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
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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 = ?
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.)
To calculate the heat transferred when 2.5 g of PCl3 is formed at constant pressure, we need to follow these steps:
Step 1: Convert mass to moles
The given mass of PCl3 is 2.5 g. To convert grams to moles, we need to divide the given mass by the molar mass of PCl3. The molar mass of PCl3 can be calculated by adding the atomic masses of phosphorus (P) and three chlorine (Cl) atoms:
Molar mass of PCl3 = (1 mol P × atomic mass of P) + (3 mol Cl × atomic mass of Cl)
= (1 × 31.0 g/mol) + (3 × 35.5 g/mol)
= 31.0 g/mol + 106.5 g/mol
= 137.5 g/mol
Moles of PCl3 = mass of PCl3 / molar mass of PCl3
= 2.5 g / 137.5 g/mol
≈ 0.018 mol
Step 2: Use stoichiometry
The balanced equation for the reaction is:
2 POCl3 → 2 PCl3 + O2
From the balanced equation, we can see that every 2 moles of POCl3 produce 2 moles of PCl3. Therefore, the moles of POCl3 consumed will be the same as the moles of PCl3 produced.
Moles of POCl3 = 0.018 mol
Step 3: Calculate heat transferred
We are given the value of the heat change (ΔH) for the reaction: ΔH = 508 kJ/mol.
Heat transferred = ΔH × (moles of POCl3 or PCl3 involved in the reaction)
= 508 kJ/mol × 0.018 mol
≈ 9.15 kJ
Therefore, the heat transferred when 2.5 g of PCl3 is formed at constant pressure for the reaction 2 POCl3 → 2 PCl3 + O2 is approximately 9.15 kJ.