equilibrium
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Equilibrium. Chapter 17. – occurs when the forward and reverse rxns happen at the same time equilibrium – chemical rxn happens continuously with no net change - PowerPoint PPT PresentationTRANSCRIPT
EquilibriumChapter 17
– occurs when the forward and reverse rxns happen at the same time
equilibrium – chemical rxn happens continuously with no net change
Ex. saturated sugar in iced tea ~ the sugar is undergoing molecular solvation at the same time that its re-crystalizing so you can’t see a net change of sugar on the bottom
Ex. Ocean ~ water is evaporating and condensing at the same time
A + B C + D rate = k[A][B]After time, C & D react together to make A & B.
( )C + D A + B rate = k’[C][D]
So aA + bB cC + dD Kc = [C]c[D]d
[A]a[B]b
Kc = conc. of products (or activities) conc. of reactants
For aqueous activity = _____________ gas activity = _____________solids & liquids = ___
the equilibrium ____________ tells the tendency of the rxn to either ______________________or ______________________
if Kc is large, ____________ __= rxn likes to form products (________________)
if Kc is small, ____________ = rxn likes to stay as reactants (________________)
if Kc = 1, then the rxn is at ____________(very rare)
Equilibrium Constant, Kc
When finding K, don’t use liquids or solids, also K has no units
Ex. 1) N2(l) + 3H2(g) 2NH3(g)
Kc =
Finding Kc expression
Ex. 2) MgCO3(s) Mg2+(aq) + CO3
2-(aq)
Kc =
Finding Kc expression
For a rxn at a specific temp, Kc is always the same value when the concentrations are at equil.
Ex. 3) 2SO2(g) + O2(g) 2 SO3(g) Exp. 1 = 0.344 M 0.172 M 0.056 MExp. 2 = 0.424 M 0.212 M 0.076 M
1st find the equilibrium expression
Ex. 3) 2SO2(g) + O2(g) 2 SO3(g) Exp. 1 = 0.344 M 0.172 M 0.056 MExp. 2 = 0.424 M 0.212 M 0.076 M
2nd – substitute the values into the expression
Kc1=
Kc2 =
Q measures the progress of the reaction, Q has the same form as Kc but involves values not necessarily at equilibrium
If…Q = Kc the rxn is at ____________
Q < Kc the rxn proceeds to the right (____________ ____________)
Q > Kc the rxn proceeds to the left (____________ ____________)
Q, the reaction quotient
Ex. 4) At a high temp, Kc = 65.0 for this rxn: 2HI(g) H2(g) + I2(g)
Using the following conc. determine if the system is at equilibrium. If not, which direction will the rxn proceed to reach equilibrium?
[HI] = 0.500 M, [H2] = 2.80 M, [I2] = 3.40 M
A rxn at equilibrium will tend to stay at equilibrium
If you disturb the rxn with a stress, the equil. will shift in a direction to ______________the stress.
3 types of stressors◦ Concentration◦ Pressure◦ Temp/heat
Le Chatelier’s Principle of shift
– adding or removing reactants/products
Ex. 5) H2(g) + CO(g) H2CO(aq)
Add more CO -
Remove H2CO -
Remove H2 -
1. Concentration
for gases only. Why? s & l are only slightly compressible.
look at the number of moles to determine which way the rxn will shift◦ If P decreases, V increases; shift to the side with the
larger # of gas moles◦ If P increases, V decreases; shift to the side with the
smaller # of gas moles
Ex. 6) 3H2(g) + N2(g) 2 NH3(g) Increase Pressure
2. Pressure
◦ If P decreases, V increases; shift to the side with the larger # of gas moles
◦ If P increases, V decreases; shift to the side with the smaller # of gas moles
Ex. 7) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Increase pressure
2. Pressure
Reminder - heat is a part of the rxn.
Endothermic – ∆H is a reactant
Exothermic – ∆H is a product
Note: K of an exothermic rxn ___________with increasing temp, and K of an endothermic rxn ___________ with increasing temp.
3. temp/heat
The magnitude of Kc is a measure of the extent to which reactions occur. For any balanced eqn. the value of Kc◦ is constant at a given temp
◦ changes if temp. changes
◦ does not depend on the initial concentrations.
Magnitude
There are basically six types of equilibrium problems in this chapter. All are approached in the same way except for the solving of x. All can be solved using RICE.
R = reaction, I = Initial conc, C = Change in conc, E = Equilibrium or ending amount
The following are examples of each kind of problem.
Notes and Example Problems Chemical Equilibrium
Ex. 8) What is the value of Kc for N2 + 3 H2 ↔ 2 NH3, if you start with 2.00M N2, 4.00M H2, and 5.00M NH3 and at equilibrium you have 2.00M H2.
R I
C
E
I. All beginning amounts and at least one equilibrium amount is given.
Ex. 9) K for A + B ↔ C + D is 49. If 0.400 moles of each A and B are placed in a 2.00 L flask, what are the concentrations of all species at equilibrium?
(see page 669 for more help)
II. Only the beginning amounts of reactants are given, and x can be solved by taking the square root of each side of the equation.
Ex. 10) For A + B ↔ C + D, the equilibrium constant is 49 at a certain temperature. If 0.600 moles of A and 0.200 moles of B are placed in a 2.00L flask, what concentration of all species are present at equilibrium?
III. Only beginning amounts of reactants are given and the quadratic equation must be used to solve for x.
Ex. 11) A .200 M HF solution has a Ka of 7.2x10-4. What is the equilibrium concentration of H+? HF(aq) ↔ H+
(aq) + F-(aq)
IV. Only beginning amounts are given, however, equation may be simplified so that the quadratic equation is not required. To be able to use this simplification, K must be small. (in the order of 1x10-4 or smaller) rule
*solids and pure liquids are not included in the equil expression
Ex. 12) A(s) ↔ B(g) + C(g) Solid A is placed in a vessel. After equilibrium is established, the total pressure is 0.75 atm. What is the value for Kp?
V. Gaseous equilibrium: worked exactly like other equilibrium problems, but the concentrations are given in atmospheres. The equilibrium expression takes on a different look since P is used in place of the brackets.
Ex. 13) H2 + I2 ↔ 2HI, K for this reaction is 50. The following concentrations are initial concentrations: H2 = 0.080 M, I2 = 0.060 M, HI =0 .79 M. What concentrations of H2, I2, and HI will be present at equilibrium?
VI. Beginning amounts are given for all reactants and products, but no equilibrium amounts are given. Q, the reaction quotient, must be used.
Kp = Kc(RT)∆n or Kc = Kp(RT) -∆n ∆n = (moles gas products) – (moles gas reactants)
Ex. 14) For the rxn 2S(g) + 3O2(g) 2SO3(g), Kp = 2.2 x 108 at 301oC. What is Kc?
Relationship between Kp & Kc
Ex.15) Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. 2NOBr(g) ↔ 2NO(g) + Br2(g)
A) What is the value of Kp? B) What is Kc ?
Thermodynamic View of Equilibrium Equilibrium point occurs at the lowest value of
free energy available to the reaction system
∆Gorxn = -RTlnK R = 8.314 J/mol K b/c of ∆Go
∆Gorxn < 0 K > 1 products favored
∆Gorxn = 0 K = 1 at equilibrium (very rare)
∆Gorxn > 0 K < 1 reactants favored
Relationship between ∆Gorxn & K
Free Energy and Equilibrium
Ex. 16) Find the Kp for N2O4(g) 2NO2(g) for the temp at 25oC.
Relationship between ∆Gorxn & K
Free Energy and Equilibrium
van’t Hoff eqn. allows us to find K at various tempsln KT2 = ∆Ho 1 - 1 KT1 R T1 T2
K at different temperatures
Ex. 17) Estimate Kp at 250. oC for the rxn of 2NO2(g) 2NO(g) + O2(g) Kp = 4.3 x 10-13
at 25oC and ∆Ho = 114kJ/mole.
K at different temperatures