2.  Calculate the equilibrium constant at 25 C  for the preparation of chlorine by the reaction of hydrochloric acid with pyrolusite, used by Scheele in 1774:
    2Cl^-(aq) + 4H₃O⁺(aq) + MnO₂(s)  <====>
                    Cl₂(g) +  Mn²⁺(aq) +  6H₂O(l)
Relevant data can be found in Appendix D.
      We know the equation for the change in Gibbs free energy to be:
                              dG  =  -RTlnK

        We already know R, the universal gas constant, and T, the temperature.  We can determine dG by using the data in Appendix D.  The dG should be the products minus the reactants.

    dG = dG(Cl₂) + dG(Mn²⁺) + 6dG(H₂O) - 2dG(Cl^-) - 4dG(H₃O⁺)
              - dG(MnO₂)

          =  0 + (-228.1) + 6(-237.18) - 2(-131.23) - 4(-237.18) - (-465.17)

          =  25.17 kJ mol-1  =  25170 J mol-1

        We now solve our initial equation for ln K and substitute in our determined values to solve the equation.

                ln K  =  -dG
                              RT

                         =          -(25170 J / mol)
                            (8.31451 J/ mol K)(298.15 K)

                 ln K =  -10.15

        We now can solve for K and find the equilibrium constant.

                     K =  e^-10.15

                         =  3.89 x 10^-5
4.  Calculate the standard enthalpy change per mole of Na₂CO₃ produced by Solvay process, using data from Appendix D.  From your calculation, what do you conclude about the overall heat requirements of that process?
        There are two different reactions which we must concern ourselves with to determine the standard enthalpy change.  The Solvay process includes the reactions:
                2NaCl + 2NH₃ + 2CO₂ + 2H₂O --->  2NaHCO₃ + 2NH₄Cl

                2NaHCO₃ --->  Na₂CO₃ + H₂O + CO₂
        We must then determine the standard enthalpy changes for each of the reactions.  From there we added them together to determine the standard enthalpy change for the entire reaction.  First the we will do the first equation.

      dH₁ = dH(2NaHCO₃ + 2NH₄Cl - 2NaCl - 2 NH₃ - 2CO₂ - 2H₂O)
              = 2(-950.81) + 2(-314.43) - 2(-411.15) - 2(-80.29)
                 - 2(-393.51) - 2(-285.83)
              = - 188.92 kJ / mol

        Now we do the second equation.

      dH₂ = dH(Na₂CO₃ + H₂O + CO₂ - 2NaHCO₃)
              =  (-1130.68) + (-241.82) + (-393.51) -2(-950.81)
              =  135.61 kJ / mol

        We can now determine the standard enthalpy change for the entire reaction.

      dH Na₂CO₃ = dH₁ + dH₂
                             = (-188.92 + 135.61) kJ / mol
                             = -53.31 kJ / mol
        From this we can determine that the reaction is endothermic and it should take place in an environment with high heat.

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