1 (404)

Prévia do material em texto

404 Chapter 18 Free Energy and Thermodynamics (b) Given: I₂(s) at 25.0 (i) = 1.00 mmHg; (ii) = 0.100 mmHg Find: Conceptual Plan: °C K and mmHg atm then T 1 atm K = 273.15 + °C = + RT In where = 760 mmHg 1 atm Solution: T = 273.15 + 25.0 °C = 298.2 K and (i) 1.00 mmHg X 760 mmHg = 0.00131579 atm then = + Q = + RT = +19.3 kJ + In = +2.9 kJ; the reaction is nonspontaneous. 1 atm (ii) 0.100 mmHg X 760 mmHg = 0.000131579 atm then = + Q = + RT = +19.3kJ + In = -2.9kJ; so the reaction is spontaneous. Check: The units (kJ) are correct. The answer is positive at higher pressure because the pressure is higher than the vapor pressure of iodine. Once the desired pressure is below the vapor pressure (0.31 mmHg at 25.0 °C), the reaction becomes spontaneous. (c) Iodine sublimes at room temperature because there is an equilibrium between the solid and the gas phases. The vapor pressure is low (0.31 mmHg at 25.0 °C), so a small amount of iodine can remain in the gas phase, which is consistent with the free energy values. 18.71 Given: CO(g) + at 25 °C, PCH₃OH = 0.855 atm, = 0.125 atm, PH₂ = 0.183 atm Find: Conceptual Plan: = - (reactants) then °C K then Pco, T AG K = 273.15 + °C = + RT In where Solution: Reactant/Product from Appendix IIB) CH₃OH(g) -162.3 CO(g) -137.2 0.0 Be sure to pull data for the correct formula and phase. = = + = + = [-137.2kJ] [-162.3kJ] = +25.1kJ T = 273.15 + = 298 K then Q = = 0.855 = 0.00489605 then = + RT In Q = + 25.1 kJ + In = +11.9kJ; so the reaction is nonspontaneous. Check: The units (kJ) are correct. The standard free energy for the reaction was positive, and the fact that Q was less than 1 made the free energy smaller. But the reaction at these conditions is still not spontaneous. 18.73 (a) Given: 2CO(g) + at 25 °C Find: K Conceptual Plan: = then °C K then T K K = 273.15 + °C = - K Copyright © 2017 Pearson Education, Inc.