Question

In: Chemistry

Consider the simple salts NaCl( s ) and AgCl( s ). We will examine the equilibria...

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 | [email protected] | Printed from www.chegg.com

Solutions

Expert Solution

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


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