HW11-key - Chemistry 111 Homework #11 Due: November 30,...

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Chemistry 111 Homework #11 Due: November 30, 2005 (Wednesday) 1. Calculate the numerical value for the equilibrium constant for the following reaction at 298 K: ( Δ f for NOCl = + 66.70 kJ/mol). The one for NO is + 86.69 kJ/mol. 2 NO (g) + Cl 2 (g) 2 NOCl (g) a) 1.02 × 10 +7 b) 2.03 × 10 -12 c) 1.02 d) 9.82 × 10 -8 e) 4.92 × 10 +11 f) None of these. Once more, in order to do this, we need to have the numerical value for the Δ G° for this reaction first. In this case, Δ G° = 2 Δ f NOCl - [ 2 Δ f NO + Δ f Cl 2 ] Substituting the values from the table (plus the one given on the problem), we get: Δ G° = 2 (66.70 kJ) - 2 (86.69 kJ) - 0 kJ Δ G° = - 39.98 kJ Finally, we can determine the numerical value for the equilibrium constant at 298 K. Δ G° = - R T ln Kp - 39,980 J = - (8.314 J/K) (298 K) ln Kp Kp = 1.02 × 10 7 . 2. Calculate the numerical value for the equilibrium constant for the following reaction at 298 K. The Δ f for N 2 O is + 104.18 kJ/mol. The one for NO can be obtained from a previous question. N 2 O (g) + 1/2 O 2 (g) 2 NO (g) a) 9.55 × 10 2 b) 1.24 × 10 12 c) 1.05 × 10 -3 d) 7.41 × 10 -13 e) 0.973 f) None of these.
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Chemistry 111 2 Homework #11 / Fall 2005 Answer Key Once more, we need to calculate the Δ G° for this reaction using the standard free energies of formation for the reactants and the products. As always, the one for oxygen gas is zero because it is an element in its standard state. Therefore, Δ G° = 2 Δ f NO - Δ f N 2 O = 2 (86.69 kJ) - 104.18 kJ = 69.20 kJ Therefore, Δ G° = - RT ln Kp 69,200 J = - (8.314 J/K) (298K) ln Kp Kp = 7.41 × 10 -13 . 3. Calculate the numerical value for the equilibrium constant for the reaction below at 298 K. The Δ f for NH 4 Cl, NH 3 , and HCl are – 202.87, - 16.45, and – 95.299 kJ/mol, respectively. NH 4 Cl (s) NH 3 (g) + HCl (g) a) 9.40 × 10 15 b) 1.82 × 10 -22 c) 1.07 × 10 -16 d) 2.25 × 10 -14 e) 1.00 f) None of these. The Δ G° for this reaction is given by: Δ G° = Δ f NH 3 + Δ f HCl - Δ f NH 4 Cl = - 16.45 kJ - 95.299 kJ - (- 202.87 kJ) 4. Determine the equilibrium constant (Kp) for this reaction. a) 4.34 b) 0.230 c) 2.59 d) 2.64 × 10 -65 e) 3638 f) None of these. In order to do this, we need to know the value for Δ G° for the reaction. Hmm. ..we’re not given this in the problem – because Dr. S wasn’t paying enough attetion when she made up this HW set! Here’s the solution. .. Δ G° = Δ H° - T Δ
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Chemistry 111 3 Homework #11 / Fall 2005 Answer Key = - 244,000 J - 298K (-831 J/K) Δ G° = + 3638 J Now, with that value, we can determine the numerical value for Kp. This is because: Δ G° = - R T ln(Kp) + 3638 J = - (8.314 J/K) (298 K) ln (Kp) Κ p = 0.230 . 5.
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This note was uploaded on 02/08/2010 for the course CHEM 111 taught by Professor Kenney during the Spring '08 term at Case Western.

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HW11-key - Chemistry 111 Homework #11 Due: November 30,...

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