ΔHf??= ΔHat??+ ΔHat??+?IE?+?EA?+ ΔHlatt?
ΔHf??= ΔH1??+ ΔHlatt?
So, if we rearrange to calculate the lattice energy, the equation becomes
ΔHlatt??= ΔHf??- ΔH1?
Calculating the lattice energy of KClGiven the data below, calculate the ΔHlatt??of potassium chloride (KCl)??
Answer
Step 1:?The corresponding Born-Haber cycle is:
Step 2:?Applying Hess’ law, the lattice energy of KCl?is:
ΔHlatt??= ΔHf??- ΔH1?
ΔHlatt??= ΔHf??- [(ΔHat??K) + (ΔHat??Cl) + (IE1?K) + (EA1?Cl)]
Step 3:?Substitute in the numbers:
ΔHlatt??= (-437) - [(+90) + (+122) + (+418) + (-349)]?= -718 kJ mol-1
Calculating the lattice energy of MgOGiven the data below, calculate the of ΔHlatt??magnesium oxide of magnesium oxide (MgO)
Answer
Step 1:?The corresponding Born-Haber cycle is:
Step 2:?Applying Hess’ law, the lattice energy of MgO is:
ΔHlatt??= ΔHf??- ΔH1?
ΔHlatt??= ΔHf??- [(ΔHat??Mg) + (ΔHat??O) + (IE1?Mg) + (IE2?Mg) + (EA1?O) + (EA2?O)]
Step 3:?Substitute in the numbers:
ΔHlatt??= (-602) - [(+148) + (+248) + (+736) + (+1450) + (-142) + (+770)]
= -3812 kJ mol-1
Make sure you use brackets when carrying out calculations using Born-Haber cycles as you may forget a +/- sign which will affect your final answer!
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