Unit 5 The 5 th planet in our solar system, Jupiter Ch. 15 Chemical equilibrium: This is based on the idea that reactions go forwards and backwards at the same conditions The Mass Action Expression describes a system undergoing a chemical change Given a reaction: rate law for forward reaction: rate law for reverse reaction: at equilibrium, rate f = rate r so aa + bb cc + dd rate f = k f [A] a [B] b rate r = k r [C] c [D] d k r [C] c [D] d = k f [A] a [B] b K eq is also called the reaction quotient at equilibrium; it is the value obtained from the equilibrium expression. When not at equilibrium, it will be a different value from k eq Given the reaction Write the equilibrium expression. K eq = 2NH 3 (g) 3H 2 (g) + N 2 (g) The products go in the numerator, the reactants in the denominator, and each variable goes to the power that the coefficient is in the BCE
When applied to aqueous solutions, k eq is referred to as k c Use molar concentrations in the expression; only molarity Given the equation: SrF 2 (s) Sr 2+ (aq) + 2F - (aq) Find k c given that the reaction began with and ended with Sr 2+ F - Initial 5.0x10-6 M 3.18x10-4 Change -2.5x10-6 -2.5x10-6 Equilibrium 2.5x10-6 M 3.16x10-4 K c = K c = 7.9x10-10 Set up an ice chart, showing the initial values, the change, and the equilibrium values. Next, calculate the change from the initial concentration of Sr 2+ to the equilibrium concentration. The concentration of F - will be the same, which you can use to find the equilibrium value. Then plug the equilibrium values into the expression and solve for k c I found the problem here: http://www.shodor.org/unchem/advanced/equ/ And I determined the initial values and equilibrium values used. When applied to gases, k eq is referred to as k p Only use partial pressures in atmospheres Given the reaction: C 2 H 6 (g) + Cl 2 (g) C 2 H 5 Cl(s) + HCl(g) Calculate k c given that at equilibria, the partial pressures of C 2 H 6, Cl 2, and HCl are 6 atm,.5 atm, and 3 atm. respectively. K p = K p =
K p =.10 Write the equation, leaving out the solid. Then plug in the numbers, and simplify. I created this question, but got the equation here: http://www.shodor.org/unchem/advanced/equ/ At equilibrium, all components are present. 0000697209 Pure liquids, solvents, and solids are not included in the equilibrium expression. Manipulating k eq : Equilibrium constant of a reaction, in the reverse reaction is the invers of the forward reaction Equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to that power Equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps LeChatlier s Principle: When a system at equilibrium is subjected to stress, the equilibrium will shift in the direction to relieve the stress Stressors: Changes in concentration Changes in temperature Changes in pressure The planet Jupiter has 67 known moons *Temperature is the only way to change k eq Given the equation: H 2 (g) + Cl 2 (g) 2 HCl (g) + heat Determine which direction the equilibrium will shift given the conditions:
Add 2.00L of H 2 The reaction will shift away from the H 2 gas towards the products side Adding heat The reaction will shift away from the heat side towards the reactants Adding.500L of HCl The reaction will shift away from the products side and towards the reactants Ch 16: Structural definition of acids and bases (Arrhenius definition): Acid: contains an H as the only cation Base: contains an OH as the anion Functional definition of acids and bases (Bronsted-Lowry): Acid: any substance that donates a proton in a reaction Base: any substance that accepts a proton in a reaction Strong acids and bases you don t have to worry about k a HI HCl HBr HNO 3 HClO 3 LiOH NaOH KOH RbOH CsOH HClO 4 Ca(OH) 2 H 2 SO 4 Sr(OH) 2 Ba(OH) 2 Acid + base salt + H 2 O Acid + carbonate salt + CO 2 + H 2 O Amphoteric substances: can be an acid or base H 2 O is an amphoteric substance When dealing with water as a part of the reaction, often the conjugate acid base pairs are written: this is the acid and base that may be formed reversibly from one another in a protolysis reaction
H 2 O + H 2 SO 4 H 3 O + HSO 4 - H 2 O/H 3 O + H 2 SO 4 /HSO 4 - The planet was named after the Roman god of the sky, although he is better known for the Greek name Zeus. Water self-ionizes 2H 2 O H 3 O + OH - The equilibrium expression is: K w = [H 3 O][OH] K w = 1 x 10-14 = x 2 X=1x10-7 Neutral ph is 7, because the exponent is 7 Equation that relates concentration and ph: [H 3 O + ] = 10 -ph ph = -log[h 3 O + ] (can also be used for poh; just substitute poh for ph and OH - for H 3 O + ) Given a substance with a ph of 6.28, what is the concentration of H 3 O + ions? [H 3 O + ]=10-6.28 [H 3 O + ]=5.2x10-7 M (only 2 sig figs, as the number before the decimal point of the ph only tells the placemnt of the decimal point, not the number) What is the poh? 14-6.28=7.72pOH What is [OH - ]? 1x10-14 = (5.2x10-7 )x x = 1.9x10-8
I created this question. K a is k eq for the dissociation of acids The same is with bases as K b is k eq for the dissociation of bases Ch. 17 A buffer solution is a weak acid with a conjugate base, or a weak base with a conjugate acid. Buffers change very little when a strong acid or base is added to them. It is an equilibrium between an acid and a base, so when an acid or base is added, it reacts, and shifts the equilibrium slightly, causing the increase in ions to be less than normal. To obtain a certain ph, prepare the buffer, then add a strong acid or base to slowly adjust the ph, although an alternative method is to use the Henderson-Hasslbach equation to determine the exact amount of acid or base needed. The Henderson-Hasselbach equation is: ph=pka+log([a ][HA]) where ph is the specific ph, pk a is the acid dissociation constant, [A-]and [HA] are the concentrations of the conjugate base and starting acid. One day on Jupiter lasts 9 hours and 55 minutes; shorter than any other planet in our solar system