Consider the simple salts NaCl( s ) and AgCl( s ). We will examine the equilibri
ID: 882257 • Letter: C
Question
Consider the simple salts NaCl( s ) and AgCl( s ). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions:
NaCl 1 s 2 Na + 1 aq 2 + Cl - 1 aq
2 AgCl 1 s 2 Ag + 1 aq 2 + Cl - 1 aq 2
(a) Calculate the value of G ° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the en-tropy term of the standard free-energy change? (c) Use the values of G ° to calculate the K sp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will G ° for the solution process of these salts change with increasing T ? What effect should this change have on the solubility of the salts? Copyright | Prentice Hall | Chemistry | Edition 13 | woodsbannecker@yahoo.com | Printed from www.chegg.com
Explanation / Answer
NaCl(s) <==> Na+(aq) + Cl-(aq)
AgCl(s) <==> Ag+(aq) + Cl-(aq)
a) delta Go = delta H - T delta S
we use free energy of formation of product and reactant to calculate delta Go of reaction.
delta Go NaCl(sol) = (-261.9 KJ/mol)+ (-131.2KJ/mol) - (-384KJ/mol) = -9.1KJ/mol
delta Go AgCl(sol) =(-77.1KJ/mol) + (-131.2KJ/mol) - (109.7KJ/mol) = + 55.6KJ/mol
b)delta S for NaCl =+43.2 J/K-mol delta H =+3.6KJ
delta S for AgCl = + 34.3 J/K-mol delta H = + 65.7KJ
entropy of NaCl & AgCl are very similar , but enthalpy of two solutions are very different.
Difference in delta Go is due to large difference in enthalpy.
C) Ksp = e^-deltaGo/RT
Ksp of NaCl = [Na+][Cl-] = e^-(- 9100/8.314x298) = e^3.7 = 40
Ksp of AgCl = [Ag+][Cl-] = e^-(55600/8.314x298) = e^-22.4 = 1.9X10^-10
d) yes -
Ksp of NaCl is high <1 so it is soluble in water.
Ksp of AgCl is very low >1 so it is insoluble in water.
e)delta S for both salt is +ve
when deltaH & delta S is constant
deltaGo become more negative when term -TdetaS is more negative.This term become more negative by increasing value of T. since Ksp =e^-deltaG/RT.
when deltaG is more negative value of KSp is high.So salt is more soluble at higher temperature