iPad 1027 AM 19 D PEARSON 882 Chapter 18 Free Energy and Th

iPad 10:27 AM @ 19% D PEARSON 882 Chapter 18: Free Energy and Thermodynamics Problems by Topic -Pearson eTextPearson Education eText 2.0 Wu MEW X 2.0 each reaction. G for BrCl(g) is-1.0 kJ/mol. a. 2 NO2(g) N2O1(g) b. Br2(g) Cl2(g)2 BrCI(9) 75, Consider the reaction: CO(9) + 2H2 (g) CH3OH(g) K, = 2.26 x 104 at 25 . C Calculate Grm for the reaction at 25°C under each of the following conditions a. standard conditions b. at equilibrium c. Pai, OH = 1.0 atm; Ro 0.010 atm 76. Consider the reaction: h(g) + Ch(g)--2 ICI(g) K, 81.9 at 25°C Calculate (r for the reaction at 25 °C under each of the following conditions a. standard conditions b. at equilibrium c. Ba = 2.55 atm;2 0.325 atm; FO, = 0.221 atrn Estimate the value of the equilibrium constant at 525 K for each reaction in Problem 730 Estimate the value of the equilibrium constant at 655 K for each reaction in Problem 74 ·AH, for Brcl is 14.6 kJ/mol.) Consider the reaction: 78. 79 H2(g) + 12(g) 2 HI(g) The following data show the equilibrium constant for this reaction measured at several different temperatures. Use the data to find and ..n for the reaction. 150 K 1.4 × 10-6

Solution

76.

(a) At standard condition,

change in Gibb\'s free energy = - R T lnKp = - 0.008314 * 298.15 * ln(81.9) = - 10.92 kJ

(b)

At equilibrium,

change in Gibb\'s free energy = 0 kJ

(c)

Equilibrium constant expression can be written as,

Kp = p_ICl² / (p_I2 * p_Cl2) = (2.55)² / (0.325 * 0.221) = 90.53

Now,

Change in Gibb\'s free energy = deltaG⁰ + R T lnKp = - 10.92 + 0.008314 * 298.15 * ln(90.53) = + 0.2488 kJ