Ammonia equilibrium reaction

The overall strategy for this calculation is to replace the partial pressures that appear in K by the molar concentrations, and thereby generate Kc. We need to keep track of the units so we write activities as Pj/bar and molar concentrations as Q]/(mol-L 1), as explained in the earlier side-notes. We consider a specific case the relation between K and Kc for the ammonia equilibrium, reaction C. 

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 reaction is an equilibrium reaction that is exothermic. Lower temperatures favor the production of ammonia. High pressures in... 

The compound also results from the reversible equilibrium reaction of sulfur with anhydrous liquid ammonia ... 

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 ... 

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 ... 

To express the relative strengths of an acid and its conjugate base (a conjugate acid-base pair ), we consider the special case of the ammonia proton transfer equilibrium, reaction C, for which the basicity constant was given earlier (Kb = [NH4+l[OH ]/ NH3]). Now let s consider the proton transfer equilibrium of ammonia s conjugate acid, NH4+, in water ... 

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. 

In an ammonia-based solution, the solution pH can vary between 7 (for an extremely diluted NH3 solution) and 11.5 (for an NH3 concentration of 1.2M).36 The equilibrium reactions suggest that, in an ammonia solution, a Zn2+ ion in the solution will stay in the form of Zn(OH)2 along with Zn (NH3)2, whereas a Cd2+ ion in an ammonia solution is predominantly Cd(NH3)4+. 

Equilibrium. Forward and reverse reactions occurring at the same rate, resulting in a concentration of reactants. A + B C + D. Ammonia synthesis is an equilibrium reaction (N2 + 3H2 2NH3). 

So far, only kinetic isotope effects have been considered, but isotopic fractionations associated with equilibrium exchange reactions have been demonstrated for the common inorganic nitrogen compounds (Letolle 1980). Of special importance in this respect is the ammonia volatilization reaction ... 

Ammonia acts as a base by stripping a proton from a water molecule, leaving an increased OH concentration. Notice in the equilibrium reaction that NH4 and NH3 are a conjugate acid-base pair, related by transferring a single proton. 

Nitric acid undergoes both wet and dry deposition rapidly and can be neutralized by ammonia, the major gaseous base found in the atmosphere. As discussed in Section E.2, the neutralization reaction is an equilibrium reaction so that by itself, this does not result in permanent removal from the atmosphere. However, as seen in this chapter and in Chapter 9, this acid-base reaction has some important implications for visibility in the atmosphere and for the nitrate concentrations found in respirable particles. 

The above are equilibrium reactions, and their successful exploitation requires that they be carried out under conditions in which the equilibrium favors the product. Specifically, this requires that the adsorbed species in Reactions (D)-(I) not be held so tightly on the catalyst surfaces as to inhibit the reaction. On the other hand, strong interaction between adsorbate and catalyst is important to break the bonds in the reactant species. Optimization involves finding a compromise between scission and residence time on the surface. Although we are especially interested in metal surfaces, those constituents known as promoters in catalyst mixtures are also important. It is known, for example, that the potassium in the catalyst used for the ammonia synthesis shifts Equilibrium (F) to the right and also increases the rate of Reaction (D) by lowering its activation energy from 12.5 kJ mole to about zero. 

The addition of ammonia and amines to unsaturated hydrocarbons, called hydroa-mination, is a desirable but difficult reaction. Only activated multiple bonds react readily in hydroamination to yield amines in an equilibrium reaction. It is necessary, therefore, to use catalysts in the transformation of nonactivated compounds. 

As example of the caculation of the equilibrium composition of a homogeneous gas reaction the ammonia synthesis reaction at 450 °C will be considered ... 

There is a lot of evidence that chemical equilibrium is dynamic. One example of this evidence is relevant to the ammonia equilibrium. Imagine carrying out two ammonia syntheses with the same starting conditions, but using D2 (deuterium) in place of H2 in one of them (Fig. 9.2). The two reaction mixtures reach equilibrium with almost exactly the same composition, except that D2 and ND3 are present in one of the systems instead of H2 and NH3. Suppose we now combine the two mixtures and... 

Consider the ammonia proton transfer equilibrium, reaction C, for which [NH4+][OH ] b INH,]... 

For example, let s write the equilibrium constant expression for the basic ionization of ammonia in water. The equation for the chemical equilibrium reaction is ... 

Both are equilibrium reactions. The formation reaction goes to virtual completion under usual reaction conditions, but the decomposition reaction is less complete. Unconverted carbon dioxide and ammonia, along with undecomposed carbamate, must be recovered and reused. 

Since the partial pressure is the mole fraction in the vapor phase multiplied by the total pressure, (i.e., p, = y, P), the equilibrium constant Keq is expressed as Keq = Ky PAn, where An = (2 - 1 - 3), the difference between the gaseous moles of the products and the reactants in the ammonia synthesis reaction. 

For ammonia (the most common weak base), the equilibrium reaction and base-dissociation constant are ... 

Ammonia, primary amines and many other compounds that contain an NH2 group as well as secondary amines add to many carhonyl compounds only to a certain extent that is, in equilibrium reactions (formation of hemiaminals formula B in Figure 9.23). This addition is almost always followed hy the elimination of an OH ion. As the result, one obtains an iminium ion (formula C in Figure 9.23). 

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