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Lattice energy of nacl vs naf1/26/2024 ![]() The enthalpy change, Δ H, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in,” positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). An endothermic reaction (Δ H positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants. An exothermic reaction (Δ H negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Average Bond Lengths and Bond Energies for Some Common Bonds For example, C–F is 439 kJ/mol, C–Cl is 330 kJ/mol, and C–Br is 275 kJ/mol. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. Average bond energies for some common bonds appear in Table 1, and a comparison of bond lengths and bond strengths for some common bonds appears in Table 2. ![]() Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Generally, as the bond strength increases, the bond length decreases. The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. The 415 kJ/mol value is the average, not the exact value required to break any one bond. Although the four C–H bonds are equivalent in the original molecule, they do not each require the same energy to break once the first bond is broken (which requires 439 kJ/mol), the remaining bonds are easier to break. The average C–H bond energy, D C–H, is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. The others are progressively more insoluble in water (K sp is 10 -10, 10 -13, and 10 -16 for AgCl, AgBr, and AgI), reflecting increasing covalency as Δχ decreases.\) Of these compounds, only AgF is soluble in water and should be thought of as an ionic compound. Our lattice energy calculation overestimates the ionic contribution in the case of the heavier silver halides, but underestimates the covalent contribution. The covalent bonding contribution to the lattice energies of AgCl, AgBr, and AgI makes these salts sparingly soluble in water.Īgain, we can interpret the fortuitous agreement between the calculated and experimentally obtained energies in terms of compensating errors. This reaction is used as a qualitative test for the presence of halide ions in solutions. Should we interpret the good agreement with values calculated from the ionic model to mean that these compounds are ionic? Clearly, this description is inappropriate for AgI, where the electronegativity difference Δχ is only 0.6 (compare this value to 0.4 for a C-H bond, which we typically view as non-polar).Ī drop of siver nitrate solution, when added to a dilute hydrochloric acid solution, results in the immediate formation of a white silver chloride precipitate. However we are still obtaining answers within about 12% error even for AgI. Looking at the table, we see that the error is small for AgF and becomes progressively larger for the heavier silver halides. It is interesting to repeat this exercise for the silver halides, which have either the NaCl structure (AgF, AgCl, AgBr) or zincblende structure (AgI). The errors in this case are only about 1% of E L. The table below shows results of more detailed lattice energy calculations for ionic fluorides in which the van der Waals term is explicitly included. We can do better by explicitly including the short-range van der Waals attractive energy between ions. The two errors partially compensate, so the overall error in the calculation is small. If we underestimate the attractive energy of the crystal lattice, the energy minimization criterion ensures that the repulsion energy is underestimated as well. This is because we used energy minimization to obtain the repulsion energy in the Born-Mayer equation. The result is promising because we neglected the van der Waals term.īut.how did we get away with neglecting the van der Waals term? Here we have to subtract 2RT to convert our cycle of energies to a cycle of enthalpies, because we are compressing two moles of gas in making NaCl(s) and PΔV = ΔnRT, where Δn = -2.Įxperimentally ΔH f for NaCl is -411 kJ/molīecause all the other numbers in the cycle are known accurately, the error in our calculation is only about 15 kJ (about 2% of E L).
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