Question

Using the information below, calculate the theoretical deltaHsoln for KCl, NaCl, and LiCl.

KCl(s)→K+(g) +Cl-(g) =701 kJ/mol

K+(g)→K+(aq) =-350.0 kJ/mol

Cl-(g)→Cl(aq) =-331.0 kJ/mol

NaCl(s)→Na+(g) + Cl-(g) =788 kJ/mol

Na+(g)→Na+(aq) =-453.0 kJ/mol

LiCl(s)→Li+(g) +Cl- (g) =834 kJ/mol

Li+(g)→Li+(aq) =-540.0 kJ/mol

Answer 1

KCl(s)    →   K+(g) +    Cl-(g)   1 = 701 kJ/mol

K+(g)     →   K+(aq)                                   2 = - 350.0 kJ/mol

Cl-(g)     → Cl-(aq)    3 = - 331.0 kJ/mol

Adding all three equations we get:

soln for KCl = 1 + 2 + 3

= 701 - 350 - 331

soln = 20 kJ/mol

For NaCl:

NaCl(s)   →   Na+(g)   + Cl-(g) 1 = 788 kJ/mol

Na+(g)    →   Na+(aq)                    2 = - 453.0 kJ/mol

Cl-(g)     → Cl-(aq)    3 = - 331.0 kJ/mol

Adding all three equations we get:

soln for NaCl = 1 + 2 + 3

= 788 - 453 - 331

soln = 4 kJ/mol

For LiCl:

LiCl(s)   →   Li+(g) +   Cl- (g)    1 = 834 kJ/mol

Li+(g)    →   Li+(aq)                    2 = -540.0 kJ/mol

Cl-(g)     → Cl-(aq)     3 = - 331.0 kJ/mol

soln for LiCl = 1 + 2 + 3

= 834 - 540 - 331

soln = - 37 kJ/mol