Position+of+Equilibrium

toc =** The equilibrium constant (K ** c ** ) **= The rate of a reaction depends on the concentration of the species involved. For example the equilibrium reaction: A + B C + D At equilibrium the rate of the forwards reaction equals the rate of the reverse reaction: So K f .[A].[B] = k r .[C].[D] Rearranging this gives: 
 * The rate of the //forwards reaction// is found by **K** ** f ****.[A].[B]** (k f  is a constant and the square brackets mean concentration so that [A] means concentration of A)
 * The rate of the //reverse reaction// is found by **k** ** r .[C].[D] **

K f  and k r  are constants at any given temperature, thus their ratio must also be constant <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is called the equilibrium constant.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The equilibrium constant (k c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">) equals <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">For example:

aA + bB + cC +…. pP + qQ + rR + ……..

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The equilibrium constant is <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">For example:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">4NH 3 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">(g) + 5O 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (g) 4NO (g) + 6H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (g) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">For a reaction occurring in aqueous solution, the concentration of water in taken not taken into account when calculating the value of K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Eg. CO(H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O) 6 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 4Cl - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) COCl 4 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (l)

//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">However, if the reaction does not occur in aqueous solution the concentration of water needs to be taken into account. //

=** The different possible values of Kc and their implications **=


 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">>1 it means that the concentration of the products is greater than that of the reactants. In other words the equilibrium lies to the right (the products side)
 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">>>1 (which means that k c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is a very large value greater than one) it means that the reaction is almost going to completion. (since the concentration of reactants is much greater than that of the products)
 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"><1 it means that the concentration of reactants is greater than the concentration of products. In other words the equilibrium lies to the left (the reactants side)
 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"><<1 (which means that K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is a very small number less than one) it means the reaction is not occurred (since the concentration of reactants will be much greater than that of products).

=** Effect of conditions on the position of equilibrium **=

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the conditions on an equilibrium mixture are changed, the equilibrium is disturbed. Thus the concentration of reactants and products changes until the rates of the forwards and reverse reactions are equal again (and a new equilibrium is reached)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">//**Le Chatelier’s principle**// allows us to predict in what direction the equilibrium will change when the conditions are altered.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">It states that: “//when the conditions on the equilibrium mixture are altered, the equilibrium moves so as to counteract the change made//.” <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">(however, you don’t need to know what the principle states, you just have to apply it!!)

Some of the factors affecting the position of the equilibrium

Concentration:
<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the concentration of a species is increased the equilibrium moves to the opposite side causing the concentration of the species to decrease.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the concentration of a species is reduced, the equilibrium moves to that side, causing the concentration of that species to increase.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">(for example if you increase the concentration of reactants, the equilibrium is disturbed so the equilibrium will shift to the right to produce more products. As a result the concentration of the reactants decreases once again)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Concentration does not affect the value of the equilibrium constant (K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">) for any reaction.

Pressure:
<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the pressure is increased, the equilibrium shifts to the side with least moles of gas, causing the pressure to fall. And vice versa. If the pressure is reduced the equilibrium shifts to the side with the most moles of gas, causing the pressure to increase.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Eg: 2SO 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (g) + O 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (g) 2SO 3 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (g)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">In the above reaction 3 moles of gas are converted to 2 moles of gas


 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Therefore if the pressure is increased, the equilibrium shifts to the right (where there are fewer moles of gas)
 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the pressure is decreased, the equilibrium shifts to the left (where there are more moles of gas)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Pressure does not have any effect on the value of equilibrium constant K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> for any reaction.

Temperature:
<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the temperature is increased, the equilibrium shifts in the direction of the endothermic reaction.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the temperature is reduced, the equilibrium shifts in the direction of the exothermic reaction.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">N 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (g) + O 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (g) 2NO (g) ΔH= +180 KJ mol -1 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (this means that the forwards reaction is endothermic)


 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the temperature is increased equilibrium shifts to the right favouring the endothermic reaction. (the value of K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> increases, because more products are formed than reactants—since the forward and reverse reactions are not speeded up by the same amount as a result of the temperature change.)
 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If the temperature is decreased equilibrium shifts to the left favouring the exothermic reaction. (the value of K c <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">decreases, because more reactants are formed than products—since the forward and reverse reactions are not speeded up by the same amount as a result of the temperature change.)

// Remember that a catalyst in the equilibrium mixture does not shift the position of the equilibrium. A catalyst only reduces the activation energies for both the forwards and the reverse reactions by the same amount. Thus both reactions are speeded up. As a result a catalyst only causes the equilibrium to be reached faster but has no effect on the position of the equilibrium. //

By the end of this lesson you should be able to:
 * Deduce the equilibrium constant from the equation of a given reaction
 * Interpret the different values of the equilibrium constant as to how far the reaction has proceeded
 * Apply Le Chatelier's principle to find the effect of changing pressure, temperature and concentration on the equilibrium and the value of the equilibrium constant
 * Explain the effect of a catalyst on the position of the equilibrium


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