a equilibrium constant K 2... 

With an aqueous fluid phase of high ionic strength, the problem of obtaining activity coefficients may be circumvented simply by using apparent equilibrium constants expressed in terms of concentrations. This procedure is recommended for hydro-metallurgical systems in which complexation reactions are important, e.g., in ammonia, chloride, or sulfate solutions. 

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