Study Guides (238,069)
United States (119,659)
Chemistry (112)
01:160:161 (46)
All (39)
Midterm

Test 3 Study Guide

9 Pages
92 Views

School
Rutgers University
Department
Chemistry
Course
01:160:161
Professor
All
Semester
Winter

Description
14.1-14.4: Chemical Equilibrium Equilibrium: balanced state of a system in which there are no observable changes w/ time Dynamic eq: balanced state of a system in which 2 opposing processes occur at = rates (physical + chemical processes [photochromic lenses]) Chemical eq: 1. chemical eq is dynamic; rate of forward reaction = rate of reverse reaction; no net change in concentration of reactants/products 2. eq is independent of direction of approach if T, V, + mass are the same 3. catalysts do not affect eq concentrations, but decrease time needed to reach eq Homogeneous eq: all reactants + products are in same phase Rate of forward = rate of reverse reaction aA+bBcC+dD k(forward)[A]^a[B]^b = k(reverse)[C]^c[D]^d k(forward)/k(reverse) = [C]^c[D]^d / [A]^a[B]^b therefore: K(c) = [C]^c[D]^d / [A]^a[B]^b k: rate constant K: eq constant [x]: moles/L (for g and solutes in dilute solutions only) ^x: stoichiometric coefficients Pure liquids, pure solids, and solvents do NOT appear in K(c) expression K(c) = [products]/[reactants] If Kc >> 1, reaction is product-favored; more product at eq If Kc << 1, reaction is reactant-favored; more reactant at eq K > 1000: reaction essentially goes to completion 1 < K < 1000: [products] > [reactants] at eq 0.001 < K < 1: [reactants] > [products] at eq K < 0.001: essentially no reaction occurs Kc of reverse reaction is reciprocal of Kc of forward reaction Kc of reaction that was multiplied by a N is the initial Kc raised to a power equal to that N Overal Kc of multistep reaction is product of steps: Kc(step 1) x Kc(step 2) The Kc for a net reaction w/ 2+ steps is the product of the eq constants for the individual steps Dilute solutions: NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq) If [H2O] >> [NH4+], [H20] >> [OH-], so [H2O] is like a constant Kc = [NH4+][OH-]/[NH3][H2O]const K’c = Kc x [H2O]const = [NH4+][OH-]/[NH3] Heterogeneous eq: not all reactants and products are in the same phase Pure liquids, pure solids, and solvents do NOT appear in K(c) expression What counts are g + aq Ideal Gas: PV = nRT For gas a: PaV = naRT; Pa = naRT/V = [a]RT Pa = partial pressure of gas a na/V = molarity = [a] [a] = Pa/RT Gas-phase reaction: aA (g) + bB (g)  cC (g) + dD (g) Kc = [C]^c [D]^d / [A]^a [B]^b Px = [x]RT Kp = PC^c PD^d / PA^a PB^b = ([C]RT)^c ([D]RT)^d / ([A]RT)^a ([B]RT)^b = [C]^c [D]^d / [A]^a [B]^b times (RT)^(c+d) / (RT)^(a+b) = Kc(RT)^(c+d-a-b) Kp = Kc(RT)^(change in n) R = 0.082 L atm/mol K T in Kelvin Change in n = moles of g product minus moles of g reactants Kc: calculated w/ concentrations in mol/L Kp: calculated with partial pressure in atm Ex: 3Fe (s) + 4H2O (g)  Fe3O4 (s) + 4H2 (g) Kc = [H2]^4/ [H2O]^4 Kp = P(H2)^4/P(H2O)^4 Calculating Kp ex: N2 (g) + 3H2 (g)  2NH3 (g), Kc = 5.8 x 10^5 at 25 degree C Kp = Kc(RT)^(change in n) T = 25 + 273 = 298 K Change n = 2 – (3+1) = -2 Kp = 5.8 x 10^5 ((0.0821 L atm/molK)(298))^-2 = 9.7 x 10^2 mol^2/L^2 atm^2 14.5-14.6: Chemical Eq Q = [products]/[reactants] Changes as a reaction moves to equilibrium Kc = [products]eq/[reactants]eq Constant! (if T is constant) At eq: Q = Kc If Q < Kc, Q increases to reach eq Amt of reactant decreases to make more product Reaction moves forward = reaction spontaneous as written If Q > Kc, Q decreases to reach eq Amt of product decreases to make more reactant Reaction moves back = reaction spontaneous in reverse Le Chatelier’s principle: don’t push me ‘cause I’ll push back If a change is applied to a system at eq, the eq will shift to reduce the effect of the change Concentration, pressure (or volume) for gas-phase eq, and T will affect eq 1. Eq shifts toward products when reactant is added or product removed Eq shifts toward reactants when product is added or reactants removed 2. For g phase ONLY: Eq shift to side w/ fewer moles when P increases or when V decreases Eq shift to side w/ more moles when P decreases or when V increases K does not change if P or V change; eq position may change If V doubles, all concentrations divide by 2 Kc is not changed by solvent addition/removal, but eq position may change 3. eq shifts in the direction that counteracts the change in T If T is lowered on endothermic eq system, eq will shift to left If T is lowered on exothermic eq system, eq will shift to right 14.8: Industrial applications of Le Chatelier’s Principle Haber-Bosch Process: N2 + 3H2  2NH3 + heat (exothermic) Increasing the pressure favors the forward reaction in which 4 mol of g molecule is converted to 2 mol Decreasing the concentration of NH3 favors the forward reaction in order to replace the NH3 that has been removed 15.3: Solubility and Equilibrium Solutions form by efficient intermolecular forces like dipole-dipole, ion-dipole, London, or H bonds The process of dissolving is favored if the solute-solute interactions are weaker than the solute- solvent interactions The solubility of a substance is the max amt that dissolves in a given amt of solvent at a specific T Ionic compounds dissolve by electrostatic attraction of opposite charges (ion-dipole interactions) like NaCl Ion electrostatic interaction (crystal structure of NaCl) < ion-dipole electrostatic interactions (hydration shells of Na+ and Cl-) The process of forming solution: 1. separating solvent molecules (endothermic) 2. separating solute particles (endothermic) 3. forming new solute-solvent associations (exothermic) Unsaturated solutions - [solute dissolved] < solubility = [solute dissolved]max at a T - Q < K - the solutions can dissolve more solute if solute is added Saturated solutions: - [solute dissolved] = const
More Less

Related notes for 01:160:161

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.