Question

WWLWUNA HOSILNILNIUSPRING 2020) Chapter #14 The following practice problems from Chang's text are not to be submitted. The numbers in brackets refer to problems appearing at the end of the chapter. Equilibrium Constant Expressions (14.8, 14.9) Calculating Equilibrium Constants (14.13 to 14.24: Do as many as you can) Multiple Equilibria (14.29, 14.31) Kinetics and Chemical Equilibrium (14.35, 14.36) Calc. of Equilibrium Conc. (14.39 - 14.48) Do as many as you can. LeChatlier's Principle (14.54, 14.55, 14.56 (a&b), 14.58) Submission date for the following problems will be announced in class. Late assignments will not be accepted. I (32pts) Consider the following reaction : SO(g) - SO2(g) + 1/2O2(g) (1) Write the Ke expression. (ii) If Kc = 1.6 x 10-10 at 300.0°C for this reaction, what is the value of Kp at this temperature? (iii) What is the value of Ke at 300.0°C for the reaction 3 SO3(g) - 3 SO2(g) + 3/2 O2(g) (iv) What is the value of Kc at 300.0°C for the reaction 2 SO2(g) + O2(g) = 2 SO3(g) II. (12pts) Using only the following information, determine the value of the missing Ke. 2 CO(g) + O2(g) = 2 CO2(g) H2O(g) = H2(g) + O2(g) CO(g) + H2O(g) = CO2(g) + H2(g) Ke = 1.21 x 1022 Ke = 7.1 x 10-12 K = ? III.(28 pts) Consider the reaction 2 NOCI(8) - 2 NO(g) + Cl2(g) (14pts) At equilibrium, and at a temperature T, a 4 Liter flask contains 2.140 moles of NOCI(g), 0.260 moles of NO(g) and 0.130 moles of Cl2(g). Calculate Ke at this temperature. (ii) (14pts)Calculate the reaction quotient Qc if a 3.00 L flask at temperature T contains 0.0380 moles of NOCI, 0.792 moles NO(g) and 0.0508 moles Cl2(g). Is the system at equilibrium? If not, will the reaction shift to the right or left to establish equilibrium?

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Answer 1

SO₃ (8) - S0₂ lg) + + O2(g) 0 Kc = - [502] [O2Jz [s03 ] 0 kp: Kc (RT) on Here I is in kelvin. given T 300.0°C = 1300.0+ 273·1) K _ =(57301 k) __ R=_gas constant = 0.0821 L-arm mol-' ker on a moles of Poodlust - moles of reactant - (1 + ķ) - 1 = % oi ke E 11.6x10-10] (0.0821x5731)** __ = 1 X 10-9 | ( If a reaction is multiplied by n ke becomes nth power of initial If reversed then new kc is /kc of initial Here, reaction is multiplied by 3. and reverse oi k (1•6 x 10-10) = 4.1 x 10-30 Kc = ki = 74.1* 10-30 = 12.44x10-29 Scannedith CS CamScanner

1 - is multiplied by Similarly the reaction 2 and reversed. .. New kc = 2 = = 13.91X10-19 | 91X19 Ics Scanned with CamScanner
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