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Chapter 18 Free Energy and Thermodynamics 417 Solution: NH₃(aq) + ATP(aq) + + ADP(aq) = -30.5kJ + C₅H₉O₃N₂⁻ (aq) + + Pᵢ(aq) = +14.2kJ NH₃(aq) + + ATP(aq) + ADP(aq) + then = K Rearrange to solve for K. K Check: The units (none) are correct. The free energy change is negative, and the reaction is spontaneous. This results in a large K. 18.105 (a) Given: + define standard state as 2 atm Find: Conceptual Plan: T new = + Q where = Solution: = and Q = = = 1 then = = = Because the number of moles of reactants and products are the same, the decrease in volume affects the entropy of both equally; so there is no change in Check: The units (kJ) are correct. The Q is 1, SO is unchanged under the new standard conditions. (b) Given: + N₂O(g), define standard state as 2 atm Find: Conceptual Plan: new = + RT Q where Solution: = +103.7kJ/mol and Q = PN₂O = = 1 then = + Q = mol + In = = The entropy of the reactants (1.5 mol) is decreased more than the entropy of the product (1 mol). Because the product is relatively more favored at lower volume, is less positive. Check: The units (kJ) are correct. The Q is less than 1, so is reduced under the new standard conditions. (c) Given: 1/2H₂(g) H(g), define standard state as 2 atm Find: Conceptual Plan: new Solution: = and Q = = 2 = then = + RT In Q = +203.3 mol kJ + = = Copyright © 2017 Pearson Education, Inc.