Exercises Equilibrium

After the equilibrium represented above is established, some pure O2(g) is injected into the reaction vessel at constant temperature. After equilibrium is reestablished, which of the following has a lower value compared to its value at the original equilibrium  

Which of the following changes alone would cause a decrease in the value of Ke for the reaction represented above  

Some N2 and H2 are mixed in a container at 200°C, and the system reaches equilibrium according to the equation above. Which of the following causes an increase in the number of moles of NH3 present at equilibrium  

In the reaction shown above, 0.50 moles of Br2 and 0.50 moles of I2 are placed in an evacuated 1.00-liter vessel and allowed to reach equilibrium. What is the value of Kc if the vessel contains 0.84 moles of IBr at 

The reaction above is used in the mining industry to extract copper from copper ore. Once the mixture is allowed to establish equilibrium at temperature T and pressure P, the equilibrium can be shifted to favor the products by 

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This exercise examines the effect of basis set on the computed equilibrium structure of N,N -Dimethylformamide. 

The dimensionless K. is regarded as a function of system T and P only and not of phase compositions. It must be exfjerimentally determined. Reference 64 provides charts of R (T,P) for a number of paraffinic hydrocarbons. K. is found to increase with an increase in system T and decrease with an increase in P. Away from the critical point, it is invariably assumed that the K, values of component i are independent of the other components present in the system. In the absence of experimental data, caution must be exercised in the use of K-factor charts for a given application. The term distribution coefficient is also used in the context of a solute (solid or liquid) distributed between two immiscible liquid phases yj and x. are then the equilibrium mole fractions of solute i in each liquid phase. 

It can be shown from a consideration of the overall stability constants of the ions [Ni( CN)4] 2 " (1027) and [ Ag( CN)2 ] (1021) that the equilibrium constant for the above ionic reaction is 1015, i.e. the reaction proceeds practically completely to the right. An interesting exercise is the analysis of a solid silver halide, e.g. silver chloride. 

The number of components in a system can change with experimental conditions, and one must exercise care in defining the system. For example, a mixture of H2, 02, and H20(g) at low temperature is a three-component mixture.11 However, heating the mixture to a high temperature causes the three species to be in rapid equilibrium through the reaction... 

Use the phase diagram for helium in Exercise 8.13 (a) to describe the phases in equilibrium at each of helium s two triple points (b) to decide which liquid phase is more dense, helium-1 or helium-II. 

EXAMPLE 9.1 Sample exercise Writing the expression for an equilibrium constant... 

EXAMPLE 9.7 Sample exercise Calculating the equilibrium composition by approximation... 

EXAMPLE 9.11 Sample Exercise Predicting the effect of compression on an equilibrium... 

Suppose that, in the same reaction as in Exercise 9.71, the total pressure at equilibrium is found to be 3.0 bar and the C1 H atom ratio is 1 3. What are the partial pressures of the three gases ... 

EXAMPLE 12.8 Sample exercise Calculating the equilibrium constant for a... 

Whilst this will be satisfactory when dealing with kinetic data in which reactions involving the solvent will not explicitly appear in the rate equations, it is not appropriate when we consider equilibrium constants. As an exercise, consider the formation of [Ni(en)3] from aqueous solutions of nickel(ii) chloride and en (en = H2NCH2CH2NH2) write the equations with the inclusion and the omission of the water molecules. Can you recognize the driving force for the formation of the chelate in each case ... 

A separate fitting exercise and a separate rate expression are needed for reactions starting on the other side of equilibrium. 

EXERCISE 3.34 Predict the position of equilibrium for the following reaction ... 

C16-0009. Draw a molecular picture illustrating the equilibrium in Exercise 16.2.1(d). 

Exercise 3.8 Partition Function, Average Energy and Equilibrium Constant... 

Exercise 3.9 Equilibrium Constants From Partition Functions... 

In this exercise we shall estimate the influence of transport limitations when testing an ammonia catalyst such as that described in Exercise 5.1 by estimating the effectiveness factor e. We are aware that the radius of the catalyst particles is essential so the fused and reduced catalyst is crushed into small particles. A fraction with a narrow distribution of = 0.2 mm is used for the experiment. We shall assume that the particles are ideally spherical. The effective diffusion constant is not easily accessible but we assume that it is approximately a factor of 100 lower than the free diffusion, which is in the proximity of 0.4 cm s . A test is then made with a stoichiometric mixture of N2/H2 at 4 bar under the assumption that the process is far from equilibrium and first order in nitrogen. The reaction is planned to run at 600 K, and from fundamental studies on a single crystal the TOP is roughly 0.05 per iron atom in the surface. From Exercise 5.1 we utilize that 1 g of reduced catalyst has a volume of 0.2 cm g , that the pore volume constitutes 0.1 cm g and that the total surface area, which we will assume is the pore area, is 29 m g , and that of this is the 18 m g- is the pure iron Fe(lOO) surface. Note that there is some dispute as to which are the active sites on iron (a dispute that we disregard here). 

This last inflammability parameter presents problems. After stating its definition it will be seen that measuring autoignition temperature proves to be a difficult exercise because its measurement is sensitive to the experimental conditions, even more sensitive than for flashpoints. Worse, this parameter seems to be controlled by kinetic factors far more complex to master than the thermodynamic factors that probably control flashpoints (in fact it is a liquid/vapour equilibrium). So whilst the influence of the nature of the cup metal on a flashpoint has never been demonstrated, this demonstration was easily made with autoignition temperatures. 

This expression for the equilibrium constant is found to contain the term V in the denominator. Since K must remain constant, an increase in V would cause % also to increase. Stated in an another form, the dissociation of AB is favoured by a reduction in the pressure. A pressure increase would bring down V, and to maintain the constant value of K, x must decrease. Thus, a pressure increase would tend to inhibit the dissociation of AB. As in the previous case, it will be of interest in this case also to examine the effects of some other factors on the equilibrium. It is left to the readers as an exercise to establish for this case the following results (i) the effect of the addition of either A or B is to suppress the degree of dissociation of AB (ii) the addition of an inert gas at constant volume does not alter the degree of dissociation of AB and (iii) the addition of an inert gas at constant pressure increases the degree of dissociation of AB. 

In principle, one may combine equilibrium and critical data in one database for the parameter estimation. From a numerical implementation point of view this can easily be done with the proposed estimation methods. However, it was not done because it puts a tremendous demand in the correlational ability of the EoS to describe all the data and it will be simply a computational exercise. 

Care should be exercised in using the coefficients from Table 4.14 to predict two-liquid phase behavior under subcooled conditions. The coefficients in Table 4.14 were determined from vapor-liquid equilibrium data at saturated conditions. 

Repeat the calculation in Exercise 8 for equilibrium conversion and equilibrium concentration, but taking into account variation of AH° with temperature. Again assume ideal gas behavior. Heat capacity coefficients for Equation 6.42 are given in Table 6.196. 

For a number of reasons, using saturation indices as measures of the mineral masses to be formed as a fluid approaches equilibrium is a futile (if commonly undertaken) exercise. First, a mineral s saturation index depends on the choice of its formula unit. If we were to write the formula for quartz as Si2C>4 instead of Si02, we would double its saturation index. Large formula units have been chosen for many of the clay and zeolite minerals listed in the llnl database, and this explains why these minerals appear frequently at the top of the supersaturation list. 

For other cases, such as La3+ where more detail is required about the nature of the species present in solution, titration data can be computer fit to more complicated multi-equilibrium models containing Mx 1 v( OR)v forms whose stoichiometry is suggested by information gained from independent spectroscopic or kinetic techniques. One must be mindful of the pitfalls of simply fitting the potentiometric data to complex multi-component models for which there is no independent evidence for the various species. Without some evidence for the species put into the fit, the procedure simply becomes an uncritical mathematical exercise of adding and removing various real and proposed components until the goodness of fit is satisfactory. 

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. 

These equations are applied in the simulation example CHROMDIFF to the case of a two-component separation with linear equilibrium. The situation of a non-linear equilibrium is considered as an exercise in the example. 

One student wanted to get a quick result. Instead of including the values obtained at equilibrium, he(she) just used the values until 180 min for parameter estimation. Compare the results with those of exercise 5 and comment. 

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