Question

a) Calculate the enthalpy of reaction H2Te(g) +2 02 (g) + 2H20(1) HTeO (aq) from the following data. H2Te(g) + 4H2O2(aq) +HTeO (aq) + 2H2O(1), AH, = -1445 kJ/mol H2(g) + 12 02(g) → H2O(1), AH", = -286 kJ/mol H2(g) + O2(g) → H2O2(aq), AH°, -191 kJ/mol - Hz Telg) + 4 H Orlag) Hs Teos (ag) +240), te= -1445 wd fuel + 4H2O(l) - 4 Malg) + 20219) , DH = +(286-4) Kynol 412(g) + 402(g) - 4162(aq), o Hi = -4.191 kJ/mol HzTelg) + 202(9) + 240(e) – Hoteoolag), OHRxn = - 1445 +(286.44) + (-4,191) = - 1065 k/mol)

Can anyone explain the reason why the third line of addition is necessary to complete the reaction? I feel like it is not necessary because we get 2H₂O from the first line.

Answer 1

Yes, the third line is necessary to complete the reaction.

We need the third line to cancel out the H_{​​​​​​2}​​​​​O_​2 (aq) in the first line.

See the following process,

1. Н. Те (9) +чно, (а) 2. К но с) чH (9) + AO, ca) - HTe o4 (а) +7н,0 (1) — 4н, с) +zo, ca) —> 4,0, сая)

As you can see,

H₂O2 from thired line cancels out H₂O₂ from firest line. 4 H₂ from thired line cancals out 4 H2 from second line. 40. from third line cancels out 20₂ from second line thereby "202 remains in the final equation. &

After the whole cancellation process,

— см 1. H, Te ca) — Но Теб, Сад,) 2 H,0 (4) — 20, c9) » Now, add 1, 2 & 3 to get the final equation, н,Te(9) + 2H,0 с) +20,49) » Н. Те са Оп, H, Te () + 24, e9) + 2H,о (1) — на Те о са ) This is the reartion given in the question it is given to calceslate the enthalpy of this reaction.

This is the final equation for the net reaction. And for this reaction you have calculated the $\Delta$H°_rxn.