Ammonia synthesis equilibrium constant

Temperature changes affect not only systems at equilibrium but also the value of equilibrium constants. In fact, equilibrium constants changing with temperature is the reason that equilibria change with temperature. For example, consider Kgq for the ammonia synthesis equilibrium. 

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 ... 

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... 

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 ... 

In our calculation we assume that the gas mixture approaches equilibrium under conditions where the pressure is constant. This situation corresponds, for instance, to a volume of gas moving through a plug flow reactor with a negligible pressure drop. (Note that if the ammonia synthesis were carried out in a closed system, the pressure would decrease with increasing conversion.)... 

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. 

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. 

If the gas mixture is considered to be an ideal gas mixture then all fugacity coefficients are 1 and since K is a constant, the effect of increasing pressure is an increase of the equilibrium mole fraction of ammonia and a decrease of the mole fractions of nitrogen and hydrogen. However, since the ammonia synthesis is a high pressure process the gas mixture is not an ideal gas and the fugacity coefficients have to be taken into account. 

By contrast, a change in temperature nearly always changes the value of the equilibrium constant. For the Haber synthesis of ammonia, which is an exothermic reaction, the equilibrium constant Kc decreases by a factor of 1011 over the temperature range 300-1000 K (Figure 13.12). 

The equilibrium constant K is independent of pressure with standard states. The effect of the pressure is shown in Equation 6-6. Ky is usually insensitive and may either increase or decrease slightly with pressure. When (r + s) > (a + b), the stoichiometric coefficients, an increase in pressure P results in a decrease in conversion of the reactants to the products (i.e., A + B <-> R + S). Alternatively, when (r + s) < (a + b), an increase in pressure P results in an increase in the equilibrium conversion. In ammonia synthesis (N2 + 3H2 <-> 2NH3), the reaction results in a decrease in the number of moles. Therefore, an increase in pressure causes an increase in equilibrium conversion due to this factor. 

The equilibrium constant Keq for ammonia synthesis is expressed as a function of the partial pressure as... 

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. 

Table 6-5 shows the conditions for which NH3 production is possible. Both low temperatures or very high pressures achieve favorable equilibrium. At 25°C, the equilibrium constant is very high, while at higher temperatures, both Keq and PNH3 decrease rapidly. Generally, ammonia synthesis reactors operate at about 350°C and 200 atm with an equilibrium conversion of about 70% in each pass. The NH3 is separated from unreacted H2 and N2, which are recycled back to the reactor. For the overall process involving the tubular reactor, separation and recycle produce about 100% ammonia conversion. 

For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 X 10-2 L2/mol2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases. 

At 450°C, Kp = 6.5 X 10-3 atm-2 for the ammonia synthesis reaction. Assume that a reaction vessel with a movable piston initially contains 3.0 mol H2(g) and 1.0 mol N2(g). Make a plot to show how the partial pressure of NH3(g) present at equilibrium varies for the total pressures of 1.0 atm, 10.0 atm, 100. atm, and 1000. atm (assuming that Kp remains constant). (Note Assume these total pressures represent the initial total pressure of H2(g) plus N2(g), where PNHl = 0.)... 

In applying these equations, the authors assumed that the adsorption of nitrogen on the iron catalyst in the presence of an ammonia-hydrogen mixture is the same as it would be when at a nitrogen pressure equivalent to the existing partial pressure of ammonia and hydrogen in the gas mixture. Thus, since the equilibrium constant for ammonia synthesis is... 

When Keq is large, the numerator of the equilibrium constant expression is larger than the denominator. Thus, the concentrations of the products will usually be greater than those of the reactants. In other words, when a reaction that has a large K q reaches equilibrium, the system s contents may be mostly products. Reactions in which more products form than reactants form are said to be favorable. Look at the first entry in Table 2, where the reaction has a large K q value. The synthesis of ammonia is very favorable at 25°C. 

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