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Chemical Equilibrium - Scf

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“<strong>Chemical</strong><br />

<strong>Equilibrium</strong>”<br />

Chapter 14<br />

Hemoglobin<br />

• Hemoglobin is a protein (Hb) found in<br />

red blood cells that reacts with O 2<br />

– enhances the amount of O 2 that can be<br />

carried through the blood stream<br />

Hb + O 2 ⇔ HbO 2<br />

– the Hb represents the entire protein – it is<br />

not a chemical formula<br />

– the ⇔ represents that the reaction is in<br />

dynamic equilibrium<br />

Hemoglobin<br />

• The concentrations of Hb, O 2 , and HbO 2 are<br />

all interdependent<br />

• The relative amounts of Hb, O 2 , and HbO 2 at<br />

equilibrium are related to a constant called<br />

the equilibrium constant, K<br />

• Changing the concentration of any one of<br />

these necessitates changing the other<br />

concentrations to reestablish equilibrium<br />

1


In the lungs, with high<br />

concentration of O 2 ,<br />

the equilibrium shifts<br />

to combine the Hb and<br />

O 2 together to make<br />

more HbO 2<br />

Hemoglobin<br />

In the cells, with low<br />

concentration of O 2 ,<br />

the equilibrium shifts<br />

to break down the<br />

HbO 2 and increase the<br />

amount of free O 2<br />

Fetal Hemoglobin<br />

Fetal hemoglobin’s<br />

equilibrium constant is<br />

larger than adult<br />

hemoglobin<br />

Because fetal hemoglobin<br />

is more efficient at<br />

binding O 2, O 2 is<br />

transferred to the fetal<br />

hemoglobin from the<br />

mother’s hemoglobin in<br />

the placenta<br />

Reactions<br />

• When a reaction starts, the reactants are consumed and<br />

products are made<br />

– In the forward reaction = reactants → products<br />

– reactant concentrations decrease and the product<br />

concentrations increase as reactant concentration<br />

decreases, the forward reaction rate decreases<br />

• Eventually, the products can react to reform some of the<br />

reactants<br />

– In the reverse reaction = products → reactants<br />

– assuming the products are not allowed to escape as product<br />

concentration increases, the reverse reaction rate increases<br />

• Processes that proceed in both the forward and reverse<br />

direction are said to be reversible<br />

– reactants ⇔ products<br />

2


Dynamic <strong>Equilibrium</strong><br />

• As the forward reaction slows and the<br />

reverse reaction accelerates, eventually<br />

they reach the same rate<br />

• Dynamic equilibrium is the condition<br />

where the rates of the forward and reverse<br />

reactions are equal<br />

• Once the reaction reaches equilibrium, the<br />

concentrations of all the chemicals remain<br />

constant<br />

<strong>Equilibrium</strong> ≠ Equal<br />

• The rates of the forward and reverse<br />

reactions are equal at equilibrium<br />

• That does not mean the concentrations of<br />

reactants and products are equal<br />

– Some reactions reach equilibrium only after almost<br />

all the reactant molecules are consumed – we say<br />

the position of equilibrium favors the products<br />

– Other reactions reach equilibrium when only a<br />

small percentage of the reactant molecules are<br />

consumed – we say the position of equilibrium<br />

favors the reactants<br />

An Analogy: Population<br />

Changes<br />

3


<strong>Equilibrium</strong> Constant<br />

• Even though the concentrations of reactants and<br />

products are not equal at equilibrium, there is a<br />

relationship between them<br />

• the relationship between the chemical equation<br />

and the concentrations of reactants and products<br />

can be seen in the <strong>Equilibrium</strong> Expression<br />

• Consider: aA + bB ⇔ cC + dD<br />

– the lowercase letters represent the coefficients of the<br />

balanced chemical equation<br />

• K is called the equilibrium constant is equal to the<br />

product concentration raised to their coefficient<br />

over the reactants raised to their coeffcients K is<br />

unitless<br />

Writing <strong>Equilibrium</strong> Constant<br />

Expressions<br />

• For the reaction<br />

• aA (aq) + bB (aq) ⇔ cC (aq) + dD (aq) the<br />

equilibrium constant expression is<br />

• For the reaction<br />

2 N 2O 5 ⇔ 4 NO 2 + O 2 the<br />

equilibrium constant<br />

expression is:<br />

What Does the Value of K eq<br />

Imply?<br />

• When the value of K eq >> 1, we know that<br />

when the reaction reaches equilibrium there<br />

will be many more product molecules present<br />

than reactant molecules<br />

• When the value of K eq


Large K<br />

Small K<br />

<strong>Equilibrium</strong> Constants for<br />

Reactions Involving Gases<br />

• The concentration of a gas in a mixture is<br />

proportional to its partial pressure<br />

• Therefore, the equilibrium constant can be expressed<br />

as the ratio of the partial pressures of the gases<br />

• For aA(g) + bB(g) ⇔ cC(g) + dD(g) the equilibrium<br />

constant expressions are<br />

5


K c and K p<br />

• In calculating K p, the partial pressures are always in<br />

atm<br />

• The values of K p and K c are not necessarily the same<br />

– because of the difference in units<br />

– K p = K c when Δn = 0<br />

• The relationship between them is: K p = K c (RT) Δn<br />

Example<br />

Heterogeneous Equilibria<br />

• Pure solids and pure liquids are materials<br />

whose concentration doesn’t change during<br />

the course of a reaction<br />

• Because their concentration doesn’t change,<br />

solids and liquids are not included in the<br />

equilibrium constant expression<br />

– the reaction aA(s) + bB(aq) ⇔ cC(l) + dD(aq)<br />

– the equilibrium constant expression is:<br />

6


Heterogeneous Equilibria<br />

Calculating <strong>Equilibrium</strong> Constants from<br />

Measured <strong>Equilibrium</strong> Concentrations<br />

• The most direct way of finding the equilibrium<br />

constant is to measure the amounts of<br />

reactants and products in a mixture at<br />

equilibrium<br />

• The equilibrium mixture may have different<br />

amounts of reactants and products, but the<br />

value of the equilibrium constant will always<br />

be the same<br />

– as long as the temperature is kept constant<br />

Calculating <strong>Equilibrium</strong><br />

Concentrations<br />

• Stoichiometry can be used to determine the<br />

equilibrium concentrations of all reactants and<br />

products if you know initial concentrations<br />

and one equilibrium concentration<br />

– suppose you have a reaction 2 A (aq) + B (aq) ⇔ 4 C (aq) with<br />

initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0.<br />

You then measure the equilibrium concentration of C as [C]<br />

= 0.50 M.<br />

7


Example<br />

The Reaction Quotient<br />

• If a reaction mixture, containing both reactants and<br />

products, is not at equilibrium; how can we determine<br />

which direction it will proceed?<br />

• The answer is to compare the current concentration<br />

ratios to the equilibrium constant<br />

• The concentration ratio of the products (raised to the<br />

power of their coefficients) to the reactants (raised to<br />

the power of their coefficients) is called the reaction<br />

quotient, Q (a snapshot of the reaction)<br />

The Reaction Quotient:<br />

Predicting the Direction of Change<br />

• Q > K, the reaction will proceed fastest in the reverse<br />

direction<br />

• Q < K, the reaction will proceed fastest in the forward<br />

direction<br />

• Q = K, the reaction is at equilibrium<br />

• Note:<br />

– if a reaction mixture contains just reactants, Q = 0, and the<br />

reaction will proceed in the forward direction<br />

– if a reaction mixture contains just products, Q = ∞, and the<br />

reaction will proceed in the reverse direction<br />

8


Example<br />

• A sample of PCl 5(g) is placed in a 0.500 L container<br />

and heated to 160°C. The PCl 5 is decomposed into<br />

PCl 3(g) and Cl 2(g). At equilibrium, 0.203 moles of<br />

PCl 3 and Cl 2 are formed. Determine the equilibrium<br />

concentration of PCl 5 if K c = 0.0635<br />

Finding <strong>Equilibrium</strong> Concentrations When<br />

Given the <strong>Equilibrium</strong> Constant and Initial<br />

Concentrations or Pressures<br />

• First decide which direction the reaction will<br />

proceed<br />

– compare Q to K<br />

• Define the changes of all materials in terms of x<br />

• Solve for x<br />

9


Example<br />

• For the reaction I 2 (g) ⇔ 2 I(g) the value of<br />

K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a<br />

2.00 L flask and heated, what will be the equilibrium<br />

concentrations of [I 2 ] and [I]?<br />

(Hint: you will need to use the quadratic formula to solve for x)<br />

Approximations to Simplify the Math<br />

• When the equilibrium constant is very small,<br />

the position of equilibrium favors the<br />

reactants<br />

• For relatively large initial concentrations of<br />

reactants, the reactant concentration will not<br />

change significantly when it reaches<br />

equilibrium<br />

– the [X] equilibrium = ([X] initial − ax) ≈ [X] initial<br />

– assuming the reaction is proceeding forward<br />

Checking the Approximation and<br />

Refining as Necessary<br />

• We can check our approximation<br />

afterwards by comparing the<br />

approximate value of x to the initial<br />

concentration<br />

• If the approximate value of x is less<br />

than 5% of the initial concentration, the<br />

approximation is valid<br />

10


Example<br />

• For the reaction 2 H 2 S(g) ⇔ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67<br />

x 10 -7 . If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2S<br />

is heated to 800°C, find the equilibrium concentrations.<br />

Disturbing and Re-establishing<br />

<strong>Equilibrium</strong><br />

• Once a reaction is at equilibrium, the<br />

concentrations of all the reactants and<br />

products remain the same<br />

• However if the conditions are changed, the<br />

concentrations of all the chemicals will<br />

change until equilibrium is re-established<br />

• The new concentrations will be different, but<br />

the equilibrium constant will be the same<br />

– unless you change the temperature<br />

Le Châtelier’s Principle<br />

• Le Châtelier's Principle guides us in<br />

predicting the effect various changes in<br />

conditions have on the position of equilibrium<br />

• It says that if a system at equilibrium is<br />

disturbed, the position of equilibrium will shift<br />

to minimize the disturbance<br />

– disturbances all involve making the system open<br />

11


Le Châtelier’s Principle<br />

• Think of this principle in term of a<br />

push/pull stress.<br />

• Whatever you do to a reaction can have<br />

one of four possible outcomes:<br />

– Shift product<br />

– Shift reactant<br />

– No Shift<br />

– Cannot be determine<br />

Le Châtelier’s Principle<br />

(Concentration Stress)<br />

Type of Stress<br />

Increase reactants or<br />

decrease products<br />

Decrease reactants or<br />

increase products<br />

Result<br />

Shift to the product<br />

side<br />

Shift to the reactant<br />

side<br />

Le Châtelier’s Principle<br />

(Pressure/Volume Stress)<br />

Type of Stress<br />

Decrease pressure or<br />

increase volume<br />

Increase pressure or<br />

decrease volume<br />

Result<br />

Shift to the side with<br />

more gas particles<br />

Shift to the side with<br />

fewer gas particles<br />

Note: This is only for reactions involving gasses (if no<br />

gas or equal moles of gas then no shift occurs)<br />

12


Type of Stress<br />

Le Châtelier’s Principle<br />

(Temperature Stress)<br />

Increase Temperature<br />

Decrease Temperature<br />

Result<br />

Shift in the endothermic<br />

direction<br />

Shift in the exothermic<br />

direction<br />

Note: You must know whether the reaction is<br />

exothermic or endothermic (If you do not know the<br />

energy exchange involved in a reaction then the shift<br />

cannot be determined)<br />

Using Le Châtelier’s Principle<br />

Example<br />

• The reaction 2 SO 2(g) + O 2(g) ⇔ 2 SO 3(g) with ΔH° = -198<br />

kJ is at equilibrium.<br />

– adding more O 2 to the container<br />

– condensing and removing SO 3<br />

– compressing the gases<br />

– cooling the container<br />

– doubling the volume of the container<br />

– warming the mixture<br />

– adding the inert gas helium to the container<br />

– adding a catalyst to the mixture<br />

13

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