Post a New Question

Chemistry College (Need to know!)

posted by on .

Am I Correct in assuming this?

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

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

Units as kJmol-1

  • Chemistry College (Need to know!) - ,

    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!) - ,

    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?

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question