Chemical equilibrium: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>DemocraticLuntz
Reverted 2 good faith edits by 120.60.22.3 using STiki
en>Dirac66
m in the concentrations → in concentrations
Line 1: Line 1:
{{Refimprove|date=March 2009}}


In a [[chemical reaction]], '''chemical equilibrium''' is the state in which both reactants and products are present at [[concentration]]s which have no further tendency to change with time.<ref name=Atkins>Peter Atkins and Julio de Paula, ''Atkins' Physical Chemistry'', 8th edition (W.H. Freeman 2006, ISBN 0-7167-8759-8) p.200-202</ref> Usually, this state results when the forward reaction proceeds at the same rate as the [[Reversible reaction|reverse reaction]]. The [[reaction rate]]s of the forward and backward reactions are generally not zero, but equal. Thus, there are no net changes in the concentrations of the reactant(s) and product(s). Such a state is known as [[dynamic equilibrium]].<ref name=aj/><ref>{{GoldBookRef|title=chemical equilibrium|file=C01023}}</ref>


== Historical introduction ==
[http://www.sweetmontreal.ca/ sweetmontreal.ca]Do you feel like you're dealing with lots of pressure? Do you want some relaxing in your daily life? Among the best strategies to chill out gets a fantastic restorative massage. Discover the things you want to look for to acquire a great therapeutic massage.<br><br>Often be delicate when giving someone else information. Even if the man or woman you might be massaging complains, you should prevent implementing too much pressure to their muscles and joint parts. Unless you have been trained in message treatment, you will probably damage them rather than to relieve their pain by being a lot more forceful.<br><br>Explore your preferences along with your specialist. Some individuals may feel a little bit uncomfortable at the prospect of receiving a restorative massage mainly because they really feel out of hand. Explain to the masseuse about any tender regions, including your rear, knees or the neck and throat, to enable them to prevent them or treat them consequently. And don't forget to make sure they know of your own choices for things such as track record tunes, place lighting and temp. The full position is for you to be able to unwind, so be sure the situations match you.<br><br>Should you practical experience stress or some other adverse inner thoughts, you should try rubbing certain things of your body. Focus on your neural endings to acquire your body to unwind. Vulnerable nerve endings can be found coupled your spinal column, the top of the your face and around your facial skin. You should try carefully tapping on top of your face, your brow, cheeks and chin before taking an in-depth air.<br><br>If you are supplying a restorative massage, try to achieve the individual you will be rubbing in as couple of content articles of garments as is possible. Should they be unpleasant, they can use a towel to feel safer constantly. Possessing parts of the body exposed is not only calming, but it additionally facilitates your career.<br><br>Stay away from positioning your breath when your massage therapist is working. Maybe you have knots which are very painful, however the specialist could not massage them out properly if you hold your breath. Try and inhale with the muscles pain, and employ a mix of deep breathing and visualization. In your thoughts, snapshot the region and imagine the knot can be a lump of butter, melting within the hot sun.<br><br>Use candle lights setting the mood. Candle lights offer simple illumination, whilst concurrently building a calming ambiance. Fragrant candle lights will also be recommended. Place them uniformly throughout the room, at diverse elevations. Additionally it is vital that you be safe when using candle lights. Keep them away from holding cloth for example drapes.<br><br>Talk with your masseuse. Your masseuse is likely not just a thoughts reader, so inform them if an area of your body demands a little extra interest. Whether or not your shoulders can be a chaos from getting hunched over a workplace or even your legs are tender from your marathon exercise routine, articulate up.<br><br>Raise the advantages of your exercise regimen by rubbing the body both before and after  escort in montreal training. Massages before training can boost blood flow to muscles. Make use of fists to make pummeling motions in your forearms and hip and legs. Carrying out massages soon after training can help accelerate muscles recovery and better waste materials eradication. You can do these by rubbing together your muscles using your palms or fists moving to your coronary heart.<br><br>Just how much should you idea a massage therapist? If you're at a motel or a community day spa, 20% is fairly common. Once they work from home, every single counselor will typically set up their own personal guidelines, but 20% is acceptable. When they operate in a massage therapy medical center, ask them with regard to their regulations - many will range from the idea within the cost.<br><br>Probably the most satisfying connecting pursuits involving infant and mother or father, or another loved one, can be a massage. Just pick up a jar of child essential oil, take off your jewellery and be sure the hands are hot, and begin at the top or underside. What you do is likely to be pleasurable if you're gentle.<br><br>Aromatherapy therapeutic massage employs fragrant therapeutic massage fats as well as mild kneading motions. These scented skin oils add more an additional measurement for the standard therapeutic massage by including the sense of aroma in the massage therapy. Aromatherapy usually employs lavender, eucalyptus  escort in montreal and chamomile. This type of restorative massage is ideal for men and women enduring for pressure related pain.<br><br>If you have to relax, it really is tough to surpass a massage. Use the suggestions previously mentioned and revel in an excellent restorative massage when you want one. Begin using these recommendations to make sure you don't neglect them.<br><br>If you enjoyed this short article and you would certainly like to obtain additional info pertaining to book escorts, [http://www.xxxmontrealescorts.com try these guys out], kindly check out the web-site.
 
[[File:Burette.svg|thumb|right|100px|[[Burette]], a common laboratory apparatus for carrying out [[titration]], an important experimental technique in equilibrium and [[analytical chemistry]].]]
The concept of chemical equilibrium was developed after [[Berthollet]] (1803) found that some [[chemical reaction]]s are [[Reversible reaction|reversible]]. For any reaction mixture to exist at equilibrium, the [[reaction rate|rates]] of the forward and backward (reverse) reactions are equal. In the following [[chemical equation]] with arrows pointing both ways to indicate equilibrium, A and B are [[reactant]] chemical species, S and T are product species, and [[Alpha (letter)|α]], [[Beta (letter)|β]], [[sigma|σ]], and [[tau|τ]] are the [[stoichiometric coefficient]]s of the respective reactants and products:
:<math> \alpha A + \beta B \rightleftharpoons \sigma S + \tau T</math>
 
The equilibrium position of a reaction is said to lie "far to the right" if, at equilibrium, nearly all the reactants are consumed. Conversely the equilibrium position is said to be "far to the left" if hardly any product is formed from the reactants.
 
[[Cato Maximilian Guldberg|Guldberg]] and [[Peter Waage|Waage]] (1865), building on Berthollet’s ideas, proposed the [[law of mass action]]:
 
:<math>\mbox{forward reaction rate} = k_+ {A}^\alpha{B}^\beta \,\!</math>
:<math>\mbox{backward reaction rate} = k_{-} {S}^\sigma{T}^\tau \,\!</math>
 
where A, B, S and T are [[activity (chemistry)|active masses]] and k<sub>+</sub> and k<sub>−</sub> are [[rate constant]]s. Since at equilibrium forward and backward rates are equal:
 
:<math> k_+ \left\{ A \right\}^\alpha \left\{B \right\}^\beta = k_{-} \left\{S \right\}^\sigma\left\{T \right\}^\tau \,</math>
 
and the ratio of the rate constants is also a constant, now known as an [[equilibrium constant]].
:<math>K_c=\frac{k_+}{k_-}=\frac{\{S\}^\sigma \{T\}^\tau } {\{A\}^\alpha \{B\}^\beta}</math>
 
By convention the products form the [[numerator]].
However, the [[law of mass action]] is valid only for concerted one-step reactions that proceed through a single [[transition state]] and is '''not valid in general''' because [[reaction rate#Rate equation|rate equations]] do not, in general, follow the [[stoichiometry]] of the reaction as Guldberg and Waage had proposed (see, for example, [[nucleophilic aliphatic substitution]] by S<sub>N</sub>1 or reaction of [[hydrogen]] and [[bromine]] to form [[hydrogen bromide]]). Equality of forward and backward reaction rates, however, is a [[Necessary and sufficient conditions|necessary condition]] for chemical equilibrium, though it is not [[Necessary and sufficient conditions|sufficient]] to explain why equilibrium occurs.
 
Despite the failure of this derivation, the equilibrium constant for a reaction is indeed a constant, independent of the activities of the various species involved, though it does depend on temperature as observed by the [[van 't Hoff equation]]. Adding a [[catalyst]] will affect both the forward reaction and the reverse reaction in the same way and will not have an effect on the equilibrium constant. The catalyst will speed up both reactions thereby increasing the speed at which equilibrium is reached.<ref name=aj>Atkins, Peter W and Jones, Loretta  ''Chemical Principles: The Quest for Insight'' 2nd Ed. ISBN 0-7167-9903-0</ref><ref>''Chemistry: Matter and Its Changes'' James E. Brady , Fred Senese 4th Ed. ISBN 0-471-21517-1</ref>
 
Although the macroscopic equilibrium concentrations are constant in time, reactions do occur at the molecular level. For example, in the case of [[acetic acid]] dissolved in water and forming [[acetate]] and [[hydronium]] ions,
:CH<sub>3</sub>CO<sub>2</sub>H + H<sub>2</sub>O {{unicode|⇌}} CH<sub>3</sub>CO<sub>2</sub><sup>−</sup> + H<sub>3</sub>O<sup>+</sup>
a proton may hop from one molecule of acetic acid on to a water molecule and then on to an acetate anion to form another molecule of acetic acid and leaving the number of acetic acid molecules unchanged. This is an example of [[dynamic equilibrium]]. Equilibria, like the rest of thermodynamics, are statistical phenomena, averages of microscopic behavior.
 
'''[[Le Chatelier's principle]]''' (1884) gives an idea of the behavior of an equilibrium system when changes to its reaction conditions occur. ''If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to partially reverse the change''. For example, adding more S from the outside will cause an excess of products, and the system will try to counteract this by increasing the reverse reaction and pushing the equilibrium point backward (though the equilibrium constant will stay the same).
 
If [[mineral acid]] is added to the acetic acid mixture, increasing the concentration of hydronium ion, the amount of dissociation must decrease as the reaction is driven to the left in accordance with this principle. This can also be deduced from the equilibrium constant expression for the reaction:
:<math>K=\frac{\{CH_3CO_2^-\}\{H_3O^+\}} {\{CH_3CO_2H\}}</math>
If {H<sub>3</sub>O<sup>+</sup>} increases {CH<sub>3</sub>CO<sub>2</sub>H} must increase and {CH<sub>3</sub>CO<sub>2</sub><sup>−</sup>} must decrease. The H<sub>2</sub>O is left out as it is a pure liquid and its concentration is undefined.
 
A quantitative version is given by the [[reaction quotient]].
 
[[Josiah Willard Gibbs|J. W. Gibbs]] suggested in 1873 that equilibrium is attained when the [[chemical potential|Gibbs free energy]] of the system is at its minimum value (assuming the reaction is carried out at constant temperature and pressure). What this means is that the derivative of the Gibbs energy with respect to [[reaction coordinate]] (a measure of the [[extent of reaction]] that has occurred, ranging from zero for all reactants to a maximum for all products) vanishes, signalling a [[stationary point]]. This derivative is called the reaction Gibbs energy (or energy change) and corresponds to the difference between the [[chemical potential]]s of reactants and products at the composition of the reaction mixture.<ref name=Atkins/> This criterion is both necessary and sufficient. If a mixture is not at equilibrium, the liberation of the excess Gibbs energy (or [[Helmholtz energy]] at constant volume reactions) is the “driving force” for the composition of the mixture to change until equilibrium is reached. The equilibrium constant can be related to the standard [[Gibbs energy|Gibbs free energy]] change for the reaction by the equation
 
:<math>
\Delta_rG^\ominus = -RT \ln K_{eq}
</math>
 
where R is the [[universal gas constant]] and T the [[temperature]].
 
When the reactants are [[solution|dissolved]] in a medium of high [[ionic strength]] the quotient of [[activity coefficient]]s may be taken to be constant. In that case the '''concentration quotient''', K<sub>c</sub>,
:<math>K_c=\frac{[S]^\sigma [T]^\tau } {[A]^\alpha [B]^\beta}</math>
where [A] is the [[concentration]] of A, etc., is independent of the [[analytical concentration]] of the reactants. For this reason, equilibrium constants for [[solution]]s are usually [[Determination of equilibrium constants|determined]] in media of high ionic strength. K<sub>c</sub> varies with ionic strength, temperature and pressure (or volume). Likewise K<sub>p</sub> for gases depends on [[partial pressure]]. These constants are easier to measure and encountered in high-school chemistry courses.
 
==Thermodynamics==
 
The relation between the Gibbs energy and the equilibrium constant can be found by considering [[chemical potential]]s.<ref name = Atkins/>
 
At constant temperature and pressure, the [[Gibbs free energy]], '''G''', for the reaction depends only on the [[extent of reaction]]: '''ξ''' (Greek letter [[Xi (letter)|xi]]), and can only decrease according to the [[second law of thermodynamics]]. It means that the derivative of '''G''' with '''ξ''' must be negative if the reaction happens; at the equilibrium the derivative being equal to zero.
:<math>\left(\frac {dG}{d\xi}\right)_{T,p} = 0~</math>: equilibrium
 
At constant temperature and volume, one must consider the [[Helmholtz free energy]] for the reaction: '''A'''; and at constant internal energy and volume, one must consider the entropy for the reaction: '''S'''.
 
In this article only the constant pressure case is considered. The constant volume case is important in [[geochemistry]] and [[atmospheric chemistry]] where pressure variations are significant. Note that, if reactants and products were in standard state (completely pure), then there would be no reversibility and no equilibrium. The mixing of the products and reactants contributes a large entropy (known as [[entropy of mixing]]) to states containing equal mixture of products and reactants. The combination of the standard Gibbs energy change and the Gibbs energy of mixing determines the equilibrium state.<ref>{{cite journal|last1=Schultz|first1=Mary Jane|title=Why Equilibrium? Understanding Entropy of Mixing|journal=Journal of Chemical Education|volume=76|pages=1391|year=1999|doi=10.1021/ed076p1391|issue=10|bibcode = 1999JChEd..76.1391S }}</ref><ref>{{cite journal|last1=Clugston|first1=Michael J.|title=A mathematical verification of the second law of thermodynamics from the entropy of mixing|journal=Journal of Chemical Education|volume=67|pages=203|year=1990|doi=10.1021/ed067p203|issue=3|bibcode = 1990JChEd..67Q.203C }}</ref>
 
In general an equilibrium system is defined by writing an equilibrium equation for the reaction
 
:<math> \alpha A + \beta B \rightleftharpoons \sigma S + \tau T</math>
 
In order to meet the thermodynamic condition for equilibrium, the Gibbs energy must be stationary, meaning that the derivative of G with respect to the [[extent of reaction]]: '''ξ''', must be zero. It can be shown that in this case, the sum of [[chemical potential]]s of the products is equal to the sum of those corresponding to the reactants. Therefore, the sum of the Gibbs energies of the reactants must be the equal to the sum of the Gibbs energies of the products.
:<math> \alpha \mu_A + \beta \mu_B = \sigma \mu_S + \tau \mu_T \,</math>
 
where [[Mu (letter)|μ]] is in this case a partial molar Gibbs energy, a [[chemical potential]]. The chemical potential of a reagent A is a function of the
[[Activity (chemistry)|activity]], {A} of that reagent.
 
:<math> \mu_A = \mu_{A}^{\ominus} + RT \ln\{A\} \,</math>, ( <math> \mu_{A}^{\ominus}~</math> is the '''standard chemical potential ).
 
Substituting expressions like this into the [[Gibbs energy|Gibbs energy equation]]:
 
:<math> dG = Vdp-SdT+\sum_{i=1}^k \mu_i dN_i </math> in the case of a [[closed system]].
Now
:<math> dN_i = \nu_i d\xi \,</math> ( <math> \nu_i~</math> corresponds to the [[Stoichiometric coefficient]] and <math> d\xi~</math> is the [[differential of a function|differential]] of the extent of reaction ).
 
At constant pressure and temperature we obtain:
 
:<math>\left(\frac {dG}{d\xi}\right)_{T,p} = \sum_{i=1}^k \mu_i \nu_i = \Delta_rG_{T,p}</math> which corresponds to the '''[[Gibbs free energy change]] for the reaction''' .
 
This results in:
 
:<math> \Delta_rG_{T,p} = \sigma \mu_{S} + \tau \mu_{T} - \alpha \mu_{A} - \beta \mu_{B} \,</math>.
 
By substituting the chemical potentials:
 
:<math> \Delta_rG_{T,p} = ( \sigma \mu_{S}^{\ominus} + \tau \mu_{T}^{\ominus} ) - ( \alpha \mu_{A}^{\ominus} + \beta \mu_{B}^{\ominus} ) + ( \sigma RT \ln\{S\} + \tau RT \ln\{T\} ) - ( \alpha RT \ln\{A\} + \beta RT \ln \{B\} ) </math>,
 
the relationship becomes:
 
:<math> \Delta_rG_{T,p}=\sum_{i=1}^k \mu_i^\ominus \nu_i + RT \ln \frac{\{S\}^\sigma \{T\}^\tau} {\{A\}^\alpha \{B\}^\beta} </math>
 
::<math>\sum_{i=1}^k \mu_i^\ominus \nu_i = \Delta_rG^{\ominus}</math>: which is the '''standard Gibbs energy change for the reaction'''. It is a constant at a given temperature, which can be calculated, using thermodynamical tables.
::<math> RT \ln \frac{\{S\}^\sigma \{T\}^\tau} {\{A\}^\alpha \{B\}^\beta} = RT \ln Q_r </math>
:::( <math>Q_r ~</math> is the [[reaction quotient]] when the system is not at equilibrium ).
 
Therefore
:<math>\left(\frac {dG}{d\xi}\right)_{T,p} = \Delta_rG_{T,p}= \Delta_rG^{\ominus} + RT \ln Q_r </math>
 
At equilibrium <math>\left(\frac {dG}{d\xi}\right)_{T,p} = \Delta_rG_{T,p} = 0 </math>
 
::<math> Q_r = K_{eq}~</math> ; the reaction quotient becomes equal to the [[equilibrium constant]].
 
leading to:
 
:<math> 0 = \Delta_rG^{\ominus} + RT \ln K_{eq} </math>
and
:<math> \Delta_rG^{\ominus} = -RT \ln K_{eq} </math>
 
<center>
'''Obtaining the value of the standard Gibbs energy change, allows the calculation of the equilibrium constant'''
</center>
 
[[File:Diag eq.svg|thumb|350px|right]]
 
===Addition of reactants or products===
For a reactional system at equilibrium: <math>Q_r = K_{eq}~</math>; <math>\xi = \xi_{eq}~</math>.
:If are modified activities of constituents, the value of the reaction quotient changes and becomes different from the equilibrium constant: <math>Q_r \neq K_{eq}~</math>
 
<math>\left(\frac {dG}{d\xi}\right)_{T,p} = \Delta_rG^{\ominus} + RT \ln Q_r~</math>
 
and
 
<math>\Delta_rG^{\ominus} = - RT \ln K_{eq}~</math>
 
then
 
<math>\left(\frac {dG}{d\xi}\right)_{T,p} = RT \ln \left(\frac {Q_r}{K_{eq}}\right)~</math>
 
*If activity of a reagent <math>i~</math> increases
 
<math>Q_r = \frac{\prod (a_j)^{\nu_j}}{\prod(a_i)^{\nu_i}}~</math>, the reaction quotient decreases.
:then
 
<math>Q_r < K_{eq}~</math> and <math>\left(\frac {dG}{d\xi}\right)_{T,p} <0~</math> : The reaction will shift to the right (i.e. in the forward direction, and thus more products will form).
 
*If activity of a product <math>j~</math> increases
:then
 
<math>Q_r > K_{eq}~</math> and <math>\left(\frac {dG}{d\xi}\right)_{T,p} >0~</math> : The reaction will shift to the left (i.e. in the reverse direction, and thus less products will form).
 
''Note'' that activities and equilibrium constants are dimensionless numbers.
 
===Treatment of activity===
 
The expression for the equilibrium constant can be rewritten as the product of a concentration quotient, ''K''<sub>c</sub> and an [[activity coefficient]] quotient, Γ.
:<math>K=\frac{{[S]} ^\sigma {[T]}^\tau ... } {{[A]}^\alpha {[B]}^\beta ...}
\times \frac{{\gamma_S} ^\sigma {\gamma_T}^\tau ... } {{\gamma_A}^\alpha {\gamma_B}^\beta ...} = K_c \Gamma</math>
 
[A] is the concentration of reagent A, etc. It is possible in principle to obtain values of the activity coefficients, γ. For solutions, equations such as the [[Debye-Hückel equation]] or extensions such as [[Davies equation]]<ref>C.W. Davies, ''Ion Association'', Butterworths, 1962</ref> [[Specific ion interaction theory]] or [[Pitzer equations]]<ref name="davies">I. Grenthe and H. Wanner, [http://www.nea.fr/html/dbtdb/guidelines/tdb2.pdf ''Guidelines for the extrapolation to zero ionic strength'']</ref>
may be used.<sup>[[#Software for chemical equilibria|Software (below)]]</sup>. However this is not always possible. It is common practice to assume that Γ is a constant, and to use the concentration quotient in place of the thermodynamic equilibrium constant. It is also general practice to use the term ''equilibrium constant'' instead of the more accurate ''concentration quotient''. This practice will be followed here.
 
For reactions in the gas phase [[partial pressure]] is used in place of concentration and [[fugacity coefficient]] in place of activity coefficient. In the real world, for example, when making [[Haber process|ammonia]] in industry, fugacity coefficients must be taken into account. Fugacity, ''f'', is the product of partial pressure and fugacity coefficient. The chemical potential of a species in the gas phase is given by
:<math>\mu = \mu^{\Theta} + RT \ln \left( \frac{f}{bar} \right) = \mu^{\Theta} + RT \ln \left( \frac{p}{bar} \right) + RT \ln \gamma </math>
so the general expression defining an equilibrium constant is valid for both solution and gas phases.
 
===Concentration quotients===
 
In aqueous solution, equilibrium constants are usually determined in the presence of an "inert" electrolyte such as [[sodium nitrate]] NaNO<sub>3</sub> or [[potassium perchlorate]] KClO<sub>4</sub>. The [[ionic strength]] of a solution is given by
:<math> I = \frac{1}{2}\sum_{i=1}^N c_i z_i^2 </math>
where ''c<sub>i</sub>'' and ''z<sub>i</sub>'' stands for the concentration and ionic charge of ion type ''i'', and the sum is taken over all the ''N'' types of charged species in solution. When the concentration of dissolved salt is much higher than the analytical concentrations of the reagents, the ions originating from the dissolved salt determine the ionic strength, and the ionic strength is effectively constant. Since activity coefficients depend on ionic strength the activity coefficients of the species are effectively independent of concentration. Thus, the assumption that [[Gamma|Γ]] is constant is justified. The concentration quotient is a simple multiple of the equilibrium constant.<ref>F.J.C. Rossotti and H. Rossotti, ''The Determination of Stability Constants'', McGraw-Hill, 1961</ref>
:<math> K_c = \frac{K}{\Gamma} </math>
However, ''K''<sub>c</sub> will vary with ionic strength. If it is measured at a series of different ionic strengths the value can be extrapolated to zero ionic strength.<ref name="davies"/> The concentration quotient obtained in this manner is known, paradoxically, as a thermodynamic equilibrium constant.
 
To use a published value of an equilibrium constant in conditions of ionic strength different from the conditions used in its determination, the value should be adjusted<sup>[[#Software for chemical equilibria|Software (below)]]</sup>.
 
===Metastable mixtures===
 
A mixture may be appear to have no tendency to change, though it is not at equilibrium. For example, a mixture of [[Sulfur dioxide|SO<sub>2</sub>]] and [[oxygen|O<sub>2</sub>]] is [[metastable]] as there is a [[Activation energy|kinetic barrier]] to formation of the product, [[Sulfur trioxide|SO<sub>3</sub>]].
:2SO<sub>2</sub> + O<sub>2</sub> <math>\rightleftharpoons</math> 2SO<sub>3</sub>
 
The barrier can be overcome when a [[Catalysis|catalyst]] is also present in the mixture as in the [[contact process]], but the catalyst does not affect the equilibrium concentrations.
 
Likewise, the formation of [[bicarbonate]] from [[carbon dioxide]] and [[water]] is very slow under normal conditions
:CO<sub>2</sub> + 2H<sub>2</sub>O <math>\rightleftharpoons</math> HCO<sub>3</sub><sup>-</sup> +H<sub>3</sub>O<sup>+</sup>
but almost instantaneous in the presence of the catalytic [[enzyme]] [[carbonic anhydrase]].
 
==Pure substances==
 
When pure substances (liquids or solids) are involved in equilibria their activities do not appear in the equilibrium constant<ref name=CEM>''Concise Encyclopedia Chemistry'' 1994 ISBN 0-89925-457-8</ref> because their numerical values are considered one.
 
Applying the general formula for an equilibrium constant to the specific case of acetic acid one obtains
:CH<sub>3</sub>CO<sub>2</sub>H + H<sub>2</sub>O {{unicode|⇌}} CH<sub>3</sub>CO<sub>2</sub><sup>−</sup> + H<sub>3</sub>O<sup>+</sup>
 
:<math>K_c=\frac{[{CH_3CO_2}^-][{H_3O}^+]} {[{CH_3CO_2H}][{H_2O}]}</math>
 
It may be assumed that the concentration of water is constant. This assumption will be valid for all but very concentrated solutions. The equilibrium constant expression is therefore usually written as
:<math>K=\frac{[{CH_3CO_2}^-][{H_3O}^+]} {[{CH_3CO_2H}]}</math>
 
where now
 
<math>K=K_c \cdot [H_2O]\,</math>
 
a constant factor is incorporated into the equilibrium constant.
 
A particular case is the [[self-ionization of water]] itself
 
:<math>H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-</math>
 
The self-ionization constant of water is defined as
 
<math>K_w = [H^+][OH^-]\,</math>
 
It is perfectly legitimate to write [H<sup>+</sup>] for the [[hydronium ion]] concentration, since the state of [[solvation]] of the proton is constant (in dilute solutions) and so does not affect the equilibrium concentrations. K<sub>w</sub> varies with variation in ionic strength and/or temperature.
 
The concentrations of H<sup>+</sup> and OH<sup>-</sup> are not independent quantities. Most commonly [OH<sup>-</sup>] is replaced by ''K''<sub>w</sub>[H<sup>+</sup>]<sup>−1</sup> in equilibrium constant expressions which would otherwise include  [[hydroxide ion]].
 
Solids also do not appear in the equilibrium constant expression, if they are considered to be pure and thus their activities taken to be one. An example is the [[Boudouard reaction]]:<ref name=CEM/>
 
:<math>2CO \rightleftharpoons CO_2 + C </math>
 
for which the equation (without solid carbon) is written as:
 
:<math>K_c=\frac{[CO_2]} {[CO]^2}</math>
 
==Multiple equilibria==
 
Consider the case of a dibasic acid H<sub>2</sub>A. When dissolved in water, the mixture will contain H<sub>2</sub>A, HA<sup>-</sup> and A<sup>2-</sup>. This equilibrium can be split into two steps in each of which one proton is liberated.
:<math>H_2A \rightleftharpoons HA^- + H^+ :K_1=\frac{[HA^-][H^+]} {[H_2A]}</math>
:<math>HA^- \rightleftharpoons A^{2-} + H^+ :K_2=\frac{[A^{2-}][H^+]} {[HA^-]}</math>
''K''<sub>1</sub> and'' K''<sub>2</sub> are examples of ''stepwise'' equilibrium constants. The ''overall'' equilibrium constant,<math>\beta_D</math>, is product of the stepwise constants.
:<math>H_2A \rightleftharpoons A^{2-} + 2H^+ :\beta_D = \frac{[A^{2-}][H^+]^2} {[H_2A]}=K_1K_2</math>
Note that these constants are [[Acid dissociation constant|dissociation constants]] because the products on the right hand side of the equilibrium expression are dissociation products. In many systems, it is preferable to use association constants.
:<math>A^{2-} + H^+ \rightleftharpoons HA^- :\beta_1=\frac {[HA^-]} {[A^{2-}][H^+]}</math>
:<math>A^{2-} + 2H^+ \rightleftharpoons H_2A :\beta_2=\frac {[H_2A]} {[A^{2-}][H^+]^2}</math>
β<sub>1</sub> and β<sub>2</sub> are examples of association constants. Clearly β<sub>1</sub> = 1/''K''<sub>2</sub> and β<sub>2</sub> = 1/β<sub>D</sub>; lg β<sub>1</sub> = pK<sub>2</sub> and lg β<sub>2</sub> = pK<sub>2</sub> + pK<sub>1</sub><ref>M.T. Beck, ''Chemistry of Complex Equilibria'', Van Nostrand, 1970. 2nd. Edition by M.T. Beck and I Nagypál, Akadémiai Kaidó, Budapest, 1990.</ref>
For multiple equilibrium systems, also see: theory of [[Response reactions]].
 
==Effect of temperature==
 
The effect of changing temperature on an equilibrium constant is given by the [[van 't Hoff equation]]
:<math>\frac {d\ln K} {dT} = \frac{{\Delta H_m}^{\Theta}} {RT^2}</math>
Thus, for [[exothermic]] reactions, (ΔH is negative) ''K'' decreases with an increase in temperature, but, for [[endothermic]] reactions, (ΔH is positive) ''K'' increases with an increase temperature. An alternative formulation is
:<math>\frac {d\ln K} {d(1/T)} = -\frac{{\Delta H_m}^{\Theta}} {R}</math>
At first sight this appears to offer a means of obtaining the standard molar enthalpy of the reaction by studying the variation of ''K'' with temperature. In practice, however, the method is unreliable because error propagation almost always gives very large errors on the values calculated in this way.
 
==Effect of electric and magnetic fields==
The effect of electric field on equilibrium has been studied by [[Manfred Eigen]]{{cn|reason=Eigen measured reaction *rates* near equilibrium by the electric-field jump method, not the effect on the *position* of equilibrium|date=July 2013}} among others.
 
==Types of equilibrium==
 
# In the gas phase. [[Rocket engine]]s<ref name="nasa">NASA Reference publication 1311 (1994), [http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19950013764_1995113764.pdf Computer Program for Calculation of Complex Chemical Equilibrium Compositions and Applications]</ref>
# The industrial synthesis such as [[ammonia]] in the [[Haber-Bosch process]] (depicted right) takes place through a succession of equilibrium steps including [[adsorption]] processes.[[File:HaberBoschProcess.png|300px|right|thumb|Haber-Bosch process]]
# [[atmospheric chemistry]]
# Seawater and other natural waters: [[Chemical oceanography]]
# Distribution between two phases
## [[LogD-Distribution coefficient]]: Important for pharmaceuticals where lipophilicity is a significant property of a drug
## [[Liquid-liquid extraction]], [[Ion exchange]], [[Chromatography]]
## [[Solubility equilibrium|Solubility product]]
## Uptake and release of oxygen by [[haemoglobin]] in blood
# Acid/base equilibria: [[Acid dissociation constant]], [[hydrolysis]], [[buffer solution]]s, [[pH indicator|indicators]], [[acid-base homeostasis]]
# Metal-ligand complexation: [[Chelation|sequestering agents]], [[chelation therapy]], [[Magnetic resonance imaging|MRI contrast reagents]], [[Schlenk equilibrium]]
# Adduct formation: [[Host-guest chemistry]], [[supramolecular chemistry]], [[molecular recognition]], [[dinitrogen tetroxide]]
# In certain [[oscillating reaction]]s, the approach to equilibrium is not asymptotically but in the form of a damped oscillation .<ref name="CEM"/>
# The related [[Nernst equation]] in electrochemistry gives the difference in electrode potential as a function of redox concentrations.
# When molecules on each side of the equilibrium are able to further react irreversibly in secondary reactions, the final product ratio is determined according to the [[Curtin-Hammett principle]].
 
In these applications, terms such as stability constant, formation constant, binding constant, affinity constant, association/dissociation constant are used. In biochemistry, it is common to give units for binding constants, which serve to define the concentration units used when the constant’s value was determined.
 
==Composition of a mixture==
 
When the only equilibrium is that of the formation of a 1:1 adduct as the composition of a mixture, there are any number of ways that the composition of a mixture can be calculated. For example, see [[ICE table]] for a traditional method of calculating the pH of a solution of a weak acid.
 
There are three approaches to the general calculation of the composition of a mixture at equilibrium.
 
#The most basic approach is to manipulate the various equilibrium constants until the desired concentrations are expressed in terms of measured equilibrium constants (equivalent to measuring chemical potentials) and initial conditions.
#Minimize the Gibbs energy of the system.<ref>This approach is described in detail in W. R. Smith and R. W. Missen, ''Chemical Reaction Equilibrium Analysis: Theory and Algorithms'', , Krieger Publishing, Malabar, Fla, 1991 (a reprint, with corrections, of the same title by [[John Wiley & Sons]], 1982). A comprehensive treatment of the theory of chemical reaction equilibria and its computation. Details at [http://www.mathtrek.com http://www.mathtrek.com/]</ref>
# Satisfy the equation of [[mass balance]]. The equations of mass balance are simply statements that demonstrate that the total concentration of each reactant must be constant by the law of [[conservation of mass]].
 
===Mass-balance equations===
 
In general, the calculations are rather complicated or complex. For instance, in the case of a dibasic acid, H<sub>2</sub>A dissolved in water the two reactants can be specified as the [[conjugate base]], A<sup>2-</sup>, and the [[hydronium|proton]], H<sup>+</sup>. The following equations of mass-balance could apply equally well to a base such as [[Ethylenediamine|1,2-diaminoethane]], in which case the base itself is designated as the reactant A:
:<math>T_A = [A] + [HA] +[H_2A] \,</math>
:<math>T_H = [H] + [HA] + 2[H_2A] - [OH] \,</math>
 
With T<sub>A</sub> the total concentration of species A. Note that it is customary to omit the ionic charges when writing and using these equations.
 
When the equilibrium constants are known and the total concentrations are specified there are two equations in two unknown "free concentrations" [A] and [H]. This follows from the fact that [HA]= β<sub>1</sub>[A][H], [H<sub>2</sub>A]= β<sub>2</sub>[A][H]<sup>2</sup> and [OH] = K<sub>w</sub>[H]<sup>−1</sup>
 
:<math> T_A = [A] + \beta_1[A][H] + \beta_2[A][H]^2 \,</math>
:<math> T_H = [H] + \beta_1[A][H] + 2\beta_2[A][H]^2 - K_w[H]^{-1} \,</math>
 
so the concentrations of the "complexes" are calculated from the free concentrations and the equilibrium constants.
General expressions applicable to all systems with two reagents, A and B would be
:<math>T_A=[A]+\sum_i{p_i \beta_i[A]^{p_i}[B]^{q_i}}</math>
:<math>T_B=[B]+\sum_i{q_i \beta_i[A]^{p_i}[B]^{q_i}}</math>
 
It is easy to see how this can be extended to three or more reagents.
 
====Polybasic acids====
 
The composition of solutions containing reactants A and H is easy to calculate as a function of [[pH|p[H]]]. When [H] is known, the free concentration [A] is calculated from the mass-balance equation in A. Here is an example of the results that can be obtained.
 
[[File:AL hydrolysis.png]]
 
This diagram, for the hydrolysis of the [[aluminium]] [[Lewis acid]] Al<sup>3+</sup><sub>aq</sub><ref>The diagram was created with the program [http://www.hyperquad.co.uk/hyss.htm HySS]</ref> shows the species concentrations for a 5×10<sup>−6</sup>M solution of an aluminium salt as a function of pH. Each concentration is shown as a percentage of the total aluminium.
 
====Solution and precipitation====
 
The diagram above illustrates the point that a [[Precipitation (chemistry)|precipitate]] that is not one of the main species in the solution equilibrium may be formed. At pH just below 5.5 the main species present in a 5μM solution of Al<sup>3+</sup> are [[aluminium hydroxide]]s Al(OH)<sup>2+</sup>, Al(OH)<sub>2</sub><sup>+</sup> and Al<sub>13</sub>(OH)<sub>32</sub><sup>7+</sup>, but on raising the pH [[aluminium hydroxide|Al(OH)<sub>3</sub>]] precipitates from the solution. This occurs because Al(OH)<sub>3</sub> has a very large [[lattice energy]]. As the pH rises more and more Al(OH)<sub>3</sub> comes out of solution. This is an example of [[Le Chatelier's principle]] in action: Increasing the concentration of the hydroxide ion causes more aluminium hydroxide to precipitate, which removes hydroxide from the solution. When the hydroxide concentration becomes sufficiently high the soluble aluminate, Al(OH)<sub>4</sub><sup>-</sup>, is formed.
 
Another common instance where precipitation occurs is when a metal cation interacts with an anionic ligand to form an electrically neutral complex. If the complex is [[hydrophobic]], it will precipitate out of water. This occurs with the [[nickel]] ion Ni<sup>2+</sup> and [[dimethylglyoxime]], (dmgH<sub>2</sub>): in this case the lattice energy of the solid is not particularly large, but it greatly exceeds the energy of [[solvation]] of the molecule Ni(dmgH)<sub>2</sub>.
 
===Minimization of free energy===
 
At equilibrium, ''G'' is at a minimum:
 
:<math>dG= \sum_{j=1}^m \mu_j\,dN_j = 0</math>
 
For a closed system, no particles may enter or leave, although they may combine in various ways. The total number of atoms of each element will remain constant. This means that the minimization above must be subjected to the constraints:
 
:<math>\sum_{j=1}^m a_{ij}N_j=b_i^0</math>
 
where <math>a_{ij}</math> is the number of atoms of element ''i'' in molecule ''j'' and ''b''<sub>i</sub><sup>0</sup> is the total number of atoms of element ''i'', which is a constant, since the system is closed. If there are a total of ''k'' types of atoms in the system, then there will be ''k'' such equations.
 
This is a standard problem in [[Optimization (mathematics)|optimisation]], known as [[constrained minimisation]]. The most common method of solving it is using the method of [[Lagrange multipliers]], also known as [[undetermined multipliers]] (though other methods may be used).
 
Define:
 
:<math>\mathcal{G}= G + \sum_{i=1}^k\lambda_i\left(\sum_{j=1}^m a_{ij}N_j-b_i^0\right)=0</math>
 
where the <math>\lambda_i</math> are the Lagrange multipliers, one for each element. This allows each of the <math>N_j</math> to be treated independently, and it can be shown using the tools of [[multivariate calculus]] that the equilibrium condition is given by
 
:<math>\frac{\partial \mathcal{G}}{\partial N_j}=0</math> &nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp; <math>\frac{\partial \mathcal{G}}{\partial \lambda_i}=0</math>
 
(For proof see [[Lagrange multipliers]])
 
This is a set of ''(m+k)'' equations in ''(m+k)'' unknowns (the <math>N_j</math> and the <math>\lambda_i</math>) and may, therefore, be solved for the equilibrium concentrations <math>N_j</math> as long as the chemical potentials are known as functions of the concentrations at the given temperature and pressure. (See [[Thermodynamic databases for pure substances]]).
 
This method of calculating equilibrium chemical concentrations is useful for systems with a large number of different molecules. The use of ''k'' atomic element conservation equations for the mass constraint is straightforward, and replaces the use of the stoichiometric coefficient equations.<ref name="nasa"/>
 
==See also==
*[[Autocatalytic reactions and order creation]]
*[[Benesi-Hildebrand method]]
*[[Determination of equilibrium constants]]
*[[Equilibrium constant]]
*[[Henderson–Hasselbalch equation]]
*[[Isotope fractionation]]
*[[Michaelis-Menten kinetics]]
*[[Standard electrode potential|Redox equilibria]]
*[[Steady state (chemistry)]]
*[[Thermodynamic databases for pure substances]]
 
==References==
{{Reflist|2}}
 
==Further reading==
{{Library resources box
|by=no
|onlinebooks=no
|others=no
|about=yes
|label=Chemical equilibrium}}
*F. Van Zeggeren and S.H. Storey, ''The Computation of Chemical Equilibria'', Cambridge University Press, 1970. Mainly concerned with gas-phase equilibria.
*D. J. Leggett (editor), ''Computational Methods for the Determination of Formation Constants'', Plenum Press, 1985.
*A.E. Martell and R.J. Motekaitis, ''The Determination and Use of Stability Constants'', Wiley-VCH, 1992.
*P. Gans, ''Stability Constants: Determination and Uses, an interactive CD, ''Protonic Software (Leeds), 2004
{{Chemical equilibria}}
 
{{DEFAULTSORT:Chemical Equilibrium}}
[[Category:Equilibrium chemistry]]
[[Category:Analytical chemistry]]
[[Category:Physical chemistry]]

Revision as of 03:10, 2 March 2014


sweetmontreal.caDo you feel like you're dealing with lots of pressure? Do you want some relaxing in your daily life? Among the best strategies to chill out gets a fantastic restorative massage. Discover the things you want to look for to acquire a great therapeutic massage.

Often be delicate when giving someone else information. Even if the man or woman you might be massaging complains, you should prevent implementing too much pressure to their muscles and joint parts. Unless you have been trained in message treatment, you will probably damage them rather than to relieve their pain by being a lot more forceful.

Explore your preferences along with your specialist. Some individuals may feel a little bit uncomfortable at the prospect of receiving a restorative massage mainly because they really feel out of hand. Explain to the masseuse about any tender regions, including your rear, knees or the neck and throat, to enable them to prevent them or treat them consequently. And don't forget to make sure they know of your own choices for things such as track record tunes, place lighting and temp. The full position is for you to be able to unwind, so be sure the situations match you.

Should you practical experience stress or some other adverse inner thoughts, you should try rubbing certain things of your body. Focus on your neural endings to acquire your body to unwind. Vulnerable nerve endings can be found coupled your spinal column, the top of the your face and around your facial skin. You should try carefully tapping on top of your face, your brow, cheeks and chin before taking an in-depth air.

If you are supplying a restorative massage, try to achieve the individual you will be rubbing in as couple of content articles of garments as is possible. Should they be unpleasant, they can use a towel to feel safer constantly. Possessing parts of the body exposed is not only calming, but it additionally facilitates your career.

Stay away from positioning your breath when your massage therapist is working. Maybe you have knots which are very painful, however the specialist could not massage them out properly if you hold your breath. Try and inhale with the muscles pain, and employ a mix of deep breathing and visualization. In your thoughts, snapshot the region and imagine the knot can be a lump of butter, melting within the hot sun.

Use candle lights setting the mood. Candle lights offer simple illumination, whilst concurrently building a calming ambiance. Fragrant candle lights will also be recommended. Place them uniformly throughout the room, at diverse elevations. Additionally it is vital that you be safe when using candle lights. Keep them away from holding cloth for example drapes.

Talk with your masseuse. Your masseuse is likely not just a thoughts reader, so inform them if an area of your body demands a little extra interest. Whether or not your shoulders can be a chaos from getting hunched over a workplace or even your legs are tender from your marathon exercise routine, articulate up.

Raise the advantages of your exercise regimen by rubbing the body both before and after escort in montreal training. Massages before training can boost blood flow to muscles. Make use of fists to make pummeling motions in your forearms and hip and legs. Carrying out massages soon after training can help accelerate muscles recovery and better waste materials eradication. You can do these by rubbing together your muscles using your palms or fists moving to your coronary heart.

Just how much should you idea a massage therapist? If you're at a motel or a community day spa, 20% is fairly common. Once they work from home, every single counselor will typically set up their own personal guidelines, but 20% is acceptable. When they operate in a massage therapy medical center, ask them with regard to their regulations - many will range from the idea within the cost.

Probably the most satisfying connecting pursuits involving infant and mother or father, or another loved one, can be a massage. Just pick up a jar of child essential oil, take off your jewellery and be sure the hands are hot, and begin at the top or underside. What you do is likely to be pleasurable if you're gentle.

Aromatherapy therapeutic massage employs fragrant therapeutic massage fats as well as mild kneading motions. These scented skin oils add more an additional measurement for the standard therapeutic massage by including the sense of aroma in the massage therapy. Aromatherapy usually employs lavender, eucalyptus escort in montreal and chamomile. This type of restorative massage is ideal for men and women enduring for pressure related pain.

If you have to relax, it really is tough to surpass a massage. Use the suggestions previously mentioned and revel in an excellent restorative massage when you want one. Begin using these recommendations to make sure you don't neglect them.

If you enjoyed this short article and you would certainly like to obtain additional info pertaining to book escorts, try these guys out, kindly check out the web-site.