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Lec-15-Chap15-5-bare

# Lec-15-Chap15-5-bare - Lec-15-Chap15-5-Complete Lec-15...

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Lec-15-Chap15-5-Complete 10/6/2011 Chemical Equilibrium Calculations. Le Chatelier’s Principle 1 Lec-15: Chemical Equilibrium Chapter 15 1 CHE 132 Second Midterm - Review Session: Friday October 14, 6:00 - 9:00 PM Harriman Hall 137 2 The value of K c for the reaction       22 H O g Cl O g 2HOCl g is 0.0900 at 25 °C. What is the concentration at equilibrium of HOCl(g) if the starting concentrations of both reactants are 0.00432 mol/L and no HOCl is present in the container? 4 3 3 2 6 (1). 5.63 10 mol/L (2). 1.13 10 mol/L (3). 8.64 10 mol/L (4). 1.23 10 mol/L (5). 1.28 10 mol/L

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Lec-15-Chap15-5-Complete 10/6/2011 Chemical Equilibrium Calculations. Le Chatelier’s Principle 2 3 Working with K p Recall that the partial pressure of a gas in a low pressure gas mixture is given by the expression   A A and similar exprresions for the other substances B, C, . .. in the mixt re A u n P T V R RT  As the partial pressures are proportional to the molar concentrations of the species, we can work in the equilibrium balance scheme directly with partial pressures.       3 3 3 NH 0 NH 0 NH atm 0.100 mol * 0.0826 * 298.15K K mol 0.955atm 2.58 n RT P V P    First we find the initial partial pressure of NH 3 (g): A sample of NH 4 HS(g) is placed in a 2.58 L flask containing 0.100 mol of NH 3 (g). What will be the total gas pressure when the equilibrium is established at 25 ºC?       32 4 NH g H S g 0.1 NH HS s 28 at 25 C p K 4 Next we consider the equilibrium balance in terms of partial pressures:
Lec-15-Chap15-5-Complete 10/6/2011 Chemical Equilibrium Calculations. Le Chatelier’s Principle 3 5         3 3 32 0 4 NH 0 NH NH HS s so NH g H S g initial 0 equilib lid solid' . P P x x Replacing in the equilibrium constant expression:   3 2 3 0 NH H NH equilibrium pS K P P P x x Given that the equilibrium constant is relatively small (0.108) and that a product is already present in a sizable amount (NH 3 is at 0.955 atm), it is reasonable to assume that 33 00 NH NH p P x K P x  6 And the total pressure at equilibrium is NH H S 1.18atm P P P Solving for x we get: 3 0 NH 0.108 0.113 0.955 p K x P   2 0 NH NH HS 1.068atm 0.113atm P P x Px  Notice that x calculated with this approximation is 10 times smaller than the initial pressure of ammonia, so the approximation is reasonable. We have:

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