Equilibrium and Acid/Base Equilibrium Practice 1 Key1. A mixture of 0.0500 mol A, 0.0100 mol B and 0.0100 mol D were combined in a 250.0 mL reaction flask.2 B (g) + 2 D (g)  4 A (s) Kc = 1.9 x 105(a) Which direction will the reaction proceed to get to equilibrium? You must explain your reasoning to get credit for your answer.[A] initial = (0.0100 mol A/0.2500 L) = 0.0400 M A[D] initial = (0.0100 mol A/0.2500 L) = 0..0400 M DQ = 1 = 3.91 x 105 (0.0400 M)2(0.0400 M)2Q > Kc There are too many products (not enough reactants), so rxn goes left(b) Calculate the amounts of all species at equilibrium. Show your work clearly.2 B (g) + 2 D (g)  4 A (s) Kc = 1.9 x 1052 B (g) +2 D (g) 4 A (s)1.9 x 105 = 1 . (0.0400 M + 2x)2(0.0400 M+ 2x)21.9 x 105 = 1 . (0.0400 M + 2x)4(1.9 x 105)1/4 = 1 . (0.0400 M + 2x)21 = 1 . (0.0400 M + 2x)1 = 0.84 + 42x42x = 0.16 x = 0.0038 MChange in moles of B:(2 x 0.0038 M) x 0.250 L = 0.0019 mol BChange in moles A0.0019 mol B x (4 mol A/ 2 mol B) = 0.0038 mol usedA = 0.0500 mol - 0.0038 mol = 0.0462 moles Ainit0.0400 M 0.0400 Mchange +2x +2xequil0.0400M+2x0.0476 M0.0400M+2x0.0476 M0.0462moles A2. The equilibrium constant for the reaction N2O4(g) <==> 2 NO2(g) is 0.212 mol/L at 100o C. What is the value of Kc at 100oC for:a. 2 NO2(g) <==> N2O4(g)reaction is reversed so Krxn = 1/(Kc) = 1/.212 = 4.72b. NO2(g) <==> 1/2 N2O4(g)reaction is halved and reversed so Krxn = 1Kc= 2.173. KP equals 0.050 for the reaction N2(g)+O2 (g)2NO(g). Calculate the equilibrium partial pressures of all species present in a mixture that has PNO2=PNO=2.5 atm, and PN2O3=0. Show your work; prove that your answer is correct.N2(g) + O2 (g)  2NO(g)   252.05.2)2(224.05.2)2(050.05.2)2(050.0050.02222222xxxxxxxPPPKONNOPinit 2.5 2.5 0change -x -x +2xequil 2.5-x 2.5-x 2xPNO=0.504=0.50 PN2=2.25=PO2  050.025.250.0:2check4. Consider the system 3CO(g)+Fe2O3(s)2Fe(l)+3CO2(g); the reaction is exothermic. Use the tablebelow to indicate if the equilibrium will shift right, shift left, or be unaffected as a result of the applied "stress". [This reaction is used in blast furnaces to reduce iron ore to iron; it accounts for the majority of oxygen used in industry.]stress shift right shift left no changeCO(g) is added XCO2(g) is removed Xthe volume of the reaction vessel is halved XAr(g) (argon) is added Xsolid Fe2O3 is added Xthe temperature of the reaction vessel is decreased by 100C X5. Write the balanced chemical equation for the reaction of K2O with water. Then calculate the pH, pOH, [H1+], and [OH1-] in a 5.49x10-4 M solution of K2O. Show your work. Your answers must have the correct number of significant digits.K2O+H2O2 KOH (aq)  121411011.910041.1114959.29594.2)001098.0log(001098.01049.52pHHpOHpHpOHOH6.Calculate the pH and equilibrium concentration of all species present in a 0.40 M solution of hydrazine, NH2NH2, which has Kb = 1.7x10-6. Show all your work. Explicitly state any assumptions you make, and justify the assumptions.NH2NH2 + H2O  OH1-NH2NH31+ 4262622132161025.840.0107.140.040.0107.1NHNHNHNH][107.1xxxassumexxOHKbinit 0.40change -x +x +xequil 0.40-x x x[NH2NH2]=0.40 [OH1-] = [ NH2NH31+]=8.25x10-4pOH=3.08 pH=10.92 [H1+]=1.2x10-11  6-4107.140.08.25x10:check7. Determine the [H+] of 0.75 M CH2ClCOOH (aq) (see text for Ka value). Draw the Lewis Structures for all species. ID the acid, base, conjugate acid, and conjugate base. Explain the difference in [H+] you determined for 0.75 M HCl (aq) and 0.75 M CH2ClCOOH (aq).CH2ClCOOH (aq) ↔ H+ (aq) + CH2ClCOO- (aq)or, if you prefer H3O+CH2ClCOOH (aq) + H2O (l) ↔ H3O+ (aq) + CH2ClCOO- (aq)acid base conjugate acid conjugate baseI 0.75 - 0 0C -x - +x +xE 0.75-x - x xKa = 1.4 x 10-3 = x2/(0.75- x)x = [H+] = 0.0324Recall [H+] = 0.75 M for 0.75 M HCl.Note the pH of this solution can easily be determined:pH = -log [H+] = -log (0.0324) =