Question

avuuuw ac mallianned at 25 L and I bar. 86. Use data from Table 7.2, together with the fact that AHº = -3509 kJ mol-1 for the complete combustion of pentane, C5H12(1), to calculate A Hº for the reac- tion below. 5 CO(g) + 11 H2(g) → C5H12(1) + 5 H2O(1) ApHº = ?

Answer 1

Given that AH .ie standard enthalpy for the reaction of combustion of pentane de AHE = -3509 KJ mout General formula for complete combustion of an alkane es CxHy + (x+4)0 & CO2 + H₂O So pentane - 5 H12 125, y = 12 combustion reaction for pentane es GH12 + (5+3) 0 - 5 CO₂ + 6H₂O Ho (1) + 8 02 (9) - 5co, (q) + 6H₂0 (1) AH = - 3509 kJ/mol

0 KT mol AiH of 029) - Oksmo to of co(g) – – 393.5 kJ/mol 4 t of $20 (1) - - 285.8 kJ/mol Ap to of CG Hw (1) - 4p tº (products) – 44 *° (reactants) – – – 3509 kJ/mon mo [16x -287-8) +(* x-393-73) -8x60) +2) = -35092 /mol - 3682.3 – 1 – – 3509 a=-173.3 kJ/mol

. 4 H of a H12 (1) = - 173.3 kJ /mol So. Now If we consider the given reaction 5 CO(g) + 11 H₂ (9) H ₂ (1) th H₂O) 4, H = 4; ° C products) - 5 44 #0 (Yeastaus co (g) – 44° = -110.5 kg mol Hz (9) - A HO = 0 kJ/mol Hi, (L) - AH = – 173.3 kJ/ mol H2o (1) - Af Ho - – 285.8 kJ/ mol - E 4f Hº (produch) = - 173.3 kJ/ mol + 5x - 285.8 ) KJ / mol - - 1602.3 KJ/ mol

SA Hº Creactants) 5x(- 110.5 KJ 1 .) + 11 XOYL mol mol – – 552.5 kJ mol 4. Ho - – 1602.3 - (-552.5) 2 – 1049.8 kJ mol . A. Ho for the given reaction is - - 1049.8 kJ/ /mol Note standard molar enthalples for H3 (9) and O2(g) ane O. (zego). To under- stand this you need to understand the defination of standard enthalpy of formation. This to the change is enthalpy when one mole of a substance sets stand and state & formed from bts elements under standard stan conditions of lat pressure and 298k.

Oxygen gas consists of los elements already in the Standard State. so there Esn't any change here. So the Standard enthalpy of formation es zero.