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- Given the dissociation constants for each deprotonation, what is the dissociation constant for the overall reaction (Ka’’)? H3PO4 + H2O H2PO4- + H3O+ Ka = 7.11 x 10-3 H2PO4- + H2O HPO42- + H3O+ Ka’ = 6.32 x 10-8 Overall: H3PO4 + 2 H2O HPO42- + 2 H3O+ Ka’’ = ?5Br-(aq) + BrO3-(aq) + 6H+(aq) 3Br2(l) + 3H2O(l) The above reaction is expected to obey the mechanism: BrO3-(aq) + H+(aq) HBrO3(aq) Fast equilibriumHBrO3(aq) + H+(aq) H2BrO3+(aq) Fast equilibriumH2BrO3+(aq) + Br-(aq) (Br-BrO2)(aq) + H2O(l) Slow(Br-BrO2)(aq) + 4H+(aq) + 4Br-(aq) products Fast Choose, from the list below, correct expressions for the overall rate law which are completely consistent with the above mechanism. d[Br-BrO2]/dt = k[Br-][BrO3-][H+]2 -d[Br-]/dt = k[Br-][BrO3-]2[H+] -d[H+]/dt = k[BrO3-][Br-][H+]2 k[Br-][H+]2[BrO3-] = -d[Br-]/dt -d[BrO3-]/dt = k[H+]2[BrO3-][Br-] -d[Br-]/dt = k[Br-][H+]2[BrO3-] Tries 0/99 For each of the given rate expressions choose the correct expression for the rate constant (k) from the list below.A. k = k1−k-1B. k = 5×([(k1k2k3)/(k-1k-2)])C. k = 6×([(k1k2k3)/(k-1k-2)])D. k = 6×[(k1+k2+k3)/(k-1+k-2)]E. k = k3F. none of the above -d[H+]/dt -d[Br-]/dt -d[BrO3-]/dtGiven the following reversible reactions and Kc values: (I) Self-ionization of water: H2O + H2O H3O+ + OH-, Kc = Kw = 1.0 x 10-14. (II) Ionization of a weak base: NH3 + H2O NH4+ + OH-, Kc (= Kb) = 1.8 x 10-5. a) Determine the Kc for the reaction: NH3 + H3O+ NH4+ + H2O b) Would you assess the Kc value as LARGE or small? Does your assessment make sense for this reaction, i.e. think about the species involved?
- Prove that the Keq for the reaction between Pb2+ and HF is 12.4. (Ksp, PbF2 = 6.76 x 10-4 and dissociation constant of HF is 3.7 x 10-8)Given the dissociation constants for each deprotonation, what is the dissociation constant for the overall reaction (Ka’’)? H3PO4 + H2O H2PO4- + H3O+ Ka = 7.11 x 10-3 H2PO4- + H2O HPO42- + H3O+ Ka’ = 6.32 x 10-8 Overall: H3PO4 + 2 H2O HPO42- + 2 H3O+ Ka’’ = ? 1.13 x 105 4.49 x 10-10 8.89 x 10-6 2.23 x 109In light of your answer to Problem 30-40, explain why a mixture of products occurs in the following reaction:
- 1. a) Give the expression for the Ka for each of the following reactions H2CO3 + H2O <=> H3O^+1 + HCO3 ^-1 Ka = H2CO3 + 2 H2O <=> 2 H3O^+1 + CO3^ -2 Ka = b) Give the expression for the Kb for each of the following reactions HCO3^ -1 + H2O <=> OH-1 + H2CO3 Kb = HPO4^ -2 + H2O <=> OH^-1 + H2PO4^ -1 Kb =the dissociation constant of 0.25 M propanoic acid C2H5COOH is 1.3 × 10-⁵. a) What is the dissociation rate of C2H5COOH in this solution?Classify each of the reactions below: a) Mg + Cl2 → MgCl2 b) 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O c) BaCl2 + Na2SO4 → BaSO4 + 2 NaCl
- 5) Ethyne (CH≡CH) has a pKa of 25 and water has pKa of 15.7 and ammonia (NH3) has a pKa of 36. Write equilibrium reactions showing arrows that indicate whether the reactants or products are the favored for reaction of Ethyne with a. HO- , and b. with NH2 - and indicate which one is better for removing proton from Ethyne. Then calculate the equilibrium constants, Ke, for each equilibrium reaction. Are calculated Ke values match the expected results from prediction you made .6. Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow: (a) O2-(aq) + H2O(1)<---> (b) CH3COOH(aq) + HS-(aq)<---> (c) NO2-(aq) + H2O(1)<---> Group of answer choices A.) (a) 2 OH-; 100% to products (b) CH3COO- + H2S; towards products (c) HNO2 + OH-; towards reactants B.) (a) 2 OH-; 100% to products (b) CH3COO- + H2S; 100% to products (c) HNO2 + OH-; towards reactants C.) (a) 2 OH-; 100% to products (b) CH3COO- + H2S; towards products (c) HNO2 + OH-; towards products D.) (a) 2 OH-; towards reactants (b) CH3COO- + H2S; towards products (c) HNO2 + OH-; towards reactantsWhich of the following reactions result in a positive ∆ Ssys? a. H2O(g) = H2O(l) b. H2(g) + I2(g) = 2 Hl(g) c. C2H2O2(g) = 2CO(g) + H2(g) d. H2O(g) + CO2 (g) = H2CO3(aq)