Use the bond energies provided to estimate Δ_rH° for the reaction below. PCl₃(g) + Cl₂(g) → PCl₅(l) Δ_rH°= ? Bond Bond Energy (kJ mol^-1) Cl-Cl 243 P-Cl 331 A) -243 kJ mol^-1 B) -419 kJ mol^-1 C) -662 kJ mol^-1 D) -67 kJ mol^-1 E) -905 kJ mol^-1

Question

Quizplus · Accepted Answer

The change in standard molar entropy for the reaction can be calculated by using the bond energies to determine the bond dissociation enthalpies of the reactants and products, and then using those values to calculate the change in standard molar entropy using the equation:

ΔS = ΣnΔS(bonds broken) - ΣnΔS(bonds formed)

Where n is the number of moles of each substance, and ΔS(bonds broken/formed) is the standard molar entropy of breaking/forming each bond.

Using the given bond energies, we can calculate the bond dissociation enthalpies as follows:

PCl3(g) → P(g) + 3Cl(g)
ΔH = 3(P-Cl) = 3(331 kJ/mol) = 993 kJ/mol

Cl2(g) → 2Cl(g)
ΔH = 1/2(Cl-Cl) = 1/2(243 kJ/mol) = 121.5 kJ/mol

PCl5(l) → P(g) + 5Cl(g)
ΔH = 5(Cl-Cl) + (P-Cl) = 5(243 kJ/mol) + 331 kJ/mol = 1446 kJ/mol

Now we can substitute these values into the equation for ΔS to get:

ΔS = (1 mol)(993 J/K/mol + 121.5 J/K/mol) - (1 mol)(1446 J/K/mol)
ΔS = -331.5 J/K

Finally, we can use the standard Gibbs free energy equation to relate ΔS to ΔH:

ΔG = ΔH - TΔS

At standard conditions (298 K), this becomes:

ΔG° = ΔH° - TΔS°
ΔG° = -1446 kJ/mol - (298 K)(-0.3315 kJ/K/mol)
ΔG° = -419 kJ/mol

Therefore, the answer is B, -419 kJ/mol.

Use the Bond Energies Provided to Estimate ΔrH° for the Reaction

Use the Bond Energies Provided to Estimate Δ_rH° for the Reaction