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How many moles POCl3 do you have? That's 2.5 g/ molar mass POCl3 = ?
Then q = heat transferred = 508 kJ/mol x #mols x 2 = ?
calculate heat transferred when 2.5g pocl3 is formed at constant pressure for the reaction 2pocl3-2pcl3+o2. delta H for 2pocl3-2pcl3+o2 =508KJ.mol-1
2 answers
Addendum note FYI => The ΔH value given in your problem is the ‘Standard Heat of Formation’ for POCl₃(s) from its basic elements in standard state. The Standard Heat of Formation is the heat gained or lost on formation of a substance from its basic elements in their standard thermodynamic state. That is, => P(s) + ½O₂(g) + Cl₂(g) => POCl₃(g) + 508Kj/mole and not Heat of Reaction as your problem notation indicates. To answer the problem given, you will need the Heat of Rxn for 2POCl₃(g) => 2PCl₃(g) + 3O₂(g) => ΔHrxn = +438Kj (endothermic) for the decomposition of POCl₃.
This requires a Hess’s Law calculation.
ΔHrxn = Σn∙ΔHᵪ(Products) - Σn∙ΔHᵪ(Reactants); ΔHᵪ = Standard Thermodynamic Heat of Formation
ΔHrxn = [2mole PCl₃(-289Kj/mole) + 3moleO₂(0.00Kj/mole)] – [2mole POCl₃(-508Kj/mole)] = +438 Kj
For your problem, you are asking for the heat flow for formation of 2.5 grams of POCl₃(g) if given the reaction 2POCl₃(g) => 2PCl₃(g) + 3O₂(g) & its Heat of Rxn.
You need to reverse this reaction to show POCl₃(g) formation as a product and use the correct Heat of Reaction value; +438Kj.
That is, use 2PCl₃(g) + 3O₂(g) => 2POCl₃(g) + 438Kj or, for simplicity, PCl₃(g) + 3/2O₂(g) => POCl₃(g) + 219Kj. Convert 2.5 grams POCl₃(g) to moles => 2.5g/152g/mole = 0.0165mole POCl₃(g) and multiply by Heat of Rxn (-219Kj/mole) => ΔH(formation 2.5g POCl₃(g)) = 0.0165mole POCl₃(g) x (-219Kj/mole) = 3.602Kj ~ 3.6Kj (2 sig.figs.)
This requires a Hess’s Law calculation.
ΔHrxn = Σn∙ΔHᵪ(Products) - Σn∙ΔHᵪ(Reactants); ΔHᵪ = Standard Thermodynamic Heat of Formation
ΔHrxn = [2mole PCl₃(-289Kj/mole) + 3moleO₂(0.00Kj/mole)] – [2mole POCl₃(-508Kj/mole)] = +438 Kj
For your problem, you are asking for the heat flow for formation of 2.5 grams of POCl₃(g) if given the reaction 2POCl₃(g) => 2PCl₃(g) + 3O₂(g) & its Heat of Rxn.
You need to reverse this reaction to show POCl₃(g) formation as a product and use the correct Heat of Reaction value; +438Kj.
That is, use 2PCl₃(g) + 3O₂(g) => 2POCl₃(g) + 438Kj or, for simplicity, PCl₃(g) + 3/2O₂(g) => POCl₃(g) + 219Kj. Convert 2.5 grams POCl₃(g) to moles => 2.5g/152g/mole = 0.0165mole POCl₃(g) and multiply by Heat of Rxn (-219Kj/mole) => ΔH(formation 2.5g POCl₃(g)) = 0.0165mole POCl₃(g) x (-219Kj/mole) = 3.602Kj ~ 3.6Kj (2 sig.figs.)
2 answers