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15 Chemical Equilibrium Solutions to Exercises The change in is (1.000 0.512) = -0.488 atm, so the change in PN₂O₄ is (b) (c) 15.41 Analyze/Plan. Follow the logic in Sample Exercise 15.9. mM = 10⁻³ M X(aq) + Y(aq) 1L XY(aq) initial 1.0 mM 1.0 mM 0 change -0.80 mM -0.80 mM +0.80 mM equil. 0.20 mM 0.20 mM 0.80 mM 15.42 The initial concentrations of drug candidate and protein are the same in the two experiments, and the two reactions have the same stoichiometry. At equilibrium, the concentration of B-protein complex is greater than the concentration of A-protein complex, so drug B is the better choice for further research. Calculation of equilibrium constants for the two reactions confirms this conclusion. A(aq) + protein(aq) 1L A-protein(aq) initial 2.00 10⁻⁶ mM 0 change -1.00 +1.00 equil. 1.00 B(aq) + protein(aq) B-protein(aq) initial 0 change equil. Applications of Equilibrium Constants (section 15.6) 15.43 (a) A reaction quotient is the result of the law of mass action for a general set of concentrations, whereas the equilibrium constant requires equilibrium concentrations. (b) In the direction of more products, to the right. 443