Question

Complete the following table of data for Ce2(SO4)3 . 9 H20. (Assume the density of water is 1.00 g/mL.) T (°C) Solubility (g/100. g) Ksp Concentration ΔG (M) (kJ/mol) 40 0.104 401,040 35.0 7.62 40 0.0850 4.0 4.0 43.0 6.21 40 0.0688 4.0 4.0 51.0 5.03 4.0 0.0561 4.0 4.0 59.0 4.10 40 0.0465 4.0 4.0 67.0 3.40 40 0.0401 4.0 4.0 2.93 75.0 40 0.0370 4.0 4.0 83.0 2.71 three significant figures.) Use a graph of AG vs. T to determine the llowing. Enter your an What is the value of AH for the system? 40-106 k/mol What is the value of AS for the system? 4.0 398 x J/mol K

Please help this is due at 1:30pm EST

Answer 1

K_sp for the equilibrium of Ce₂(SO₄)₃:

$Ce_{2}(SO_{4})_{3}_{(s)}\leftrightharpoons 2\, Ce^{3+}_{(aq)}+3\, SO_{4(aq)}^{2-}$

Is given by:

$K_{sp}=[Ce^{3+}]^{2}[SO_{4}^{2-}]^{3}$

Where these are equilibrium concentrations, which are equivalent to the highest possible concentration of the ions in the aqueous phase in contact with the solid precipitate. This concentrations are given by the solubility of the compound, bearing in mind that, due to the stoichiometry of the salt, the concentration of Ce³⁺ will be double (there are two moles Ce³⁺ per each mole of salt) the concentration of the salt (given by its molar solubility) and the concentration of SO₄^2- will be triple the concentration of salt. With this, we can re-write the expression for Ksp as:

$K_{sp}=(2s)^{2}(3s)^{3}=4s^{2}\cdot 27s^{3}=108s^{5}$

And this value can be calculated for each temperature, since you have already calculated the molar solubility. For example, for 35.0 °C:

$K_{sp}=108\cdot 0.104^{5}=0.00131$

The rest of the values are on the following table, which I built to plot the graph that comes after (please bear in mind that commas are decimal separators):

T(°C) 324 molar solubility Kps 308 0,104 0,00131399 316 0,085 0,0004792 0,0688 0,00016648 332 0,0561 6,0012E-05 340 0,0465 2,3479E-05 348 0,0401 1,1198E-05 356 0,037 7,4891E-06

Delta G can be calculated using the expression:

$\Delta G=-RTln(K_{sp})$

Where R is the universal constant and T is the temperature in Kelvins. The values are added to the previous table:

T(°C) 324 332 molar solubility kps Delta G (J/mol) 308 0,104 0,00131399 16989,53202 316 0,085 0,0004792 20080,89472 0,0688 0,00016648 23437,19067 0,0561 6,0012E-05 26832,27725 340 0,0465 2,3479E-05 30131,5192 348 0,0401 1,1198E-05 32982,61653 0,037 7,4891E-06 34931,53535 356

And now, since another expression for delta G is:

$\Delta G=\Delta H-T\Delta S$

A plot of delta G vs T should yield a straight line, with a slope equivalent to -delta S and an intercept equivalent to delta H. The plot in presented here, with the best linear fit:

40000 35000 30000 25000 20000 y = 385,37x - 101461 R2 = 0,9951 delta G (J/mol) 15000 10000 5000 300 310 320 340 350 360 330 T(K)

And from the fit, we can determine that delta S = -385 J/(mol.K) and delta H = -101000 J/mol