Chemistry 2000 Slide Set 12: Temperature dependence of the - PowerPoint PPT Presentation

Chemistry 2000 Slide Set 12: Temperature dependence of the equilibrium constant Marc R. Roussel February 6, 2020 Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 1 / 15

Temperature dependence of the equilibrium constant So far, all the problems we have solved have been at 25 ◦ C because our tables of free energies of formation are at this temperature. ∆ H and ∆ S vary relatively slowly with temperature. However, ∆ G = ∆ H − T ∆ S has a strong dependence on T . At any temperature, ∆ r G ◦ m ( T ) = − RT ln K ( T ). Suppose we have two temperatures, T 1 and T 2 . Let K 1 = K ( T 1 ) and K 2 = K ( T 2 ). Then ∆ r G ◦ m ( T 1 ) = ∆ r H ◦ m − T 1 ∆ r S ◦ m = − RT 1 ln K 1 ∆ r G ◦ m ( T 2 ) = ∆ r H ◦ m − T 2 ∆ r S ◦ m = − RT 2 ln K 2 Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 2 / 15

∆ r H ◦ m − T 1 ∆ r S ◦ = − RT 1 ln K 1 m ∆ r H ◦ m − T 2 ∆ r S ◦ = − RT 2 ln K 2 m Divide both sides by RT : ∆ r H ◦ − ∆ r S ◦ m m = − ln K 1 RT 1 R ∆ r H ◦ − ∆ r S ◦ m m = − ln K 2 RT 2 R If we assume that ∆ r H ◦ m and ∆ r S ◦ m are independent of temperature and subtract the two equations, we get � 1 ∆ r H ◦ − 1 � m = ln K 2 − ln K 1 R T 1 T 2 or � 1 � K 2 � = ∆ r H ◦ − 1 � m ln K 1 R T 1 T 2 Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 3 / 15

Exponentials and logarithms ln e x = x e ln x = x = ⇒ the exponential function and the natural logarithm are inverse functions. Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 4 / 15

Example: Temperature dependence of K w K w is the equilibrium constant for the autodissociation of water: H 2 O (l) ⇋ H + (aq) + OH − (aq) K w = 1 . 01 × 10 − 14 at 25 ◦ C and ∆ r H ◦ = 55 . 8 kJ mol − 1 . What is K w at 37 ◦ C? Answer: 2 . 4 × 10 − 14 Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 5 / 15

For neutral water, a H + = a OH − so ( a H + )( a OH − ) = ( a H + ) 2 K w = � = ∴ a H + K w At 37 ◦ C, � 2 . 4 × 10 − 14 = 1 . 5 × 10 − 7 = a H + − log 10 (1 . 5 × 10 − 7 ) = 6 . 81 ∴ pH = The pH of neutral water is 7 only at 25 ◦ C! Similarly, the rule pOH = 14 − pH only applies at 25 ◦ C. Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 6 / 15

Effect of temperature on reaction of CO 2 with water For the reaction 3(aq) + H + CO 2(g) + H 2 O (l) ⇋ HCO − (aq) ∆ f H ◦ / kJ mol − 1 CO 2(g) − 393 . 5 HCO − − 691 . 11 3(aq) H 2 O (l) − 285 . 840 We previously calculated K = 1 . 47 × 10 − 8 at 25 ◦ C for this reaction. The average surface temperature of the sea is 16 . 1 ◦ C. Calculate ∆ r H ◦ K at 16 . 1 ◦ C Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 7 / 15

Effect of temperature on reaction of CO 2 with water (continued) Answers: ∆ r H ◦ = − 11 . 8 kJ mol − 1 K = 1 . 70 × 10 − 8 at 16 . 1 ◦ C Note that the increase in K with decreasing temperature is consistent with Le Chatelier’s principle. Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 8 / 15

Boiling point as a function of pressure A liquid boils when its vapor pressure equals the atmospheric pressure. The normal boiling point is the boiling point at 1 atm pressure. 1 atm = 1.013 25 bar Lethbridge is about 940 m above sea level. The atmospheric pressure here is about 90 kPa on a typical day. What is the boiling point of water in Lethbridge? Answer: 96 . 7 ◦ C Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 9 / 15

Boiling-point elevation What effect does a solute have on the boiling point of a solvent? Again consider K = p / p ◦ H 2 O (l) → H 2 O (g) , X H 2 O � 1 � K 2 � = ∆ vap H ◦ − 1 � m ln K 1 R T 1 T 2 Take T 1 = boiling point of pure solvent, K 1 = p / p ◦ , and note that ∆ vap H ◦ m > 0. Since X H 2 O < 1 for a solution, K 2 > K 1 . Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 10 / 15

If K 2 > K 1 , then in the formula � 1 � K 2 � = ∆ vap H ◦ − 1 � m ln K 1 R T 1 T 2 the left-hand side is positive, which means that 1 − 1 > 0 T 1 T 2 (since ∆ vap H ◦ m > 0), or T 1 < T 2 The boiling point of a solution is higher than the boiling point of the pure solvent. Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 11 / 15

Example: What is the boiling point of a solution made by dissolving 0.5032 mol NaCl in 1.01 kg of water at (exactly) 1 atm pressure? Data: The enthalpy of vaporization of water at the normal boiling point is 40.66 kJ/mol. Note: This is roughly the salt concentration you would use to boil pasta. Answer: 100 . 51 ◦ C Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 12 / 15

Freezing-point depression What effect does a solute have on the freezing point of a solvent? The process is H 2 O (s) ⇋ H 2 O (l) with K = a l = X H 2 O 1 a s � 1 � K 2 � = ∆ fus H ◦ − 1 � m ln K 1 R T 1 T 2 Take T 1 = freezing point of pure solvent, K 1 = 1, and note that ∆ fus H ◦ m > 0. Since X H 2 O < 1 for a solution, K 2 < K 1 = 1. Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 13 / 15

If K 2 < K 1 , then in the formula � 1 � K 2 � = ∆ fus H ◦ − 1 � m ln K 1 R T 1 T 2 the left-hand side is negative, which means that 1 − 1 < 0 T 1 T 2 (since ∆ fus H ◦ m > 0), or T 1 > T 2 The freezing point of a solution is lower than the freezing point of the pure solvent. Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 14 / 15

Example: What is the freezing point of a solution made by dissolving 100 g of sucrose (C 12 H 22 O 11 ) in 500 g of water? Data: The enthalpy of fusion of ice at the normal freezing point is 6007 J/mol. Answer: − 1 . 08 ◦ C Marc R. Roussel Temperature dependence of equilibrium February 6, 2020 15 / 15