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Chemical Equilibrium. Chemistry 100. The concept. A condition of balance between opposing physical forces A state in which the influences or processes to which a thing is subject cancel one another and produce no overall change Oxford English Dictionary. Static and Dynamic.
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Chemical Equilibrium Chemistry 100
The concept • A condition of balance between opposing physical forces • A state in which the influences or processes to which a thing is subject cancel one another and produce no overall change Oxford English Dictionary
Static and Dynamic • A book sitting on a desk is in static equilibrium; • The book remains at rest; its position is constant. • The moon circles the earth. • There is movement but the (average) distance between the two is unaltered. This is dynamic equilibrium.
Equilibrium • The molecules of A are able to turn into molecules of B • The rate at which this happens is proportional to [A]. Ratefor = kfor[A] • Likewise, if B can turn into A, then Raterev = krev[B]
The Equilibrium Condition • Start with pure A. • [A] decreases and [B] increases as A turns into B • What happens to the rate at which A turns into B, and the rate at which B turns into A? • The rate of A B decreases, while BA increases • What eventually happens? • Rate of A B = Rate B A Ratefor = Raterev kfor[A] = krev[B]
We have as an equilibrium condition kfor[A] = krev[B] And then what? Keq the thermodynamic equilibrium constant
The Meaning of PB/PA= K • K is a constant number such as 2.3, 0.65, etc • What the equilibrium expression means is: • No matter how much A or B we start with, when the system reaches equilibrium
Reversible reactions • If these two reactions are possible • A B and B A, • we have a reversible reaction A ⇌ B • Here is a real reversible reaction N2(g) +3H2(g) ⇌ 2NH3(g)
Equilibrium can be reached from either side At start PH2 = 3; PN2 = 1; PNH3 = 0 At start PH2 = 0; PN2 = 0; PNH3 =2
For the reaction aA (g) + bB (g) ⇌pP (g) + qQ (g) Law of Mass ActionExpression for K
Examples of Keq N2(g) +3H2(g) ⇌2NH3(g) Br2(g) +Cl2(g) ⇌ 2BrCl(g) SO2(g) +½O2(g) ⇌ SO3(g)
Magnitude of Keq • 2 HI(g) ⇌H2(g) + I2(g) Keq = 0.016 • The magnitude (size) of Keq provides information • K >> 1 the products are favoured • K << 1 the reactants are favoured • CO(g) + Cl2(g) ⇌COCl2(g)Keq= 4.57109 • Equilibrium lies far to the right - there is very little CO and Cl2 in the equilibrium mixture.
Heterogeneous Equilibrium • When the substances in the reaction are in the same phase (e.g., all gases), reactions are termed homogeneous equilibria. • When different phases are present, we speak of heterogeneous equilibrium. • We will look at reactions involving gases and solids, and gases and liquids
Solids do not appear in Keq • Examine the reaction CaCO3(s) ⇌CaO(s) + CO2(g) For a pure solid X (or liquid X) [X] = density/molar mass. Note molar mass and density are intensive properties!! [X] = constant
Heterogeneous Equilibrium • At a given temperature, the equilibrium between CaCO3(s), CaO(s), and CO2(g) yields the same concentration (same partial pressure) of CO2(g). • True as long as all three components are present. Note that it does not matter how much of the two solids are present; we just need some.
More heterogeneous equilibria CO2(g) + H2(g) ⇄CO(g) + H2O(l) SnO2(s) + 2CO(g) ⇄ Sn(s) + 2CO2(g)
Keq values for forward and reverse reactions • For the reaction 2 HI(g) ⇌H2(g) + I2(g), Keq = 0.016 • What is Keq for H2(g) + I2(g) ⇌2HI(g) ? • Call the first reaction (F) and the second (R)
An aside • For the equilibrium H2O(l) ⇌ H2O(g), write down the expression for Keq. When liquid water and water vapour are in equilibrium the vapour has a fixed pressure at a given temperature!
Applications • Obtaining the equilibrium constant from the measured equilibrium concentrations • Calculating the composition of the equilibrium system • have the concentration of all but one component at equilibrium and the value of Keq • given initial amounts of reactants and the equilibrium constant
Applications (II) • For a given reaction, Keq has a set value for a given temperature • Q depends on the experimental conditions • Q = Keq at equilibrium
Summary of the Q and Keq story • When Q > Keq • reaction shifts left • When Q = Keq • equilibrium • When Q < Keq • reaction shifts right
Equilibrium is approachable from either side of the reaction. Applications (III)
Le Châtelier’s Principle • Perturb a system at equilibrium • Change in temperature, pressure, or the concentration of a component • The system will shift its equilibrium position so as to counteract the disturbance. • The effect of the last two disturbances can also be be predicted by the Law of Mass Action
Changing concentration (I) • Examine the system 2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g) • Introduce a small amount of substance X that reacts with Cl2 to make XCl. • The value of PCl2 has been decreased • Le Châtelier’s Principle predicts that more NO2Cl will react to increase PCl2
Changing concentration (II) At equilibrium Remove Cl2 - the new value of PCl2 is 0.05 atm Q is now smaller than Keq. The reaction moves to the right to increase Qeq. Same prediction!
Changing concentration (III) CaCO3(s) ⇌ CaO(s) + CO2(g) If we have this system in equilibrium and add either CaCO3(s) or CaO(s), there will be no effect on the equilibrium.
Changing pressure (I) 2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g) • Increase the pressure in the system by making the vessel smaller. • Note that there are a total of 3 moles on the right of the reaction and 2 on the left. • The left “takes up less space” • Le Châtelier’s Principle predicts that the species on the left will react to form more NO2Cl.
Changing pressure (II) 2 NO2Cl (g) 2 NO2 (g) + Cl2 (g) • System is initially at equilibrium. • Increase the pressure by making the vessel smaller. • We could use the Keq expression to predict what happens but Le Châtelier’s Principle is much easier to use!
Cautionary Note • Le Châtelier’s Principle predicts what occurs when we change the partial pressure of one or more of the species in the reaction • Change the total pressure by adding/removing an inert gas (not involved in the reaction) NO EFFECT ON THE EQUILIBRIUM
More pressure changes Predict what happens • N2(g) + 3H2(g) ⇌ 2NH3(g); total pressure is decreased Reaction shifts to left; more moles on left • H2(g) + I2(g) ⇌2HI(g); total pressure is increased No effect; 2 moles on each side
What is Heat - not a substance! • Some textbooks • Heat is treated as a chemical reagent when applying Le Châtelier’s Principle to change of temperature problems. • Treating heat as a substance can lead to confusion. There is a better way!
Changing the temperature • An exothermic reaction causes an increase in temperature. The reverse causes cooling. • Warm up an exothermic reaction • Le Châtelier’s Principle predicts the system will move in the direction that will bring the temperature back down • The direction that cools. So the reverse reaction () occurs
Changing Temperature N2O4(g) 2 NO2(g) H = 58.0 kJ • The forward reaction is endothermic • Absorbs heat. • Decrease the temperature - reaction shifts to the left • Brings T back up. • If we increase the temperature, opposite effect • Reaction takes in heat and lower the temperature
Temperature and Keq • Endothermic reactions – increasing temperature increases the value of the equilibrium constant! • Exothermic reactions – increasing temperature decreases the value of the equilibrium constant! Temperature changes are the only stresses on the systems that change the numerical values of Keq
Temperature and Keq (II) Co(H2O)62+(aq)+ 4 Cl- (aq) ⇌ CoCl42- (aq) + 6 H2O (l) ∆H> 0
Catalyst does NOT change K • A catalyst speeds up a reaction by providing and alternate reaction pathway with a lower Ea. • Reversible reaction • the forward and backward reactions have their Ea’s changed by the same amount. • Keq is not altered. A catalyst cannot alter K!! Otherwise we would be able to build a perpetual motion machine!!