Question

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Complete the following table of data for Na2HASO4. (Assume the density of water is 1.00 g/mL.) Concentration (M AG (kJ/mol) 40] 49 49 49 T(°C) Solubility (g/100. g) 22.0 29.3 32.0 39.2 42.0 49.0 52.0 58.9 62.0 68.7 72.0 82.0 88.4 49 40 49 78.6 1442 49 40 Use a graph of AG vs. T to determine the following. (Enter your answers to three significant figures.) What is the value of AH for the system? kJ/mol What is the value of As for the system? J/mol K

Answer 1

* Given salt Na, HASDAL Gram molecular weight of this salt V=2312) 41(1) +35(1) +614) - 186 grams med Fox Ist reading, T: 2280 solubility = 29.3 g salt/100g water concentration : nouo moles of salt un e volume of water. n = w lt 29.39 - 0.158 mo gmw 186 g/mol V=100 me water -0.9 litre water. LOCAL concentrations n = 0.158 = 1:58 : S (say) V 0.1 Na HASO 2Nat+ HASO Ksp = (ne) (Hasoj] - (25)2 (5) 1: There are 2 Nations). = 453 k sp - 4 (1.58)3 - 15.636. AG = RT enkso R = 8.314 J/molk T: 298 2956(2ąc) E-8316 (295) es 15.636). = -502-9 J/mol 2 -0 503 kJ/mol The values of 'in, 'concentration, Ksp and DG are calculated and tabulated in an Erces Spreadsheet.

E T(degree c) Solubility(g/100g) No. of moles concentration(M) Ksp delta G(kJ/mol) 0.158 42 29.3 39.2 49 58.9 68.7 78.6 88.4 0.211 0.263 0.317 0.369 0.423 0.475 1.575 2.108 2.634 3.167 3.694 4.226 4.753 15.636 37.444 73.132 127.019 201.554 301.848 429.416 -0.503 -0.964 -1.499 -2.094 -2.735 -3.418 -4.133 72 82 delta G vs T 0.000 0 10 20 30 40 50 60 70 80 90 -0.500 -1.000 -1.500 -2.000 -2.500 -3.000 -3.500 y = -0.0608x + 0.9713 -4.000 24 -4.500 25 Sheet1

from the graph of AG vs To the equation of the line is y=-0.0608x + 0.9713 - 0 EV Iwe know that BAAG = AH-TAS The above equatim is of the form Су: mx + c whenen en m = slope 3 - AS BET les intercept = OH comparing 6) & we get LAS = 0.0608 kJmdk = 60-8 J/molka AH = 0-9713 kJ/mol