Question

5 pts Question 4 A mixture of 0.75 moles of H2S. 0.40 moles of H2, and 1.25 moles of S2 are placed in a 10.0L container. Which of the following statements is true if the following reaction has a Ke value of c.40 x 108 at a temperature of 750°C? 2 H25 (8)2H2(g) + S2 (8) The reaction shifts to the right to establish equilibrium The reaction is already at equilibrium The value of K will change as the reaction proceeds toward equilibrium The concentrations of H2S and H2 will decrease since they are larger than the concentration of 52 The concentrations of the products will decrease, while the concentration of the reactants will increase to establish equilibrium 5 pts Question 5 A hypothetical reaction is shown below. 2A (8) + B2 (8) 22 (8) A 1.0 L flask is charged with 0.382 M of A and 0.0952 M of B2 and allowed to come to equilibrium. At equilibrium, the vessel is found to contain 0.0624 M of C. What is the value of Kc for this reaction? O 0.494 O 0.280 O 0.149 O 0.190 O 0.596

Answer 1

4) 2 H2S (g) = 2H2 cg + Szeg) here H2S - 0.75 mol 0.075 mol/l LOL H2 - O. 40 mol 0.040 mol/lo TOL S2 1025 mol 0.125 molli lo L Q = [Hz] [5] - [ H25]2 (0.040) x (0.125) (0.0752 oooo2 0.005625 a = 0.0355 > Kc =400xl The concen teation of the products will decrease cechile the concentration of the reactant will increase to establish equilibrium -is the correct answer L- Because reaction will backward go

5) 2Aig) + B2 2 cg) - 20 Ccg) Given concentration of A = concentration of B2 = Oo 382 m 1 lil Oogg 52 m 1 eit 0.0624th 1 lit concentreation of C = B2 2 - 0.0624 M) (0.382M em) ² (0.0952M) - 0003893 0.145920 X o o og 52 0.003893 - 0.013891 kc 6.280 end option is is the correct answer.

6 . Given 0.002 55M [OH] (Base concen Keation) = we have 60 [ht] [or] = IX1014 THt] =) [ht] = 1x1014 concentration acid 1x1016 0,00 2550 il 392.156 X10-14 3.9215x 10/2 PH = -log[ht] 3.9215x10 xide) log pH 11.407 5th option is correct answer

Given [OH] Base concentration = 0.00318M. Here Clou is conjugate base. pH = -log [H] [HD is a concentration of acid we' have [ht] [oH] = 1x1014 Here [04'] [] [ht] Ixlo14 Гон) 1x1014 Oo 00318 -14 314.46 xl = -12 3.1446 x 10 - log (3.144681012) pH [PH 11.502 5th option is is correct answer.

Given kb - 107x109 (5 H5 N + H2O C5 H5NH T oH H2O C5 H5N 6=25x10m C5 H5NHT OH I O O C ta fa F x x 6.25*104_x K6= [CSHS NH COA [c5 HsN] lo 7x109 = (x) (x) 60 25x10 4 1.7 X 109 x 6.258104 - 13 10.625 X10 2 - ✓ 106.25 x 1014 -7 10.307810 x= [OH = 10.3078 107

pH = -log [OH -log [ 10.307x107 PH 5.987 optron 3rd is correct answer