Answer :
The enthalpy change ΔH for the given reaction is calculated to be -1 kJ. Option E is correct.
The enthalpy change for a reaction is calculated using the bond energies. The bond energies of the reactants are added and the sum of the bond energies is subtracted from the products. The difference represents the change in enthalpy.
Given bond energies:
Cl-Cl = 243 kJ/mol
F-F = 160 kJ/mol
Cl-F = 255 kJ/mol
The total bond energy of the reactants (Cl₂ and 3F₂):
Reactants: Cl₂ (g) + 3 F₂ (g)
Total bond energy of Cl₂ = 2 * Cl-Cl = 2 * 243 kJ/mol = 486 kJ/mol
Total bond energy of 3F₂ = 3 * (2 * F-F) = 3 * (2 * 160 kJ/mol) = 960 kJ/mol
Sum of reactant bond energies = 486 kJ/mol + 960 kJ/mol = 1446 kJ/mol
The total bond energy of the products (2ClF₃):
Total bond energy of 2ClF₃ = 2 * (3 * Cl-F) = 2 * (3 * 255 kJ/mol) = 3060 kJ/mol
The sum of product bond energies = 3060 kJ/mol
ΔH is calculated by subtracting the sum of the product bond energies from the sum of the reactant bond energies:
ΔH = Sum of reactant bond energies - Sum of product bond energies
ΔH = 1446 kJ/mol - 3060 kJ/mol
ΔH = -1614 kJ/mol
ΔH is approximately -1614 kJ/mol. Therefore, the correct option is E. -1 kJ.
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