Question

October 16, 2019 CHEM 1252 Problem Set 12 5. Each part below refers to the balanced equation shown. 2Cl (g) +2H20(e) 4HCI(g) + Oz(g) (a) If you start with only reactants and no products, tell whether the total pressure will increase, decrease, or stay the same as the reaction approaches equilibrium, and explain why. (b) Under certain conditions at equilibrium, the total pressure is 0.81 bar, and the mole fractions for each gas are: 2/9 (Cla); 2/9 (H2O); 4/9 (HCI); and 1/9 (O2). Calculate Kp. 6. The equilibrium constant for the reaction in question 2c is K, (a) Assuming this reaction is at equilibrium, calculate [I-] when [Ia] 721 M- [I ] =0.0010 M. (b) Assuming this reaction is at equilibrium, find [I ] when [I2] 0.0025 M and [I ] 0.010 M. (c) Assuming this reaction is at equilibrium, find [I2] when [I] 0.00030 M and [I-]- 0.10 M. (d) Before the reaction reaches equilibrium, a solution has [Ia ] 0.0020 M, and [I-] = [I] = 0 M. Calculate the concentrations of all three species after the solution reaches equilibrium. (Assume that x M of triiodide dissociates to iodide and iodine, and solve for x.)

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Answer 1

5 2cle ) + 2 H20 @ = 440@ +0, @) According to Le chatelier's principle if a cauilibrium iss disturbed by changing the condition, the position of eauilibrium moves to counter act the change. (a) According to le chatelier's frinciple when réactants concentration are high & product concentrations are less the eauilibrium of reaction mill shift to product formation. So, when only reactants are present no products are present then cauilibrium aus shift to froduct formation. So, more forod veth aill be formed. Here number of moles of products ares' & number of moler of reactants are 2. So, bartial pressure of the gan molicies acull in creare. So, total pressure of the system aull increare. &) 2 Ch 6 + 2H20 67 – 4 HCl ( + 02 (9) partial pressure (p.) of the species - mole fraction of ith species (2) х тона о ws (1) Pengt er og - 1.8 bar Pho 2 xoo 1.8 bar Pra- x el 3.6 bar 0.9 -0.9 bar Pongress

(576) the PHOI the canilibrium constant of reaction Kpo Pelux find (3.6) 4 x 0.9 (1.8) (1.8) . kp = 14.4 .. (kg = 144