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Chemical reactions and why Featured content. Free courses. All content. We should always check any assumptions we make by calculation.
I will illustrate the point with the following example. Given initially that 1. Because the value K is at least two orders of magnitude smaller than the activity 12, we can try the following assumption. We assume that the extent of reaction is small since K is small and hence x is small. We assume, in fact that x is so small that the denominator x can be approximated by Thus, we can now write. This technique of assuming that there was a amount of reaction small x because of a small K value made the problem quite a bit easier to solve.
This is valid most of the time if the K constant is at least two order of magnitudes smaller than the activities. You can see that in this case, we were borderline. Our answers were accurate to about 2 sig. If we had need more precision on our numerical answers, we could not have made the assumption. When the K constant is large, we can simplify the calculations by adding one extra step before using the Ice table.
In this case, the amount of reaction done the normal way would be almost exactly the same as the amount of the limiting reagent and our calculations would be very difficult. If we first, pretend the reaction goes to completion and then allow it to reverse towards equilibrium, we get results that are easier to work with. Consider the combustion reaction of kPa of methane in excess oxygen 80kPa to produce carbon monoxide and hydrogen gas.
Since only two sig. But we know that there is a measurable K constant so there must be a measurable amount of CH 4 at equilibrium. In this case, the limiting reagent calculation would show that all the methane is used up. We can quickly figure out the amount of oxygen used up and the products created using stoichiometry. Once the limiting reageant calculation is done, we can do the ICE table.
The value of x is at least 13 orders of magnitude smaller than the constants; obviously, our assumption about x being small is valid. Here, we assume that x is small. In other words, we assume that 1. Thus we were able to simplify the expression. Now, we multiply by kPa to convert the equilibrium activity values into pressures so we can now report the amounts pressures of all the chemicals.
A chemical system at equilibrium has many factors affecting the position of that equilibrium. The amount of each of the components, the temperature, the pressure in gas phase reactions especially , presence of radiation, etc. We must be able to analyze these conditions and determine if one of them is changed what affect that will have on the position of the equilibrium.
If a change stress is applied to a system at equilibrium the system will adjust to try to reduce the change.
Consider again the Haber process for the production of NH 3. Experience shows that the pressure and temperature both affect the equilibrium. Higher pressures give higher amounts of NH 3 and higher temperatures give lower amounts of NH 3 at equilibrium. Unfortunately, at low temperatures, the rate of reaction is slower. A balance must be reached between the thermodynamic requirements for high yield and the kinetic requirements for achieving the results rapidly.
We will look at kinetics later. We will focus on the thermodynamic properties of the mixture as an example. What effect will the addition of N 2 to the system such that its concentration raises instantaneously by 1.
We can guess from LP that if addition of N 2 is the change then the equilibrium will shift to try to use up some of the extra N 2. The equilibrium will shift to the right.
Calculation of Q will confirm this quantitatively. Now, we can calculate the equilibrium constant from the equilibrium concentrations. In this case, we are merely doing a comparison of K with Q so we will just do the calculations using the concentration values rather than trying to convert to activities.
In other words, we are going to compare K c with Q c. We calculate the reaction quotient using the non-equilibrium conditions that result from the sudden change in concentration of the N 2 concentration to 1. This is just what LP predicted qualitatively.
The second will have no effect on a gas phase reaction according to the assumptions made in deriving the ideal gas law. Since the partial pressures or concentrations of the individual gas components of the reaction do not change, an inert gas will not change the equilibrium.
If the volume of the container is changed, the individual pressures will change and so will their concentrations. Qualitatively, we can use LP to predict the effect on an equilibrium if we re word the original definition slightly. If a change in pressure is applied to a gaseous system at equilibrium the system will adjust to try to reduce the change in pressure. Thus, if the equilibrium shifts to the right the pressure will drop and if it shifts to the right the pressure will rise.
If we squeeze the container so as to increase the pressure of each partial pressure we will change the equilibrium. LP predicts that the equilibrium will shift to the right to reduce the extra pressure.
The new equilibrium will be reached where more products are present than before. What will happen to the following reactions if the pressure is increased by reducing the volume. The changes discussed so far have changed the equilibrium position but have not had effect on the equilibrium constant itself. Temperature changes will almost always cause a change in the equilibrium constant.
It is exceedingly rare to find a chemical reaction where the energy absorbed is exactly equal to the energy released. In general we can assume that there is always some net change in energy involved in a chemical reaction. We need to consider this net change as a reactant or product to use LP. Consider again the Haber process but this time note that energy is released during the process exothermic. Since nothing else was changed, the new equilibrium position will have a different equilibrium constant than the old one.
Since the new equilibrium has less products than the old one, the new K constant will be smaller than the old one as T rises, K drops. A rigorous derivation will be given for this equation later in this course. What is the average enthalpy of reaction for the process in that temperature range? In reality, there are few gases which behave in an ideal way except at pressures which are relatively low and temperatures that are not very low.
Higher pressures result in properties of gases which deviate from ideal behavior quite appreciably. Since the observed pressure deviates from ideal pressure noticeably at high pressure, this calculated value will be in some error and the error will increase as pressure increases.
See the table below for a set of pressures and observed equilibrium constants to illustrate this point. The direction of reversible reactions can be altered by changing the reaction conditions. Ammonia is made by the Haber process. Reversible reactions. Many chemical reactions are reversible reactions.
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