{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW11-key - Chemistry 111 Homework#11 Due(Wednesday 1...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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. 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
Background image of page 2
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. Let’s say that we mix initially in a flask at 298 K the gases HA, O 2 , and A 2 O 6 , at partial pressures of 0.555, 1.00, and 0.100 atm, respectively, together with a substantial amount of liquid water. Will the reaction proceed spontaneously toward the formation of A 2 O 6 ? Type the word yes , no , or cannot tell in Blackboard. Ditto for this problem. It’s actually linked to #4. The reaction is 4 HA (g) + 7 O 2 2 A 2 O 6 (g) + 2 H 2 O (l) We want to know the sign of Δ G at a given point of the reaction. That is what we get from the following equation: Δ G = Δ G° + R T ln Q But what is Q? Q is defined as: Q = P A 2 O 6 2 P HA 4 ! P O 2 7 Substituting the values, we get: Q = (0.100) 2 (0.555) 4 ! (1.00) 7 Q = 0.105 Therefore, Δ G = + 3638 J + (8.314 J/Kmol) (298 K) ln (0.105)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Chemistry 111 4
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}