Thermodynamics PRE-LAB QUESTIONS 1. Using your previous general chemistry knowle

Question

Thermodynamics PRE-LAB QUESTIONS 1. Using your previous general chemistry knowledge, give some guidelines for predicting the sign of ??' in terms of breaking and forming bonds. 2. Which scenario from Table 8-1 is consistent with what we know about the melting and forming of ice as a function of temperature? 3. Write the balanced chemical equation for the reaction of hydrochloric acid with calcium hy- droxide. What are the total ionic and net ionic equations for the reaction? Balanced chemical equation Total ionic equation: Net ionic equation: When adding the 25 mL of DI water to your saturated solution of cakcium hydroxide, you are changing the ion concentration. Why does this not matter? 4. Saturated solutions of barium hydroxide are prepared at two different temperatures, equilibrat ed, filtered, and then 15.00 mL portions of the solution are titrated with 0.180 M hydrochloric acid: 5. Solution Temperature Titration Volume of HCI 17 25 65 what are the values of, and ?G° at each temperature and what are the values of ??"and for the dissociation of barium hydroxide? 123

Explanation / Answer

1. Brekaing of bonds need energy (dH = +ve), whereas, formation of bonds releases energy (dH = -ve)

when dHrxn = -ve, it is an exothermic reaction

when dHrxn = +ve, it is an endothermic reaction

2. Table not shown here.

3. reaction of Ca(OH)2 with HCl

balanced reaction,

Ca(OH)2 + 2HCl --> CaCl2 + 2H2O

total ionic,

Ca2+ + 2OH- + 2H+ + 2Cl- --> Ca2+ + 2Cl- + 2H2O

net-ionic,

H+ + OH- --> H2O

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4. when water was added to Ca(OH)2 solution, the ion concentration changes (molarity), however the total moles of ions remained the same in solution and moles of Ca(OH)2 reacted is equal to 2 moles of HCl. therefore no effect on the titration is seen.

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5. Ba(OH)2 + 2HCl --> BaCl2 + 2H2O

at 25 oC

moles HCl used = 0.180 M x 17 ml = 3.06 mmol

moles Ba(OH)2 reacted = 3.06 mmol/2 = 1.53 mmol

molarity [Ba2+] = 1.53 mmol/15 ml = 0.102 M

molarity [OH-] = 2 x 0.102 M = 0.204 M

Ksp = (0.102)(0.204)^2 = 4.24 x 10^-3

dG = -RTlnKsp = -8.314 x (25 + 273) ln(4.24 x 10^-3)

      = 13.54 kJ

Similarly calculation for T = 55 + 273 = 328 K can be done

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