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Equm U ving reaction for Questions 1 - 3 and answer the questions based upon th sed upon this Consider the following reaction reaction: co (g) + 3 H, (g) + CH (9) + H2O (g) 1. At 25°C. Kega 4.0. What must be Kes 4.0. What must be the coat equilibrium If the rulibrium concentrations on HJE 2.0M, (CH) = 5.OM, and [HO] - 3.OM? Answer: 0.469 M a way to keep track of what is happening in the reaction vessel. The initial concentrations if given are the species (either reactants, products or both) that are put in to begin the reaction. The initial concentrations will change based upon the stoichiometry of the reaction. If we do not know the exact amount of change, we can assume a general amount of "X". If the reactants are moving to make the products, we would say the reaction is going in the forward direction and the amount of change for the reactants would be a decrease based upon the stoichiometry. For example, in a reaction in which 3 A 2 B, then the amount of A would decrease by 3x -3x) and the amount of B would increase by 2x (+2x). We can then say that the initial concentration plus the change in concentration will equal the final or equilibrium concentration. 2. 4.0 moles of CO and 2.0 moles of H, were mixed in a 1.0 L container, and kept at constant temperature until equilibrium was reached. At equilibrium, it was found that 3.4 moles of CO were still in the container. Complete the ICE table below. CÓ (g) + 3H (9) = 1 CH(9) + 1 HL0 (g) : Substance Initially (moll) . Change @ Equilibrium (mol/L) Answer. x = 0.6M; [CO] = 3.4 M: [H.]= 0.2M, [CH] = 0.6M, and [H0] = 0.6M 3. In another experiment, 2.0 M. of CO.0.5 M. of H. 1.7 M OF CH, and 2.0 M. Of H,O were mixed and equilibrated at constant temperature in a container. At equilibrium, the concentration of CO is 2.8 M. a. Complete an ICE table with the information above. D. Solve for the equilibrium concentrations of each species Answer: [Co] =2.8 M; [H.]= 3.2M, (CH) = 0.9M, and [H0] = 1.7M

Answer 1

Ans-consider the following seaction - cocg2 + 3H2(g) = CH4 692 + Hoo cg) Apply the law of chemical equation- Keq = [city cgs] [reo go] [60 g)] [H₂ (g) 33 Given data- 4 5 X 3 [cly] =5M [co g] *(2) 3 [120 (g]]= 3 M [H₂] =2M [C0g) 3= 15m 4x8 IS M -0.4687 M 32 [co G)] = 0.469 M Ang @ Giren equation - Volume of container - 1L co cg) + 3Hq g) - CMU (g) + H O cg) Initially 4 2- moles Initial y = 4moll 2 mol/l - conc. 1 = 4 mol/L - 2 mol/L 1 SL x - 30 tr ta Change

At equill. 4 x 2-3x x x (mol IL) AĽ equillibriun , the moles of co = 3.4 At equill., concentrat of 10 = 3.4 = 3.4 moll" [Co] – 3.4 M7 According to the equation . Conc. of co at equilliboriwn = 4-X 4 x = 3.4 4-3.4 = x n = 0.6 molt n = 0.6M 2-3x0.6 - 2-18 = 0.2 M [H] = 2-3x = T.: [H.] = 0.2M [Cry] = x = 0.6M [120] = x = 0.6M Ang 3 @ Initial Change C09+ 3 H2 G) = RM 0.5M .ax -3x CHU Cg) + Ho Gay 1.7M 25 M to the conc. - At equill conco go x 0.5-32 17th 25th

At equillibrium , concentration of co = 2.8M 2-X = 28 - 2-28 = -0.8 M = 0.5 - 3x = 0.5-3 X (-0.8) -0.5 +2.4 = 2.9 M (CH4] - 1.7+ = 1.7 -0.8 - 0.9 M E 2.5 + x = 2.5-0.8 – 1.7 M - © k for the steaction is - k= [city go] [ Hgoog] [Cogo] [, 93] K = 0.9 X 1.7 2.8 X (2.9)2 1.53 23.548 K = 0.0649 @ The value K is different from K in problem I as the equillibrin con cenderation of alle ewoctant and products are different.

Ans A mixture of A, B and C is placed in 1L container. AC) = 3B (g) + 26 g) 3.31 moles 4.33 moles 595 mods. Initial moles Initial con. 3.31 M 4:33M 5.95 M At equille 3.1-x Conci 4.33474 5.75 + 2x At eqwllibrium; No. of moles of B = 6.16 conc. - 6.16 = 6.16 M 6.16 = 4.33 + 3x 3x = 1.83 x = 0.617 [A] = 3.31 -76 = 3.31 - 0.61 = 2.7 M BJ = 4.33 +3% = 4.33 + 3x0.G1 = 4.33 + 1.83 - 6.16 M [B] = 6.267 - 5.95 + 2x = 5.95 2x0.61 = 5.95 + 1.22 = 7.17 M [c] = 7.17 M

Kc = [B (g)] 'x. [c G)]? [Ag] = (6.1613 (7.17)2 (207) = 233.74 x 51.40 ::2.7 = 12014.236 2.7 = I kc = 4449.71 4.449 x103 4.45 × 103 1