Homework Help: Chemistry College (Need to know!)

Posted by Ben on Thursday, June 11, 2009 at 5:18pm.

Am I Correct in assuming this?

dH of solution = -dH of lattice enthalpy + dH of hydration

E.g:
dHsolution NaCl = -dHLE NaCl(s) + dHhyd Na+(g) + dHhyd Cl-(g)

Units as kJmol-1

Chemistry College (Need to know!) - GK, Thursday, June 11, 2009 at 5:43pm
Your signs on the right side are reversed. Bond breaking is endothermic and bond formation is exothermic. Some bond formations are endothermic (i.e. hydrogen bonding) that must be taken into account in conjunction with hydration which is also bond formation.
∆H(soln) = |∆H(lattice)| - ∆H(solute-solvent)
Chemistry College (Need to know!) - Ben, Sunday, June 14, 2009 at 4:30pm
How is this Possible GK.
Everywhere I have looked, including mark schemes and chemistry websites dH(solution) is worked out via this equation:

dH(solution) = dH(LE) - dH(hyd) + dH(hyd)

What is it you are working out with solute and solvent?

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