For the dissolving of ammonium nitrate in water, delta H = 25.7 KJ/mol and delta

Question

For the dissolving of ammonium nitrate in water, delta H = 25.7 KJ/mol and delta S = 108.7 J/mol K. Calculate delta G and Ksp for the process at 25 degree C. b. Is the process endothermic or exothermic? c. Is the sign of the entropy change consistent with your expectations? Explain your answer. 2. 100 mL of a saturated aqueous SrCrO4 solution contains 0.12 g of strontium chromate at 25 degree C. Calculate Ksp and delta G for the dissolving of strontium chromate in water. 16.5 mL of 0.1000 M HCl are required to titrate a 10.00 mL aliquot of a saturated strontium hydroxide solution (20 degree C). Calculate Ksp and delta G for the dissolving of strontium hydroxide in water.

Explanation / Answer

[SrCrO4] = 0.12g/100ml = 1.20g/L = 1.2/203.63 = 5.89 * 10^-3mol/L
Ksp = [Sr2+][CrO4^2] = (5.89 * 10^-3)^2 = 3.47 * 10^-5M^2
dG = -RTlnKsp = -8.314 * 298 * ln3.47 * 10^-5 = 25,441.62J = 25.44kJ

Sr(OH)2 + 2HCl = SrCl2 + 2H2O

1 mole of Sr(OH)2 = 2 moles of HCl
No. of moles of HCl = 0.10 * 0.0165 = 0.00165 Hence moles of Sr(OH)2 = 0.00165/2 = 8.25 x 10^-4
Now Sr(OH)2 = Sr2+ + 2OH- Then [Sr] = 8.25 * 10^-4M and [OH-] = 2 * 8.25 * 10^-4 = 1.65 * 10^-3M. Ksp = [Sr2+][OH-]^2= (2 * 8.25 * 10^-4)^2=2.72*10^-6

dG = -RTlnKsp = -8.314 * 298 * ln2.72*10^-6= 20337.86J=20.34 KJ

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