Equilibrium constant exothermic reaction

In a lower temperature range, the equihbrium constant Ky is larger, and the temperature has a marked effect on the reaction rate constant ki. However, as the temperature is gradually increased, the reversible exothermic reaction equilibrium constant Ky is decreased. The value in the square brackets in equation (8.5) is reduced, with the effect of temperature on reaction rate decreased too. When the temperature reaches a certain point, the effect of temperature on reaction rate becomes zero. Under continuously rising temperature, the impact of temperature on the equilibrium constant reverses, the reaction rate reduced when the temperature increases. That is, for a given composition of reactants, at a low temperature range )y > 0. When the temperature reaches a certain value with -)y = 0, the reaction rate reaches a maximum, which is the optimum temperature under this certain composition. After that, )y < 0 with the temperature continues increasing. 

For each reaction in a surface chemistry mechanism, one must provide a temperature dependent reaction probability or a rate constant for the reaction in both the forward and reverse directions. (The user may specify that a reaction is irreversible or has no temperature dependence, which are special cases of the general statement above.) To simulate the heat consumption or release at a surface due to heterogeneous reactions, the (temperature-dependent) endothermicity or exothermicity of each reaction must also be provided. In developing a surface reaction mechanism, one may choose to specify independently the forward and reverse rate constants for each reaction. An alternative would be to specify the change in free energy (as a function of temperature) for each reaction, and compute the reverse rate constant via the reaction equilibrium constant. 

From Eqn. (14) it follows that with an exothermic reaction - and this is the case for most reactions in reactive absorption processes - decreases with increasing temperature. The electrolyte solution chemistry involves a variety of chemical reactions in the liquid phase, for example, complete dissociation of strong electrolytes, partial dissociation of weak electrolytes, reactions among ionic species, and complex ion formation. These reactions occur very rapidly, and hence, chemical equilibrium conditions are often assumed. Therefore, for electrolyte systems, chemical equilibrium calculations are of special importance. Concentration or activity-based reaction equilibrium constants as functions of temperature can be found in the literature [50]. 

Equation (13.14) gives the effect of temperature on tlie equilibrium constant, and hence on the equilibrium conversion. If AH" is negative, i.e., if the reactionis exothermic, the equilibrium constant decreases as the temperature increases. Conversely, K increases with T for an endothennic reaction. 

As previously stated, WGS is an equilibrium-limited reaction that is moderately exothermic. The equilibrium constant depends on temperature as reported in Eqn (1.2) (Haring, 2008) ... 

Thermodynamics. The paraffins isomerization reactions have equilibrium constants of the order of one. For example, for the isomerization of n-hexane to 2-methylpentane and to 3-methylpentane the equilibrium constants are 1.14 and 0.76, respectively. The equilibrium conversion of n-hexane to dimethylbutanes is in the order of 30-35% of the isomers. These dibranched hexanes have higher octane number than the monobranched hexanes isomers. Since the heat of reaction is slightly exothermic, the equilibrium constant has a minor decrease as the reaction temperature increases. There is no effect of the hydrogen partial pressure or the total pressure on the equilibrium conversion. 

Because the reaction is exothermic, the equilibrium constant is larger for lower temperatures. But the reaction proceeds too slowly at room temperature to be practical, even in the presence of the best available catalysts. The optimum choice of temperature, found experimentally to be about 450°C, is a compromise between an increased rate of reaction at higher temperature and an increased yield of ammonia at lower temperature. Because the formation of ammonia decreases the moles of gases, the yield of product is improved by high pressures. The equihbrium constant is only 0.159 at 450°C, so higher pressures (up to 600 atm) are required for an economical yield of ammonia. Ammonia from the Haber reactor is removed from the reaction mixture by cooUng the compressed gases until NH3 liquefies. Unreacted N2 and H2 circulate back to the reactor. 

Reactions 1 and 3 are highly exothermic and therefore have equilibrium constants that decrease rapidly with temperature. Reaction 2 is moderately exothermic, and consequently its equilibrium constant shows a moderate decrease with temperature. Reaction 4 is moderately endothermic, and its equilibrium constant increases with increasing temperature. The relationship between temperature and equilibrium constant for these four reactions is depicted in Figure 1 where carbon is assumed to be graphite. Thermodynamic data were taken from Refs. 1 and 2. 

We can see from Table 9.2 that the equilibrium constant depends on the temperature. For an exothermic reaction, the formation of products is found experimentally to be favored by lowering the temperature. Conversely, for an endothermic reaction, the products are favored by an increase in temperature. 

Cyclohexane (C) and methylcyclopentane (M) are isomers with the chemical formula C6H12. The equilibrium constant for the rearrangement C M in solution is 0.140 at 25°C. (a) A solution of 0.0200 mol-L 1 cyclohexane and 0.100 mol-I. 1 methylcyclopentane is prepared. Is the system at equilibrium If not, will it will form more reactants or more products (b) What are the concentrations of cyclohexane and methylcyclohexane at equilibrium (c) If the temperature is raised to 50.°C, the concentration of cyclohexane becomes 0.100 mol-L 1 when equilibrium is reestablished. Calculate the new equilibrium constant, (d) Is the reaction exothermic or endothermic at 25°C Explain your conclusion. 

L-mol 1 -min 1 and the rate constant for the reverse reaction is 392 L-mol 1 -min. The activation energy for the forward reaction is 39.7 kj-mol 1 and that of the reverse reaction is 25.4 kj-mol" (a) What is the equilibrium constant for the reaction (b) Is the reaction exothermic or endothermic (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant ... 

Experimental studies on how temperature affects equilibria reveal a consistent pattern. The equilibrium constant of any exothermic reaction decreases with increasing temperature, whereas the equilibrium constant of any endothermic reaction increases with increasing temperature. We can use two equations for A G °, Equations and, to provide a thermod3mamic explanation for this behavior AG = -RT x Teq AG° — AH°-TAS°... 

In order to illustrate this principle, let the effect of temperature on the equilibrium constant of an exothermic reaction, involving the oxidation of a metal to its oxides, be considered. Upon increasing the temperature of this reaction some of the metal oxides will dissociate into the metal and oxygen and thereby reduce the amount of heat released. This qualitative conclusion based on Le Chatelier s principle can be substantiated quantitatively from the Varft Hoff isochore. 

For reversible exothermic reactions, the situation is more complex. Figure 6.5a shows the behavior of an exothermic reaction as a plot of equilibrium conversion against temperature. Again, the plot can be obtained from values of AG° over a range of temperatures and the equilibrium conversion calculated as discussed previously. If it is assumed that the reactor is operated adiabatically, and the mean molar heat capacity of the reactants and products is constant, then for a given starting temperature for the reaction Tin, the temperature of the reaction mixture will be proportional to the reactor conversion X for adiabatic operation, Figure 6.5a. 

The reactor brings the reaction mixture to equilibrium at the outlet temperature. The reaction is exothermic and the equilibrium constant K is given by ... 

For cases where AH0 is essentially independent of temperature, plots of in Ka versus 1/T are linear with slope —(AH°/R). For cases where the heat capacity term in equation 2.2.7 is appreciable, this equation must be substituted in either equation 2.5.2 or equation 2.5.3 in order to determine the temperature dependence of the equilibrium constant. For exothermic reactions (AH0 negative) the equilibrium constant decreases with increasing temperature, while for endothermic reactions the equilibrium constant increases with increasing temperature. 

The reaction favours the formation of ozone with a significant equilibrium constant. Appendix C also lists the enthalpies of formation and the standard enthalpy of the reaction ArH° can be calculated. The answer for the enthalpy calculation is ArH° = —106.47 kJ mol, showing this to be an exothermic reaction, liberating heat. The entropy change at 298 K can also be calculated because ArG° = ArH° — T ArS°, so ArS° = 25.4 Jmol-1 K-1, indicating an increase in the entropy of the reaction as it proceeds by creating one molecule from two. 

Does this agree with LeChatelier s Principle that for an exothermic reaction the system will shift to form more reactants at higher temperatures Yes, it does. Recall, that that only temperature will affect the value of the equilibrium constant. More reactants are formed because the value of the equilibrium constant decreased when the temperature increased. 

B We expect the value of the equilibrium constant to increase as temperature decreases since this is an exothermic reaction and it should become more spontaneous (shift right) at lower temperatures.Thus, we expect to be larger than 1000, which is its value at 4.3 x 102 K. 

This behavior can be shown graphically by constructing the rD-7 -/A relation from equation 5.3-16, in which kp kr, and Keq depend on T. This is a surface in three-dimensional space, but Figure 5.2 shows the relation in two-dimensional contour form, both for an exothermic reaction and an endothermic reaction, with /A as a function of T and ( rA) (as a parameter). The full line in each case represents equilibrium conversion. Two constant-rate ( -rA) contours are shown in each case (note the direction of increase in (- rA) in each case). As expected, each rate contour exhibits a maximum for the exothermic case, but not for the endothermic case. 

There is an important difference in this behavior between an exothermic reaction and an endothermic reaction. Fran equation 3.1-5, the van t Hoff equator, the equilibrium constant (Keq) decreases with increasing T for an exothermic reaction, and increases for an endothermic reaction. The behavior of f eq(T) corresponds to this. 

Heat effects accompanying chemical reaction influence equilibrium constants and compositions as well as rates of reaction. The enthalpy change of reaction, AHr, is the difference between the enthalpies of formation of the participants. It is positive for endothermic reactions and negative for exothermic ones. This convention is the opposite of that for heats of reaction, so care should be exercised in applications of this quantity. Enthalpies of formation are empirical data, most often known at a standard temperature, frequently at 298 K. The Gibbs energies of formation, AGfl likewise are empirical data. 

Changing the temperature changes the value of the equilibrium constant. It also changes the amount of heat in the system and can be treated as a concentration effect. To treat it this way, one must know which reaction, forward or reverse, is exothermic (releasing heat). One last time, let s consider the Haber reaction ... 

One of the issues of the industrial process design is related to the heat released by this reaction. A temperature rise will decrease the acetic acid yield, not only because the equilibrium constant becomes lower (the reaction is exothermic see section 2.9) but also because it will reduce the enzyme activity. It is therefore important to keep the reaction temperature within a certain range, for instance, by using a heat exchanger. However, to design this device we need to know the reaction enthalpy under the experimental conditions, and this quantity cannot be easily found in the chemical literature. 

Let us suppose that the acetic acid content of the final aqueous solution is 5%, corresponding to a ratio of approximately 1 mol of CH3COOH to 60 mol of H2O. As the yield of reaction 2.1 will be near 100% (recall that reaction 2.2 is rather exothermic, implying a very high equilibrium constant see section 2.9), the same value will be used for the molar ratio (H2 O) / n (C 2115OII), despite the increased total amount of substance of water in the reaction products. In the present case, the difference of 1 mol of water between the product and the reactant mixtures has a negligible enthalpic effect. The enthalpies associated with the solution of ethanol and acetic acid in 60 mol of water are derived from literature data [17] as Asin//(1) = -10.0 0.1 kJ mol-1 and Asin//(3) = —1.0 0.1 kJ mol-1. This calculation will be detailed in section 2.5. 

In Figure 2.4, data for the equilibrium constants of esterification/hydrolysis and transesterification/glycolysis from different publications [21-24] are compared. In addition, the equilibrium constant data for the reaction TPA + 2EG BHET + 2W, as calculated by a Gibbs reactor model included in the commercial process simulator Chemcad, are also shown. The equilibrium constants for the respective reactions show the same tendency, although the correspondence is not as good as required for a reliable rigorous modelling of the esterification process. The thermodynamic data, as well as the dependency of the equilibrium constants on temperature, indicate that the esterification reactions of the model compounds are moderately endothermic. The transesterification process is a moderately exothermic reaction. 

The related formation equilibrium constant should have a small value if it is zero no XZ would form if, on the other hand, it is very large the stable XZ would not re-decompose and no X re-deposition could be obtained. Notice that if the synthesis of XZ is endothermic, the equilibrium will be displaced to the right with increasing temperature (the opposite is true if the reaction corresponds to an exothermic formation of XZ). Therefore, in order to have the transport of X, (synthesis of the intermediate at one end of the tube and re-decomposition with deposition of X at the other end) this must be placed at the hot end if the formation of XZ is endothermic (or the cold end, if exothermic). 

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