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Chemical Equilibrium - Part 1: Forward and Reverse Reactions - Biology

Chemical Equilibrium - Part 1: Forward and Reverse Reactions - Biology



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Chemical equilibrium - Part 1: Forward and Reverse Reactions

The concept of chemical equilibrium is critical to following several of the discussions that we have in Bis2a and indeed throughout biology and the sciences. Let us, rather, begin developing our understanding of equilibrium by considering the reversible reaction below:

Hypothetical reaction #1: A hypothetical reaction involving compounds A, B and D. If we read this from left-to-right we would say that A and B come together to form a larger compound D. Reading the reaction from right-to-left we would say that compound D breaks down into smaller compounds A and B.

We first need to define what is meant by a “reversible reaction”. This term “reversible” simply means that a reaction can proceed in both directions. That is, the things on the left side of the equation can react together to become the things on the right of the reaction AND the things on the right of the equation can also react together to become the things on the left side of the equation. Reactions in which the likelihood of the reaction happening in one of the two directions is vanishingly small - and thus only proceed in one direction small - are called irreversible reactions.

To start our discussion of equilibrium we begin by considering a reaction that we posit is readily reversible. In this case, it is the reaction depicted above: the imaginary formation of compound D from compounds A and B. Since it is a reversible reaction we could also call it the decomposition of D into A and B. Let us, however, imagine an experiment in which we watch the reaction proceed from a starting point where only A and B are present.

Example #1: Left-balanced reaction

Hypothetical Reaction #1: time course

At time t = 0 (before the reaction starts) the reaction has 100 concentration units of compounds A and B and zero units of compound D. We now allow the reaction to proceed and observe the concentration of the three compounds over time (t=1, 5, 10, 15, 20, 25, 30, 35, and 40 time units). Notice that as A and B react D forms. In fact, one can see D forming from t=0 all the way to t=25. After that time, however, the concentrations of A, B and D stop changing. Once the reaction reaches the point where the concentrations of the components stop changing we say that the reaction has reached equilibrium. Notice that the concentrations of A, B and D are not equal at equilibrium. In fact, the reaction seems left balanced so that there is more A and B than D.

Note

****Common student misconception warning****

Many students fall victim to the misconception that the concentrations of a reaction’s reactants and products must be equal at equilibrium. Given that the term equilibrium sounds a lot like the word “equal” this is not surprising. But as the experiment above tries to illustrate, this is NOT correct!

Example #2: Right-balanced reaction

We can examine a second hypothetical reaction, the synthesis of compound J from the compounds E and F.

Hypothetical reaction #2: A hypothetical reaction involving compounds E, F and J. If we read this from left-to-right we would say that E and F come together to form a larger compound J. Reading the reaction from right-to-left we would say that compound J breaks down into smaller compounds E and F.

The structure of the hypothetical reaction #2 looks identical to the that of hypothetical reaction #1 that we considered above - two things come together to make one bigger thing. We just need to assume, in this case, that E, F and J have different properties from A, B and D. Let’s imagine a similar experiment to the one described above and examine this data:

Hypothetical Reaction #2: time course


In this case, the reaction also reaches equilibrium. This time, however, equilibrium occurs at around t=30. After that point the concentrations of E, F and J do not change. Note again, that the concentrations of E, F and J are not equal at equilibrium. By contrast to hypothetical reaction #1 (the ABD reaction), this time the concentration of J, the thing on the right side of the arrows, is at a higher concentration than E and F. We say that this for this reaction, equilibrium lies to the right.

Four more points need to be made at this juncture.

Point 1: Whether equilibrium for a reaction lies to the left or the right will be a function of the properties of the components of the reaction and the environmental conditions that the reaction is taking place in (e.g. temperature, pressure, etc.).

Point 2: We can also talk about equilibrium using concepts of energy and we will do this soon - just not yet.

Point 3: While hypothetical reactions #1 and #2 appear to reach a point where the reaction has “stopped” you should imagine that reactions are still happening even after equilibrium has been reached. At equilibrium the “forward” and “reverse” reactions are just happening at the same rate. That is, in example #2, at equilibrium J is forming from E and F at the same rate that it is breaking down into E and F. This explains how the concentrations of the compounds aren’t changing despite the fact that the reactions are still happening.

Point 4: From this description of equilibrium we can define something we call the equilibrium constant. Typically the constant is represented by an uppercase K and may be written Keq. In terms of concentrations Keq is written as the mathematical product of the reaction product concentrations (stuff on the right) divided by the mathematical product of the reactant concentrations (stuff on the left). For example, Keq1 = [D]/[A][B] and Keq2 = [J]/[E][F]. The square brackets "[]" indicate “concentration of” whatever is inside the bracket.


Chemical Equilibrium - Part 1: Forward and Reverse Reactions - Biology

GENERAL CHEMISTRY TOPICS

Kinetics and equilibrium

Chemical equilibrium is the state of constant composition attained when opposing reaction rates become equal. There is an essential relationship between reaction rates reaction rates ---> and chemical equilibrium, one that we can describe quantitatively. At first thought, the connection may seem obscure - do we not need to be far from equilibrium to properly measure reaction rates? The dynamic nature of chemical equilibrium means that both forward and reverse reactions can be taking place at significant - even high - rates, but since these rates are equal, no change in concentration is observed over time.

Let us once again use the hypothetical example of a simple, reversible reaction, A = B (such as an isomerization) to show the approach to equilibrium from either pure A or pure B. The interconversion of A and B via two opposing reaction rates is shown by two different progress curves (see figure below). In (i), a reaction vessel starts out with 100% A, and it begins to undergo conversion to B. At very early times (leftmost region of the graph, shaded pink), only a negligible amount of B is present, so the observed rate is that corresponding to the forward reaction, A &rarr B. As B begins to accumulate (middle portion of graph, lavender), some of it undergoes the reverse reaction, B &rarr A. The net rate still favors production of B, so that the concentration of B continues to rise and [A] still drops, but more slowly. The conversion from A to B is still faster than the conversion of B to A, and this is indicated in the figure by the relative length of the arrows in the chemical equation. Finally, enough B is present so that the forward and reverse reaction rates are in balance, and the net rate is zero (right side of graph, purple). These equilibrium conditions, once attained, persist indefinitely.

If the starting conditions are 100% B, as in (ii), the kinetics at very early times (leftmost region of the graph) are dominated by the reverse reaction, B &rarr A. Once some A forms, the rate of the forward reaction begins to pick up. The net rate still favors production of A for a short time (middle graph), but its concentration levels off by

60 s, having attained equilibrium level. Notably, these equilibrium conditions are the same as those in (i) - this assumes both experiments are carried out at the same temperature.

Kinetics, equilibrium, and the reaction coordinate diagram

To further illustrate the relationship between reaction rate and chemical equilibrium, let's consider the reaction coordinate diagram for our simple, reversible reaction.

This diagram, at right, shows a graph (purple line) of chemical potential energy (vertical axis) versus a "reaction coordinate" (horizontal axis), which here represents the progress of a reaction in which one bond must break and a new bond forms. Examples of such reactions are provided by the many isomerization reactions, such as the interconversion of cis-2-butene and trans-2-butene.

Let us first consider the forward reaction, A &rarr B. Reactant molecules (A) in the system with enough energy to reach the energetic peak, the transition state (symbolized by the double dagger, ‡) can continue forward along the reaction coordinate to conversion to product B.

This minimum energy required is the activation energy. For the forward reaction, this is symbolized as Ea, fwd, and is the energetic cost required to advance the bond breaking in a molecule of A to the point where formation of the new bond in B can begin to provide an energetic payback. The increase in potential energy must be provided by the kinetic energy of the molecules in the system, which varies by temperature, according to a Maxwell-Boltzmann distribution. For molecules, the kinetic energy can reside not only in their motion through space, but also in bond stretching, bending, and twisting. Kinetic energy in these forms is constantly being redistributed by collisions between the molecules. We can easily imagine a bond breaking in molecules where an excess of kinetic energy is momentarily transferred into these bond motions. The reverse reaction, B &rarr A, proceeds in a similar manner, but more conversion of kinetic to potential energy at the molecular level is required. Molecules of B in the system with enough energy (the activation energy for the reverse reaction, Ea, rev) to reach the transition state can continue backward along the reaction coordinate to conversion back to A. Since the forward reaction activation energy is less than the reverse reaction activation energy (Ea, fwd < Ea, rev ), the reverse reaction will be slower than the forward reaction when the concentrations of A and B are equal. In order for this reaction to reach equilibrium, the concentration of B will have to increase beyond that of A up to the point where the rates of the forward and reverse reactions become equal. Qualitatively, since Ea, fwd < Ea, rev in this case, Keq > 1.


We can conclude that the net change in chemical potential energy is related to the equilibrium constant (Keq) for the reaction, whereas the magnitude of the activation energy must be related to the rate constants for the reaction. Let us specify now that the forward reaction, A &rarr B, and the reverse reaction, B &rarr A, both follow first-order rate laws. rate laws . ---> Then we can readily derive the relationship between rate constants and the equilibrium constant for our simple first-order case. The rate laws for the forward and reverse reactions in this case are

These are just first-order rate laws, with k1 as the forward first-order rate constant, and k&minus1 as the first-order rate constant for the reverse reaction. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. Setting these expressions equal when equilibrium concentrations are reached, we have

which can be rearranged to

The right side expression is of course the equilibrium constant for this reaction, so we have shown in this case that


11.1: The Concept of Dynamic Equilibrium

In the last chapter, we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibrium, the point at which the composition of the system no longer changes with time.

Figure (PageIndex<1>): Dinitrogen tetroxide is a powerful oxidizer that reacts spontaneously upon contact with various forms of hydrazine, which makes the pair a popular propellant combination for rockets. Nitrogen dioxide at &minus196 °C, 0 °C, 23 °C, 35 °C, and 50 °C. (NO2) converts to the colorless dinitrogen tetroxide (N2O4) at low temperatures, and reverts to NO2 at higher temperatures. (CC BY-SA 3.0 Eframgoldberg).

Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide ((ce)) to nitrogen dioxide ((ce)). You may recall that (ce) is responsible for the brown color we associate with smog. When a sealed tube containing solid (ce) (mp = &minus9.3°C bp = 21.2°C) is heated from &minus78.4°C to 25°C, the red-brown color of (ce) appears (Figure (PageIndex<1>)). The reaction can be followed visually because the product ((ce)) is colored, whereas the reactant ((ce)) is colorless:

The double arrow indicates that both the forward reaction

occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates.

Figure (PageIndex<2>) shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of (ce) were zero, then it increases as the concentration of (ce) decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no (ce) but an initial (ce) concentration twice the initial concentration of (ce) (Figure (PageIndex<2a>)), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure (PageIndex<2b>)). Thus equilibrium can be approached from either direction in a chemical reaction.

Figure (PageIndex<2>): The Composition of (ce)/(ce) Mixtures as a Function of Time at Room Temperature. (a) Initially, this idealized system contains 0.0500 M gaseous (ce) and no gaseous (ce). The concentration of (ce) decreases with time as the concentration of (ce) increases. (b) Initially, this system contains 0.1000 M (ce) and no (ce). The concentration of (ce) decreases with time as the concentration of (ce) increases. In both cases, the final concentrations of the substances are the same: [(ce)] = 0.0422 M and [(ce)] = 0.0156 M at equilibrium. (CC BY-SA-NC Anonymous by request)

Figure (PageIndex<3>) shows the forward and reverse reaction rates for a sample that initially contains pure (ce). Because the initial concentration of (ce) is zero, the forward reaction rate (dissociation of (ce)) is initially zero as well. In contrast, the reverse reaction rate (dimerization of (ce)) is initially very high ((2.0 imes 10^6, M/s)), but it decreases rapidly as the concentration of (ce) decreases. As the concentration of (ce) increases, the rate of dissociation of (ce) increases&mdashbut more slowly than the dimerization of (ce)&mdashbecause the reaction is only first order in (ce) (rate = (k_f[N_2O_4]), where (k_f) is the rate constant for the forward reaction in Equations ( ef) and ( ef)). Eventually, the forward and reverse reaction rates become identical, (k_f = k_r), and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.

Figure (PageIndex<3>): The Forward and Reverse Reaction Rates as a Function of Time for the (N_2O_<4(g)>)𕇺(NO_<2(g)>) System Shown in Part (b) in Figure (PageIndex<2>). (CC BY-SA-NC Anonymous by request)

The rate of dimerization of (ce) (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of (ce) is zero, the rate of the dissociation reaction (forward reaction) at (t = 0) is also zero. As the dimerization reaction proceeds, the (ce) concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of (ce) and (ce) no longer change.

At equilibrium, the forward reaction rate is equal to the reverse reaction rate.

The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation:

[2A ightleftharpoons B onumber]

where the blue circles are (A) and the purple ovals are (B). Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?

Given: three reaction systems

Asked for: relative time to reach chemical equilibrium

Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.

In systems 1 and 3, the concentration of A decreases from (t_0) through (t_2) but is the same at both (t_2) and (t_3). Thus systems 1 and 3 are at equilibrium by (t_3). In system 2, the concentrations of A and B are still changing between (t_2) and (t_3), so system 2 may not yet have reached equilibrium by (t_3). Thus system 2 took the longest to reach chemical equilibrium.

In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?


Introduction to Equilibrium

This is a series of lectures in videos covering Chemistry topics taught in High Schools. How to describe equilibrium in terms of the rates of the forward and reverse reactions as well as in terms of the concentrations of reactants and products.

Introduction to Equilibrium Part 1 of 3
Learn about reversible reactions and what it means for a reaction to "go to equilibrium".

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Contents

Le Chatelier's principle describes the qualitative behavior of systems where there is an externally induced, instantaneous change in one parameter of a system it states that a behavioral shift occurs in the system so as to oppose (partly cancel) the parameter change. The duration of adjustment depends on the strength of the negative feedback to the initial shock. Where a shock initially induces positive feedback (such as thermal runaway), the new equilibrium can be far from the old one, and can take a long time to reach. In some dynamic systems, the end-state cannot be determined from the shock. The principle is typically used to describe closed negative-feedback systems, but applies, in general, to thermodynamically closed and isolated systems in nature, since the second law of thermodynamics ensures that the disequilibrium caused by an instantaneous shock must have a finite half-life. [5] The principle has analogs throughout the entire physical world.

While well rooted in chemical equilibrium and extended into economic theory, Le Chatelier's principle can also be used in describing mechanical systems in that a system put under stress will respond in such a way as to reduce or minimize that stress. Moreover, the response will generally be via the mechanism that most easily relieves that stress. Shear pins and other such sacrificial devices are design elements that protect systems against stress applied in undesired manners to relieve it so as to prevent more extensive damage to the entire system, a practical engineering application of Le Chatelier's principle.

Effect of change in concentration Edit

Changing the concentration of a chemical will shift the equilibrium to the side that would counter that change in concentration. The chemical system will attempt to partly oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent, and yield of products will be altered corresponding to the impact on the system.

This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.

Suppose we were to increase the concentration of CO in the system. Using Le Chatelier's principle, we can predict that the concentration of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to "fill the gap" and favor the side where the species was reduced. This observation is supported by the collision theory. As the concentration of CO is increased, the frequency of successful collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if the desired product is not thermodynamically favored, the end-product can be obtained if it is continuously removed from the solution.

The effect of a change in concentration is often exploited synthetically for condensation reactions (i.e., reactions that extrude water) that are equilibrium processes (e.g., formation of an ester from carboxylic acid and alcohol or an imine from an amine and aldehyde). This can be achieved by physically sequestering water, by adding desiccants like anhydrous magnesium sulfate or molecular sieves, or by continuous removal of water by distillation, often facilitated by a Dean-Stark apparatus.

Effect of change in temperature Edit

The effect of changing the temperature in the equilibrium can be made clear by 1) incorporating heat as either a reactant or a product, and 2) assuming that an increase in temperature increases the heat content of a system. When the reaction is exothermic (ΔH is negative and energy is released), heat is included as a product, and when the reaction is endothermic (ΔH is positive and energy is consumed), heat is included as a reactant. Hence, whether increasing or decreasing the temperature would favor the forward or the reverse reaction can be determined by applying the same principle as with concentration changes.

Take, for example, the reversible reaction of nitrogen gas with hydrogen gas to form ammonia:

Because this reaction is exothermic, it produces heat:

If the temperature were increased, the heat content of the system would increase, so the system would consume some of that heat by shifting the equilibrium to the left, thereby producing less ammonia. More ammonia would be produced if the reaction were run at a lower temperature, but a lower temperature also lowers the rate of the process, so, in practice (the Haber process) the temperature is set at a compromise value that allows ammonia to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable.

In exothermic reactions, an increase in temperature decreases the equilibrium constant, K, whereas in endothermic reactions, an increase in temperature increases K.

Le Chatelier's principle applied to changes in concentration or pressure can be understood by giving K a constant value. The effect of temperature on equilibria, however, involves a change in the equilibrium constant. The dependence of K on temperature is determined by the sign of ΔH. The theoretical basis of this dependence is given by the Van 't Hoff equation.

Effect of change in pressure Edit

The equilibrium concentrations of the products and reactants do not directly depend on the total pressure of the system. They may depend on the partial pressures of the products and reactants, but if the number of moles of gaseous reactants is equal to the number of moles of gaseous products, pressure has no effect on equilibrium.

Changing total pressure by adding an inert gas at constant volume does not affect the equilibrium concentrations (see Effect of adding an inert gas below).

Changing total pressure by changing the volume of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations (see §Effect of change in volume below).

Effect of change in volume Edit

Changing the volume of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations. With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable [6] and with a pressure decrease due to an increase in volume, the side with more moles is more favorable. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical equation.

Considering the reaction of nitrogen gas with hydrogen gas to form ammonia:

N2 + 3 H2 4 moles ⇌ 2 NH3 2 moles ΔH = -92kJ mol −1

Note the number of moles of gas on the left-hand side and the number of moles of gas on the right-hand side. When the volume of the system is changed, the partial pressures of the gases change. If we were to decrease pressure by increasing volume, the equilibrium of the above reaction will shift to the left, because the reactant side has a greater number of moles than does the product side. The system tries to counteract the decrease in partial pressure of gas molecules by shifting to the side that exerts greater pressure. Similarly, if we were to increase pressure by decreasing volume, the equilibrium shifts to the right, counteracting the pressure increase by shifting to the side with fewer moles of gas that exert less pressure. If the volume is increased because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.

Effect of adding an inert gas Edit

Effect of a catalyst Edit

A catalyst increases the rate of a reaction without being consumed in the reaction. The use of a catalyst does not affect the position and composition of the equilibrium of a reaction, because both the forward and backward reactions are sped up by the same factor.

For example, consider the Haber process for the synthesis of ammonia (NH3):

In the above reaction, iron (Fe) and molybdenum (Mo) will function as catalysts if present. They will accelerate any reactions, but they do not affect the state of the equilibrium.

Le Chatelier's principle refers to states of thermodynamic equilibrium. The latter are stable against perturbations that satisfy certain criteria this is essential to the definition of thermodynamic equilibrium.

It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

For this, a state of thermodynamic equilibrium is most conveniently described through a fundamental relation that specifies a cardinal function of state, of the energy kind, or of the entropy kind, as a function of state variables chosen to fit the thermodynamic operations through which a perturbation is to be applied. [7] [8] [9]

In theory and, nearly, in some practical scenarios, a body can be in a stationary state with zero macroscopic flows and rates of chemical reaction (for example, when no suitable catalyst is present), yet not in thermodynamic equilibrium, because it is metastable or unstable then Le Chatelier's principle does not necessarily apply.

A body can also be in a stationary state with non-zero rates of flow and chemical reaction sometimes the word "equilibrium" is used in reference to such states, though by definition they are not thermodynamic equilibria. Sometimes, it is proposed to consider Le Chatelier's principle for such states. For this exercise, rates of flow and of chemical reaction must be considered. Such rates are not supplied by equilibrium thermodynamics. For such states, it has turned out to be difficult or unfeasible to make valid and very general statements that echo Le Chatelier's principle. [10] Prigogine and Defay demonstrate that such a scenario may or may not exhibit moderation, depending upon exactly what conditions are imposed after the perturbation. [11]

In economics, a similar concept also named after Le Chatelier was introduced by American economist Paul Samuelson in 1947. There the generalized Le Chatelier principle is for a maximum condition of economic equilibrium: Where all unknowns of a function are independently variable, auxiliary constraints—"just-binding" in leaving initial equilibrium unchanged—reduce the response to a parameter change. Thus, factor-demand and commodity-supply elasticities are hypothesized to be lower in the short run than in the long run because of the fixed-cost constraint in the short run. [12]

Since the change of the value of an objective function in a neighbourhood of the maximum position is described by the envelope theorem, Le Chatelier's principle can be shown to be a corollary thereof. [13]


Procedure

Materials and Equipment

Equipment: 10 small test tubes, test tube rack, test tube holder, Bunsen burner, 2 medium-sized beakers (for stock solutions), 10-mL graduated cylinder, wash bottle, stirring rod, and scoopula.

All of the acids and bases used in this experiment ((ce), (ce), (ce) and (ce)) can cause chemical burns. In particular, concentrated 12 M (ce) is extremely dangerous! If any of these chemicals spill on you, immediately rinse the affected area under running water and notify your instructor. Also note that direct contact with silver nitrate ((ce)) will cause dark discolorations to appear on your skin. These spots will eventually fade after repeated rinses in water. Finally, in Part 4 you will be heating a solution in a test tube directly in a Bunsen burner flame. If the solution is overheated it will splatter out of the tube, so be careful not to point the tube towards anyone while heating.

Experimental Procedure

Record all observations on your report form. These should include, but not be limited to, color changes and precipitates. Note that solution volumes are approximate for all reactions below. Dispose of all chemical waste in the plastic container in the hood.

Part 1: Saturated Sodium Chloride Solution

  1. Place 3-mL of saturated (ce) (aq) into a small test tube.
  2. Carefully add concentrated 12 M (ce) (aq) drop-wise to the solution in the test tube until a distinct change occurs. Record your observations.

Part 2: Acidified Chromate Solution

  1. Place 3-mL of 0.1 M (ce) (aq) into a small test tube.
  2. Add an equal amount of 6 M (ce) (aq) to this solution. Record your observations.
  3. Now add 10% (ce) (aq) drop-wise until the original color is returned. Record your observations.
  4. Here the added sodium hydroxide is effectively removing acidic hydrogen ions from the equilibrium system via a neutralization reaction:

Part 3: Aqueous Ammonia Solution

Instructor Prep: At the beginning of lab prepare a stock solution of aqueous ammonia. Add 4 drops of concentrated 15 M (ce) (aq) and 3 drops of phenolphthalein to a 150-mL (medium) beaker, top it up with 100-mL of distilled water, and mix with a stirring rod. Label the beaker and place it on the front desk. The entire class will then use this stock solution in Part 3.

  1. Place 3-mL of the prepared stock solution into a small test tube.
  2. Add a medium scoop of (ce) powder to the solution in this test tube. Record your observations.

Part 4: Cobalt(II) Chloride Solution

  1. Place 3-mL of 0.1 M (ce) (aq) into 3 small test tubes. Label these test tubes 1-3.
  2. The solution in test tube #1 remains untouched. It is a control for comparison with other tubes.
  3. To the solution in test tube #2, carefully add concentrated 12 M (ce) (aq) drop-wise until a distinct color change occurs. Record your observations.
  4. To the solution in test tube #3, first add a medium scoop of solid (ce). Then heat this solution directly in your Bunsen burner flame (moderate temperature). Firmly hold test tube #3 with your test tube holder, and waft it back and forth through the flame (to prevent overheating and &ldquobumping&rdquo) for about 30 seconds, or, until a distinct change occurs. Record your observations. Then cool the solution in test tube #3 back to room temperature by holding it under running tap water, and again record your observations.

Part 5: Iron(III) Thiocyanate Solution

Instructor Prep: At the beginning of lab prepare a stock solution of iron(III) thiocyanate. Add 1-mL of 0.1 M (ce) (aq) and 1-mL of 0.1 M (ce) (aq) to a 150-mL (medium) beaker, top it up with 100-mL of distilled water, and mix with a stirring rod. Label the beaker and place it on the front desk. The entire class will then use this stock solution in Part 5.

  1. Place 3-mL of the prepared stock solution into 4 small test tubes. Label these test tubes 1-4.
  2. The solution in test tube #1 remains untouched. It is a control for comparison with other tubes.
  3. To the solution in test tube #2, add 1-mL of 0.1 M (ce) (aq). Record your observations.
  4. To the solution in test tube #3, add 1-mL of 0.1 M (ce) (aq). Record your observations.
  5. To the solution in test tube #4, add 0.1 M (ce) (aq) drop-wise until all the color disappears. A light precipitate may also appear. Record your observations. Here the added silver nitrate is effectively removing thiocyanate ions from the equilibrium system via a precipitation reaction:

First, consider a factor that does not affect equilibrium: pure substances. If a pure liquid or solid is involved in equilibrium, it is considered to have an equilibrium constant of 1 and is excluded from the equilibrium constant. For example, except in highly concentrated solutions, pure water is considered to have an activity of 1. Another example is solid carbon, which may be form by the reaction of two carbom monoxide molecules to form carbon dioxide and carbon.

Factors that do affect equilibrium include:

  • Adding reactant or product or a change in concentration affects equilibrium. Adding reactant can drive equilibrium to the right in a chemical equation, where more product forms. Adding product can drive equilibrium to the left, as more reactant forms.
  • Changing the temperature alters equilibrium. Increasing temperature always shifts chemical equilibrium in the direction of the endothermic reaction. Decreasing temperature always shifts equilibrium in the direction of the exothermic reaction.
  • Changing the pressure affects equilibrium. For example, decreasing the volume of a gas system increases its pressure, which increases the concentration of both reactants and products. The net reaction will see to lower the concentration of gas molecules.

Le Chatelier's principle may be used to predict the shift in equilibrium resulting from applying a stress to the system. Le Chatelier's principle states that a change to a system in equilibrium will cause a predictable shift in equilibrium to counteract the change. For example, adding heat to a system favors the direction of the endothermic reaction because this will act to reduce the amount of heat.


Chemical Equilibrium - PowerPoint PPT Presentation

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Acid/Base Equilibrium

Acids and bases have a chemical equilibrium in solution. At chemical equilibrium, the products and reactants have reached a state of balance. Reactions may still be taking place within the sample, but the forward and reverse reactions are taking place at the same rate, so the concentrations of the products and reactants are not changing with time.

Concepts related to acid base equilibrium include designing buffers and plotting pH curves.

Additionally, determining the acid base equilibrium can help in predicting the products of a reaction and the relative concentration of the products, as well as aiding in the identification of the weaker acid and base.

Weak acids and bases are common in nature. For example, many insects of the order hymenoptera, including bees, wasps, and ants, use a weak acid (formic acid) as a weapon when they bite or sting. As a result, many animals avoid contact with these insects. In order for the weaponized acid to be effective, it must produce enough hydrogen ions to wound the bee's enemy, but not enough to dissolve the bee's stinger and organs. Formic acid

This defense occasionally backfires, as other animals have adapted their own pH in response to formic acid. Anteaters, a species well known for feasting on anthills, do not produce the hydrochloric acid (a strong acid) that most animals rely on for digestion. The anteaters are attracted to the ants, because they need the formic acid to break up their stomach contents.

Giant Anteater [1]


Watch the video: Chapter 5 Gases - Part 3 u0026 Chapter 13 Chemical Equilibrium - Part 1 (August 2022).