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| From KWOK The Chem Teacher |
I am certain all of you are aware that pressure affects equilibria which has changes in the number of gaseous molecules. Hence, to ensure a constant pressure to be created is more difficult than to create an environment for constant temperature. Hence, please ponder on the above question. The solution isn't complicated but it is an application of what we have learnt.
Lastly, ensure your answers are clear and concise. This solution does not need excessive elaboration.
Background
The background of the question came from two different questions found in separate preliminary examination paper. The questions raised a curious thoughts, how is it possible to ensure a constant pressure for gaseous reaction system with temperature being kept constant?
Let's take Haber Process as the example. From its equation, you can tell that the number of particles at equilibrium will be less than the number of particles at the start of the reaction. Hence, if pressure was kept constant (i.e. inital P = final P), with a decrease number of particles, the volume must decrease. Therefore a contraction occurs.
Thus, the curious question. How is it possible for constant pressure, constant volume and constant pressure be maintained for the reaction for Haber Process? Fundamentally, the pressure exerted by the system has to be that exerted by the gas particles.
Hence, if initially we only have the reactants and the total pressure is 200 atm, how can a reduction in gas particles result in the pressure to still remain the same; when volume and temperature are constant? We would actually expect that pressure decreases.
Therefore, my suggestion is to add inert gas such as He into the reaction mixture. The inert gas added will ensure that the total pressure remains, in addition, this addition will not keep causing the equilibrium position to shift.
However, Jing Yew made an excellent suggestion of using a vortex. Personally, I am clueless to what it is. However, a check with a Physics teacher I learnt that it is a machine that can alter the kinetic energy of the gas particles. Hence, if this machine is feasible, it can alternate the speed of the particles at equilibrium to ensure that the total pressure remains the same as the initial pressure.
The other question
The simplest way is to cool the reaction mixture quickly and separate the species. Quickly because you do not want equilibrium to have any time to respond to the change in temperature. In addition, cooling ensure that the particles do not have enough energy to overcome Ea, hence hardly any forward or backward reaction can take place.
Subsequently, extract a small portion of the sample (to ensure only small amount of NH3 is present), then titrate against standardised concentration of HCl with a suitable indicator The small amount ensure that you will not need to have to use an exaggerated amount of HCl to attain full neutralisation.
Misconceptions
Some glaring misconceptions include
(1) Thinking that removing gases (especially NH3) will do the trick. That is a BIG NEVER. Since, the number of gas particles will decrease as we compare the initial with the equilibrium, removing particles will make the situation worse.
(2) Adding more gas reactants. Will that will ensure that you can get a constant temperature, but can you maintain the equilibrium? If you add a reactant, chances are equilibrium position will shift right. Hence, you will constantly have a situation where equilibrium position shifts and hence you can't achieve the equilibrium.
(3) Using a pH to determine the concentration of ammonia present. That is extremely inaccurate! We never do that!
(4) Apply PV = nRT. I am now sure how are u going to separate the ammonia from the mixture and hence measure the partial pressure of ammonia. Subsequently, using the ideal gas equation to calculate number of moles of ammonia.
Conclusion
Personally, I am not sure if this is what happen in industries. However, from my extensive search on the web, my sense is probably Jing Yew may be more accurate than I am. Since, the gases in Haber process are actually injected into the reaction vessel at a particular speed to create the desired pressure (Note: More KE the gas particles has, the greater force they exert when the collide of the vessel's wall). The notion of speed causing the pressure gives me that suspicion that it could be a voltex or at least some instructment that can adjust the speed of the particles and yet ensuring the volume, pressure and temperate are kept constant.
22 comments:
To the first question, i think to maintain the equlibrium it is require that we cool the gases to condense the NH3 into liquid form. As NH3 has a lower melting point than N2 and O2, it will condense into liquid before them. When the NH3 is in liquid form, it will exert negligible pressure on the system. Hence, the remaining gas particles are those of the N2 and H2. By Le charteir's principle, the equilibrum will hence shift to the right and more NH3 will be produced, which will then further condense into liquid form. This causes the equilibrium to further shift to the right and will cause the equation to move towards completion. As the total number of moles of gases does not increase since NH3 is in liquid form, the pressure in the system is mantained as the number of moles in the system belongs to the gases of H2 and N2.
As to the second question, i'm not so sure but i think that the percentage yeild could be calculated by taking the finding the number of moles of liquid ammonia formed and divide it by the theoratical yeild of ammonia, which could be calculated through stoichiometric calculations.
i. Ammonia can be constantly removed by channeling the equilibrium mixture of gases into another vessel to be cooled, while the leftover gases are channeled back to the reaction vessel. Because 4 moles of reactants produce 2 moles of products, the pressure would decrease as ammonia is formed. To maintain the pressure, the equilibrium mixture of gases is thus constantly removed, and are replaced with more reactants at 200 atm.
ii. To know percentage yield, the equilibrium mixture of gases is removed from the vessel (in a known volume), and at the same pressure (to prevent shift of equilibrium due to changing pressure), cooled in another connected vessel to the extent such that ammonia condenses into liquid and removed, while the n2 and h2 remains as leftover gases. Thus the amount of condensed ammonia can be calculated, and then compared to the total volume of the gases removed such that percentage yield can be obtained.
i) Two pipes should be attached to the sealed vessel, which pump in Hydrogen and Nitrogen gas separately at a pressure of 200 atm. A lid which can flip open should cover both pipes. Whenever pressure in the sealed vessel drops as a result of the formation of Ammonia, the two lids will flip open due to the difference in pressure between the pipes and the vessel, allowing Hydrogen and Nitrogen to enter the vessel.
The lids should close when there is no longer a pressure difference, which occurs when the pressure in the sealed vessel reaches 200 atm. Therefore, a pressure of 200 atm is maintained in the vessel. Additionally, to ensure that the right proportions of Nitrogen and Nitrogen enter the vessel, the surface area of the pipe containing Hydrogen should be three times larger than the one containing Nitrogen.
ii) The sealed vessel should be filled up with Nitrogen and Hydrogen gas in a 1:3 ratio, at 200 atm and 450°C. Using the formula, PV = nRT, the number of moles of Nitrogen and Hydrogen can be calculated, and thus the number of moles of Ammonia formed if the reactants fully react.
During the reaction, an inert gas should be pumped into the vessel using the pipes from (i) to maintain the pressure at 200 atm. When equilibrium is reached, the sealed vessel should be quickly cooled to allow the Ammonia to condense into liquid form. A titration should then be carried out to determine the actual yield in moles of Ammonia. This value should be divided by the theoretical yield from earlier, and then multiplied by 100% to get the percentage yield of Ammonia at 200 atm and 450°C.
(i): Place the vessel in a vortex machine. This will supply the particles with more kinetic energy. After the reaction has reached equilibrium, remove the ammonia by dissolving it in water, pump in more hydrogen and nitrogen, and resume vortexing.
(ii): To measure the percentage yield, first stop the vortexing and add a small amount of water to the vessel, and some universal indicator. Ammonia is alkaline and would have a pH value above 7. How much ammonia has been produced, in other words the percentage yield at that point of time, will determine how concentrated the ammonia solution is. This in turn will affect the pH of the solution. The colour of universal indicator can be measured using a colour chart.
i) Pressure is maintained simply by adding more reactant gas as pressure is directly proportional to number of mol.
As the forward reaction is favoured, number of mol of gas decreases from 4 to 2. In this case, adding 2 mol of reactants maintains pressure. Or if you're actively removing NH3, add 4 mol o reactants.
Temperature is maintained simply by heating the thing to the correct temperature:D
ii) He can know the equilibrium yield by the kc. S/He should know the kc value of the process at 450 degrees Celsius ( if s/he doesn't, fire him/her).
As the concentration of the reactants should be known as s/he knows the volume of the container and number of mol of reactants s/he adds in, simply using the kc value and its equation to calculate the concentration of NH3 produced should do the trick.
Kp can be used in lieu of Kc as well.
(i) To keep the pressure constant, the number of gaseous particles have to remain constant as pressure is directly proportional to the number of gaseous particles present. So in order to keep the pressure constant, pipes can be used to inject H2 and N2 molecules into the system at the same rate at which H2 and N2 are consumed to produced NH3. The temperature inside the vessel is kept such that only NH3 condenses into liquid ammonia(NH3). By doing so, the total number of gaseous particles at all times remain constant and thus an environment of constant pressure is created.
(ii)Making use of the ideal gas equation PV=nRT, the manager can calculate the number of moles of H2 and N2 when a fixed amount of both is injected into a vessel with a fixed temperature of 450 degrees and 250 atm. Through this,the balanced equation can be used to find the theoratical yield(in moles) of NH3 to be produced with that given amount of H2 and N2. When equilibrium is reached, the gaseous NH3 is condensed into Liquid NH3 and a titration experiment is peformed on it, using a standard solution of HCl and an indicator of phenolphthalein. Through this the manager can then calculate the actual yield(in moles) of NH3 produced. The manager can then divide the actual yield(in moles) of NH3 by the theoratical yield(in moles) of NH3 to obtain the percentage yield value.
(i) The pressure is maintained in the vessel through the use of pumping of inert gas into the vessel. This increases the total volume of gases in the vessel of constant volume, this increasing the pressure to 200atm. This does not affect the equilibrium position as this gas does not participate in the reaction, thus not affecting the partial pressures of Nitrogen, Hydrogen and Ammonia. Therefore the pressure of 200atm can be obtained by a constant addition of inert gas, thus increasing the yield of Ammonia.
Through the use of 200atm in a sealed vessel, this higher pressure will cause the position of equilibrium to shift to the right, as number of gas molecules goes from 4 to 2. Thus in an attempt to minimize the effect of disturbance, the forward reaction is favoured. This increases the amount of ammonia collected. Furthermore, the rate of forward reaction is increased as well, due to particles being closer together, increasing the chance of effective collisions to occur.
This equation is exothermic, thus having a lower temperature favours the forward reaction as they system tries to minimize the effect of the disturbance. However, having too low a temperature lowers the rate of reaction, thus the temperature is compromised at 450oC to obtain the most economical yield.
However, high yield is still able to be obtained due to the presence of a catalyst in the equation. This allows more particles to have sufficient energy for effective collision by providing a pathway requiring lesser activation energy. Thus the rate of the forward reaction is faster, allowing a temperature of 450oC to produce most economical amount of ammonia.
ii) The equilibrium constant of the equation at 450oC would be known. This allows for the calculation of pressure of ammonia. This is possible as the mixture of gases is passed through a cooler, whereby Nitrogen and Hydrogen are recycled in the equation. Through this recycling passage, the total pressure can be obtained, which can then be divided according to the stoichiometric ratio, and used to find the partial pressure of ammonia. This value will then be used to calculate the number of moles of ammonia and hence the percentage yield.
(i) Add a inert gases to the container to maintain the partial pressure of the reacting gases as the ammonia is being formed.
(ii) In the plant, the ammonia will be piped out and the amount obtained measured. That would be the actual yield of the reaction. The theoratical yield is calculated by an ICE table (one must also know the initial amounts of reactants). Dividing the actual yield by the theoratical yield would give the percentage yield.
i) As the reaction proceeds, inert gases are continually pumped into the vessel to keep the total volume of gases constant. The individual partial pressure of the ases involved in the reaction remains the same in this way, resulting in equilibrium position remaining the same and neither the forward nor backward reaction being favoured.
ii) By drawing an ICE table, the pressure of NH3 at equlibrium can be found. Since pressure is proportional to number of moles of gases, the theoretical yield of NH3 can be determined.
In order to obtain the actual yield, the mixture of gases is passed through a cooler to liquefy and remove ammonia, allowing the remaning nitrogen and hydrogen to be recycled. The actual yield can then be obtained through the liquefied ammonia.
Percentage yield = (actual yield)/(theoretical yield) x 100%
(i)at low temperature, exothermic (forward) reaction is favoured as it liberates energy and hence yield of ammonia is increased. however, the rate of reaction is slow at low temperatures due to low frequency of effective collision between molecules. therefore, an optimum temperature of about 450 degrees Celsius.
Moreover, at high pressure, the forward reaction is favoured as the number of moles of product < moles of reactions. hence, yield of ammonia would be higher. in addition, rate of reaction is faster at higher pressure as well, causing the equilibrium to reach faster. however, cost and maintenance of equipment used to maintain high pressure is too high and hence a compromise is made and an optimum pressure of 250 atm is used to produce an economical yield.
(ii) The closed vessel as a fixed volume to ensure that non of the reactants/products escape frm the vessel. hence, a known amount of reactants could be used, and ensure that the amount of each reactant corresponds to the stiochometric equation to ensure that there is no excess reactants. after all the reactants have been reacted to form ammonia, find the concentration of the ammonia and compare it with the initial concentration of reactants and calculate % yield
(i) Constantly remove ammonia from the reaction chamber which according to Le Chatelier's principle will cause the position of equilibrium to shift to the right, favouring the forward reaction and therefore increasing the yield of the ammonia. Also, adding of catalyst would reduce the amount of activation energy required for the process, hence conserving electricity whereby the unreacted reactants can now be recycled again and again until the yield is almost 100%
(ii) Condense the ammonia gas into liquid. Using the Haber process equation, calculate the percentage yield.
(i) Add inert gasses as the reaction proceeds. The number of gasses in the reaction decreases from 4 moles to 2. Thus by adding inert gas, the total number of moles of gas would be maintained. This ensures that the pressure is kept constant throughout.
(ii) Use Kc to calculate percentage yield.
Haber process is a continuous process, in which the product ammonia is constantly removed by cooling which condenses the ammonia to liquid. In order to maintain pressure, more moles of reactant gases have to be added continuously to compensate for the removal of ammonia molecules. Any unreacted gases are recycled, resulting in an overall conversion of 98% eventually.
Maintaining the pressure in the vessel can be done by adding more moles of reactant gas, which is nitrogen and hydrogen gas. A constant feed of gases into the reaction chamber at the right pressures would enable the pressure to maintain the same with the removal of NH3 molecules.
When equilibrium is established, the ammonia gas can be cooled, condensed and collected as a liquid. The ammonia can then be removed from the system in a liquid form and the system responds to this to form more ammonia and the equilibrium position would shift to the right. The manager can then use the amount of ammonia collected as a liquid to calculate the percentage yield.
(i)ammonia gas formed is constantly taken out. This would allow the pressure to be kept constant and induce a forward reaction(by LCP).
the forward reaction is exothermic. So, the gases are constantly cooled so as to ensure a constant temp and keeping it a forward reaction.
(ii) first he should measure the total yield of ammonia gas from a give amount of nitrogen and hydrogen by constantly recycling them. Next, he should measure the yield of ammonia gas at equilibrium.
equilibrium yield/total yield x 100 = percentage yield of ammonia.
(i)As the reaction proceeds, number of moles of gaseous molecules will decrease as the mole ratio of products to reactants is 4:2. Hence pressure will decrease.
To maintain the constant pressure at 200atm for optimum yield of ammonia, we may add inert gases into the vessel or pump in more reactants at the same rate as NH3 is being produced and taken out of the system to keep internal conditions constant.
(ii) He may only know the percentage yield when he collects and measures the ammonia produced by first cooling it into a liquid which then makes it easier to store and measure.
(i)This can be done by adding an innert gas into the system so that the pressure can be kept at a constant of 200 atm. As, as the reaction progresses to the right, less no of moles of gasses would be produced hence pressure would drop, hence this will cause a change in equilibrium and the presure cant be kept constant. Hence by adding an inert gas, it wont cause a change in the partial pressures of the molecules allowing the equilibrium to kept constant.
(ii) The manager can know the percentage yield by using the partial pressures of the gasses hence knowing the position of the equilibrium of the reaction and therefore can caculate its percentage yield. ANd the addition of the inner gasses also wont change the partial pressure hence this method can still be used.
i) To maintain a constant pressure of 200atm in the vessel of fixed volume, the N2 and H2 are first added in a fixed volume ratio of 1:3 at an initial pressure of 200atm.
The unreacted nitrogen and oxygen are re-circulated back into the vessel and the volume of gases that has been used up, that is, reacted to form ammonia at equilibrium ( 3 unit volumes of hydrogen to 1 unit volume of nitrogen) is replenished by pumping a continuous supply of these fixed volumes of gases back into the system.
Since ammonia is removed from the system in liquid form (as ammonia has the highest boiling point among H2, N2 and NH3), the pressure exerted by ammonia is negligible in the vessel once it has been cooled to a liquid and the total pressure of the gases exiting the vessel to be re-circulated is equal to the volumes of unreacted N2 and H2 (in the proportion of 1:3).
By adding N2 and H2 in the volumes required to replace the reacted N2 and H2 such that the total volume and proportion of gases re-entering the vessel is equal to the original, the pressure is maintained at 200atm.
ii) The plant manager is able to find the percentage yield by calculating the amount of moles of NH3 obtained (mass of NH3 collected / molar mass of NH3) from the plant reaction and dividing it by the expected amount of moles of NH3 which should be produced if the H2 and N2 had been completely reacted.
(i) Add inert gas as reaction progresses. This will help to maintain the pressure in the vessel at optimum pressure of 200atm without interfering with the reaction itself since the inert gas will not take part in reaction.
(ii) Using the initial volumes of the gases to calculate no. of moles of gas, and then calculating the theoretical yield of NH3, compare with the actual yield of NH3 obtained from the reaction after cooling the gases to form aq NH3 to calculate percentage yield of ammonia.
-eileen
(i) NH3 is constantly cooled and removed, resulting in a decrease in number of gaseous particles in the vessel which in turn causes a decrease in pressure in the vessel. Maintaining the pressure constant can be done by keep adding reactants to the reactant vessel.
(ii) Percentage yield of the process can be obtained by taking percentage of the actual amount of ammonia collected to the amount of ammonia "supposed to be collected" based on practical calculation.
(i) Ammonia is removed from the equilibrium mixture of gases (hydrogen and nitrogen present) leaving the reaction vessel. The hot gases are cooled enough, whilst maintaining a high pressure, for the ammonia to condense and be removed as liquid.
(ii) Since PV=nRT, n=PV/RT, where V, R and T are constants. When partial pressure of NH3 is obtained from the reaction, number of moles of NH3 produced can be calculated. The mass of NH3 produced can be derived by multiplying the Mr of NH3. Lastly, the yield of NH3 in this reaction is divided by the theoritical yield, multiplied by 100%.
(i)When the forward reaction takes place, the number of mole of gases present decreases. Since the reaction takes place in a vessel with a constant volume, the pressure would decrease since P = nRT/V. Thus in order to maintain the pressure, the reactants should be fed into the reaction chamber in order to increase the number of moles of gas particles, thereby increasing the pressure to what it used to be.
(ii)Percentage yield can be calculated by passing all the gases present in the reaction vessel through a liebig condenser filled with cold water. This would cause only the ammonia formed to condense. Calculating the number of moles of ammonia formed over the theoretical amount (found by the stiochiometric equation if the reaction goes to completion)
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