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19 Chemical Thermodynamics Solutions to Exercises moles ATP = -2878.8 kJ 1 mol ATP / (-30.5 kJ) = 94.4 mol ATP / mol glucose Note that this calculation is done at standard conditions, not metabolic conditions. A more accurate answer would be obtained using G values that reflect actual concentration, partial pressure, and pH in a cell. 19.104 (a) The equilibrium of interest here can be written as: K⁺ (plasma) 1L K⁺ (muscle) Since an aqueous solution is involved in both cases, assume that the equilibrium constant for the above process is exactly 1, that is, = 0. However, G is not zero because the concentrations are not the same on both sides of the membrane. Use Equation [19.16] to calculate G: G = + RT [K⁺ (muscle)] [K⁺ (plasma)] = (8.314) In (5.0 (0.15) (b) Note that G is positive. This means that work must be done on the system (blood plasma plus muscle cells) to move the ions "uphill," as it were. The minimum amount of work possible is given by the value for G. This value represents the minimum amount of work required to transfer one mole of ions from the blood plasma at 5 10⁻³ M to muscle cell fluids at 0.15 M, assuming constancy of concentrations. In practice, a larger than minimum amount of work is required. 19.105 (a) To obtain from the equilibrium constant data, graph InK at various temperatures vs 1/T, being sure to employ absolute temperature. The slope of the linear relationship that should result is thus, is easily calculated. (b) Use = TAS° and = -RT In K. Substituting the second expression into the first, we obtain - RT In K = In = RT R = RT + R Thus, the constant in the equation given in the exercise is AS°/R. 19.106 S = k In W (Equation [19.5]), k = R/N, W (The number of particles, m, is the same in both states.) 605