equil_solutions-1 - Solutions to the Practice Problems on...

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Unformatted text preview: Solutions to the Practice Problems on Equilibrium ¡ 2 a2 PCO PCO P 2 CO aC aCO2 1 PCO2 P P PCO2 aH2 CO3 H2 CO3 c P H2 CO3 (b) K aCO2 aH2 O PCO2 P XH2 O c PCO2 XH2 O It is usually the case that the mole fraction of the solvent is very close to 1. It is then a reasonable approximation to write P H2 CO3 K c PCO2 ¤ ¥¤ ¦ ¥¤ OH ¢¤ ¥¦¡  Ba2 3 c 2 ¦ £ 2 ¥ c 2 PNH3 PCO2 PH2 O P4 ¢¤ ¢¤ ¡ ¢ ¤ £¦  ¢¤ £ ¡ ¡ ¢ ¡ ¢ ¥ ¢ ¤ £¦  ¡ 1 ¡ OH £ c PH2 O P ¢¤ ¢ ¡ ¡ ¡ ¢ ¢ Ba2 PCO2 P 1 ¡ ¥¡ £ ¡ ¡ ¡ ¥ ¡ ¢ © 2 2 § ¢¤ ¡ ¢ aBa2 aOH aBa OH 2 PNH3 P ¤ ¡ £ ¤ £¦ ¤ ¢¤ £ ¢ ¡ ¨ © ¤ ¨ ¢ ¢ (c) K (d) K aNH3 2 aCO2 aH2 O a NH4 2 CO3 ¦ ¢¤ (a) K 1. ¤ Because P 1 atm and c 1 mol L, we don’t normally write down these factors. However, it then becomes essential to remember to express concentrations in mol/L and pressures in atm. ¡ ¡ §   £  § ¢ ¡ ¡ 0 56 0 17 1 8 10 3 £¢  § ¢ §  ¢ ¡ aPCl3 aCl2 aPCl5 53. ¢ ¡  § § § 4. We can’t do anything in these problems without a balanced reaction. The reaction is SO3 g ©¨  © ¨  © ¨ The corresponding equilibrium constant is aSO3 aSO2 aO2 12 § ¢ ¡ ¢ ¡ K § 1 O 2 2g SO2 g Note that the data are given to us in the wrong form: The activities of gases are computed from the pressure, not from the number of moles in a litre. However, the ideal gas law lets us calculate pressures n from these concentrations: P RT V . I like to do calculations with the ideal gas law in SI units, because it’s less likely that I will make mistakes if I’m completely consistent about this. There are 1000 L in 1 m3 , so the concentration of (for instance) sulfur dioxide in SI units is 3 77mol m3 . Thus 0 309atm § 31 3 kPa § § £ ¡ § ¢ 1 ¢ 1000K 3 77mol m3 ¡ ¢ 1  mol § 1 £  8 314510J K ! § ¡ PSO2 ¢ 3. The equilibrium constant is K £ 2. The reaction ratio for this reaction is Q a2 aI2 . According to the measurements, aI2 0 87atm 1 atm I 0 87 and aI 8 2 10 7 . The reaction ratio therefore has the value Q 7 7 10 13 . Since this is much smaller than the equilibrium constant, the reaction is not in equilibrium. In fact, the dissociation of iodine molecules is spontaneous under these conditions. 0 353atm. The equilibrium constant is therefore 12 1 85 ¢ § 0 339 0 309 0 353 § ¡§ § ¢ § § ¡ K 0 339 and PO2 Similarly, PSO3 § Note that we needed the balanced reaction to determine the form of the equilibrium constant. We could have balanced the reaction differently. For instance, we could have avoided the use of fractions by multiplying the reaction by 2: O2 g 2SO3 g ©¨ § ©¨ 2SO2 g © ¨ The equilibrium constant would then have been 3 41 ¢ § a2 3 SO aSO2 2 aO2 § ¡ ¡ ¢ K The value of the equilibrium constant depends on how we balance the reaction. In fact, an equilibrium constant not accompanied by a balanced reaction is meaningless. In some cases to be discussed later (e.g. solubility, acid dissociation), there is a convention about how we should write the reaction. Otherwise, you must always give a balanced reaction when stating an equilibrium constant. 0 098mol § 25g 253 808g mol £ § nI2 nRT V ), £ 5. We need to know the initial pressure of iodine. We can get this from the ideal gas law (P but first we need to know the number of moles. § The initial pressure of molecular iodine is therefore 68kPa # 1000K 0 67atm § 1 § ¡ ¢  ¡ 0 098mol 8 314510J K 1 mol 12 10 3 m3 ¢  §   ¢ § ¡ PI2 initial " We want to determine the stoichiometric relationships among the activities (which enter into the equilibrium ratio). Since pressure is proportional to number and activity is proportional to pressure, we have Final aI2 2 0 67 aI2 ¢ ¡ $ aI2 aI Initial 0.67 0 § The last entry in the table is obtained by stoichiometry: The number of iodine molecules reacted is calculated by subtracting the final amount from the initial. The number of iodine atoms formed is twice as large, by stoichiometry. Using the data in the table, we see that the equilibrium relationship is ¥ % 4a22 I §  5 36aI2 § § $ § § 2 2 0 67 aI2 aI2 2 § 1 80 ¡ 0 31aI2 a2 I aI2 $ 0 31 ¦ &¢ K § ' ¢ $ ¡§¡§ § $ 5 672 4 4 1 80 24 1 83 § ¡ ¢ ) 0( § 5 67 § 5 67 24 0 94 or 0 48 § ¡ ¢ ( ' § § ' aI2 1 80 ¢ 5 67aI2  4a22 I 0 § The first solution is wrong: It corresponds to a larger I2 pressure than we started with. The second solution is the physically correct one. The equilibrium pressures are therefore PI2 0 48atm and PI 2 0 67 0 48atm 0 38atm. ¢ § 0 92 § $ ¡ $ § 2 0( ¢ § $ § § 1 $ § $ § 0 165atm 0 34atm § § § § §  § § 3 0 165 § § 0 005  § ¡ 0 005 ¡ § § " $ ' '  PBr2 4 3 4¢ § ¢ ' § §  PCO x 5 42 § ¡ § 0 17 x2 0 17 x $ ¡ ¢ §  PCOBr2 0 92 1 54 2 0 165 ' The total pressure is P § § ¡ aCOBr2 x $ ¢ aBr2 § §   aCO 5 4x x 0 17atm §  x  0 ¢ 5 4 0 17 2 § x2 17kPa Final 0 17 x x x aCO aBr2 aCOBr2 54 346 15K ¢ Initial 0.17 0 0 aCOBr2 aCO aBr2 K 1 # ¡ 0 015mol 8 314472J K 1 mol 2 5 10 3 m3 ¡ $ PCOBr2 initial § 6. ...
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