Explain the solubility product for barium sulfate

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Reference no: EM13519013

1) The solubility product for barium sulfate is 1.10x10-10. Calculate ?G° in kJ/mol at 298 K. Report your answer with one decimal point.

2) The vapor pressure of water is the same as the equilibrium constant (Kp) for the following reaction. Report your answer with three decimal points.
H2O (l) ? H2O (g) ?Ho = 40.7 kJ/mole
At 20oC, Kp has a value of 0.036 atm. Calculate Kp at 10oC. (R= 8.314 J/K)

3) Calculate the standard free energy change at 298K for the conversion of diamond to graphite and determine whether or not the reaction is spontaneous under standard conditions:
C(diamond) ? C(graphite);
?Hf°(diamond) = 1.895 kJ/mol; S°(diamond) = 2.337 J mol-1K-1; S°(graphite) = 5.740 J mol-1K-1.

A. 2.2 kJ; not spontaneous
B. -1.9 kJ; spontaneous
C. -2.9 kJ; spontaneous
D. 1.9 kJ; not spontaneous
E. 4.0 kJ; not spontaneous

Reference no: EM13519013

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