Question

1c. The Ksp for AgCl at 25.0oC is 1.782x10–10. The Ksp at 35.0oC Is 4.159x10–10. What are ÆHo and ÆSo for the dissolution reaction of AgCl(s).

d. Ka for acetic acid is1.754x10-5 at 25oC. At 50.0oC,Kais1.633x10–5. What are ÆHo and ÆSo for the ionization of acetic acid?

Answer 1

Apply vant Hoff equation

ln(K) = -H/R*(1/T) + dS/R

substitute both points:

ln(K1) = -H/R*(1/T1) + dS/R

ln(K2) = -H/R*(1/T2) + dS/R

substitute

ln(1.782*10^-10) = -H/(8.314)*(1/(25+273)) + dS/(8.314)

ln(4.159*10^-10) = -H/(8.314)*(1/(35+273)) + dS/(8.314)

substract both:

ln(1.782*10^-10) - ln(4.159*10^-10) = -H/(8.314)*(1/(25+273)) + H/(8.314)*(1/(35+273))

-0.84753 = H (-1/(8.314)*(1/(25+273)) + 1/(8.314)*(1/(35+273))

-0.84753 = H * (-0.000013104)

H = 0.84753 /0.000013104 = 64677.197 J/mol

H = 64.677 kJ/mol

This is endothermic, so as T increses, K increases

solve for S:

ln(1.782*10^-10) = -H/(8.314)*(1/(25+273)) + dS/(8.314)

ln(1.782*10^-10) = -64677.197 /(8.314)*(1/(25+273)) + dS/(8.314)

ln(1.782*10^-10) + 64677.197 /(8.314)*(1/(25+273)) = dS/(8.314)

3.6569 * 8.314 = dS

dS = 30.4034666 J/molK

d)

Apply same logic as previously:

ln(K1) = -H/R*(1/T1) + dS/R

ln(K1/K2) = - H/R*(1/T1-1/T2)

substitute

ln((1.754*10^-5)/(1.633*10^-5)) = - H/(8.314)*(1/(25+273)-1/(50+273)))

H = -ln((1.754*10^-5)/(1.633*10^-5)) * 8.314 / (1/(25+273)-1/(50+273)))

H = -2288.0938 J/mol

H = --.28 kJ/mol

Now, get dS:

ln(K1) = -H/R*(1/T1) + dS/R

ln(1.754*10^-5) = 2288.0938 /(8.314)*(1/(25+273)) + dS/(8.314)

dS = ln(1.754*10^-5) -  2288.0938 /(8.314)*(1/(25+273))

dS = - 11.874(8.314) = -98.720436 J/molK