1 / 21

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM. Dr. Saleha Shamsudin. What is chemical equilibrium?. A dynamic state in which the rates of forward (f) and reverse (r) reaction are identical. Chemical Equilibrium – A Dynamic Equilibrium.

imelda
Download Presentation

CHEMICAL EQUILIBRIUM

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHEMICAL EQUILIBRIUM Dr. Saleha Shamsudin

  2. What is chemical equilibrium? • A dynamic state in which the rates of forward (f) and reverse (r) reaction are identical.

  3. Chemical Equilibrium – A Dynamic Equilibrium • Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium. • Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

  4. 1. Chemical Reactions 1.1 Introduction • Chemical reactions Involves chemical change. Chemical change: the atoms rearrange themselves in such a way that new substances are formed. Components decrease in quantity are called reactants and those increase in quantity are called products. The chemical reaction is: aA + bB ↔ cC + dD

  5. The rate of reaction = constant x concentration each species raised to the power of the number of molecules participating in the reaction. Forward reaction: kf= rate constant [A] and [B]= molar concentrations of A and B Forward and reverse rates are equal kf[A]a[B]b = kb[C]c[D]d

  6. Molar equilibrium constant, K, kb[C]c[D]d = kf = K [A]a[B]b kb • Rearranging these equation gives molar equilibrium constant (K) : [C]c[D]d = kf = K [A]a[B]b kb When the two rates become equal, the system is in a state of equilibrium. Forward and reverse rates are equal kf[A]a[B]b = kb[C]c[D]d

  7. Type of equilibria • We can write equilibrium constants for many types of chemical processes. The equilibria may represent : • Dissociation (acid/base, solubility) • Formation of products (complexes) • Reaction (redox) • Distribution between 2 phases (water and non-aqueous solvent)

  8. Types of equilibria

  9. Equilibrium constant of reverse reaction: aA + bB  cC + dD cC + dD  aA + bB

  10. Equilibrium constants for dissociating and associating species • For weak electrolyte • Slightly soluble substances • Example: 1) AB  A + B • Equilibrium constant written as: [A][B] = Keq [AB]

  11. Stepwise dissociation A2B  A + AB K1 = [A][AB] [A2B] AB  A + B K2 = [A][B] [AB] Overall dissociation: A2B  2A + B Keq = [A]2[B] [A2B]

  12. 3) Heterogeneous equilibria • Involving more than one physial phase, example, solids, water and pure liquids. • Concentrations of any solid reactants and products are omitted from the equi. cons. expression. • Molar concentrations of water (or of any liquid reactant or product) is omitted from the equi. cons. expression.

  13. E.g. Write out the equilibrium expression for K using the reaction below: N2(g) + 3H2(g)  2NH3(g) Kc = ?

  14. Calculating Equilibrium Constants Try this example: An aqueous solution of ethanol and acetic acid, each with a concentration of 0.810 M, is heated to 100oC. At equilibrium, the acetic acid concentration is 0.748 M. Calculate K at 100oC for the reaction C2H5OH(aq) + CH3CO2H(aq)  CH3CO2C2H3(aq) + H2O(l) ethanol acetic acid ethyl acetate • Write the equilibrium constant expressions in terms of concentrations: • Calculate K at 100oC for the reaction:

  15. The equilibrium constant expressions is: • K = [CH3CO2C2H3] notice that liquid • [C2H5OH][CH3CO2H] water does not • 2. The calculation of K is: appear in equi. • = 0.062 = 0.11 expressions. • (0.748)(0.748)

  16. Calculating an Equilibrium from an Equilibrium Constant Problem: K= 55.64 for H2(g) + I2(g)  2HI(g) Has been determined at 425oC. If 1.00 mol Each of H2 and I2 are placed in 0.5 L flask at 425oC, what are the concentrations of H2, I2 and HI when equilibrium has been achieved?

  17. Solution: • Write the equi. Cons. Expression. • Set up an ICE table • Assume x is equal to the quantity of H2 or I2 • consumed in the reaction

  18. Substitute the equi. Con. Into K expression. 55.64 = (2x)2 = (2x)2 (2.0 – x)(2.0 – x) (2.0 – x)2 55.64 = 7.459 = 2x 2.0 – x x = 1.58 H = I = 2.0 – x = 0.42 M HI = 2x = 3.16 M

  19. Calculating Equilibrium Concentrations • Using initial concentrations, stoichiometry and Kc, equilibrium concentrations of all components can be determined. E.g.2 [H2]o = [I2]o = 24 mM, were mixed and heated to 490oC in a container. Calculate equil. composition. Given that Kc = 46. Solution: set up an equilibrium table and solve for unknown after substitution into equilibrium expression. H2(g) +I2(g)2HI(g)

  20. Equilibrium concentrations of each component are in last row. Substitute into equilibrium expression and solve for x. • It may be necessary to rearrange so that quadratic equation can be used. • . • Rearrange to ax2 + bx + c = 0; determine a, b, c and substitute into quadratic equation:

More Related