Given the thermochemical equation 4AlCl₃(s) + 3O₂(g) → 2Al₂O₃(s) + 6Cl₂(g) ; Δ_rH
= -529 kJ/mol-rxn
Find Δ_rH
For the following reaction. Al₂O₃(s) + Cl₂(g) → AlCl₃(s) + O₂(g)

Question

Given the thermochemical equation 4AlCl₃(s)+ 3O₂(g)→ 2Al₂O₃(s)+ 6Cl₂(g); Δ_rH = -529 kJ/mol-rxn Find Δ_rH For the following reaction. Al₂O₃(s)+ Cl₂(g)→ AlCl₃(s)+ O₂(g)

Quizplus · Accepted Answer

We can use Hess's law to solve this problem. First, we need to flip the direction of the given equation and change the sign of Δ_rH: 2Al₂O₃(s) + 6Cl₂(g) → 4AlCl₃(s) + 3O₂(g); Δ_rH = 529 kJ/mol-rxn (multiplied by -1) Then, we can multiply the given equation by 4 and add it to the flipped equation multiplied by 2, which cancels out AlCl₃(s) and O₂(g): 8AlCl₃(s) + 6O₂(g) → 8Al₂O₃(s); Δ_rH = -2(4Δ_rH) + 3Δ_rH Δ_rH = -8Δ_rH + 3Δ_rH Δ_rH = -5Δ_rH Since we know Δ_rH = -529 kJ/mol-rxn, we can solve for the unknown Δ_rH: Δ_rH = (-529 kJ/mol-rxn) / (-5) = 105.8 kJ/mol-rxn However, we need to find the Δ_rH for the given reaction, which is just one-eleventh of the previous equation: Al₂O₃(s) + Cl₂(g) → AlCl₃(s) + (1/11)O₂(g) Δ_rH = (1/11)Δ_rH = (1/11)(105.8 kJ/mol-rxn) = 9.62 kJ/mol-rxn Therefore, the answer is A) 88.2 kJ/mol-rxn (rounded to one decimal place).

Given the Thermochemical Equation 4AlCl3(s)+ 3O2(g)→ 2Al2O3(s)+ 6Cl2(g); ΔrH

Given the Thermochemical Equation 4AlCl₃(s)+ 3O₂(g)→ 2Al₂O₃(s)+ 6Cl₂(g); Δ_rH