(b) The original expression burns 2 mol isooctane. We want to burn 100. mol isooctane. So, Multiplier100. mol2 mol50.050.0 [2 C8H18() + 25 O2(g)16 CO2(g) + 18 H2O()] H° = 50.0 (–10,992 kJ) 100. C8H18() + 1250 O2(g)800. CO2(g) + 900. H2O() H° = –5.50 105kJ (c) The original expression burns 2 mol isooctane. We want to burn 1.00 mol isooctane. So, Multiplier1.00 mol2 mol0.5000.500 [2 C8H18() + 25 O2(g)16 CO2(g) + 18 H2O()] H° = 0.500 (–10,992 kJ) 1.00 C8H18() + 12.5 O2(g)8.00 CO2(g) + 9.00 H2O() H° = –5.50 103kJ Reasonable Answer Check:It makes sense that when more moles are involved, the H° is larger, and when fewer moles are involved, the H° is smaller. 41. Answer: –1.45 103kJ/molStrategy and Explanation: Given a chemical equation for the combustion of a fuel, the mass of fuel burned, and the thermal energy evolved at constant pressure for the reaction, determine the molar enthalpy of combustion of the fuel. Molar enthalpy change (H°) is identical to the thermal energy released at constant pressure per mol of substance. So, convert the mass into moles, then divide the thermal energy by the moles to get the molar enthalpy of combustion. The balanced thermochemical expression tells us that exactly 1 mol C2H5OH burns to produce heat energy. q = –3.62 kJ (It is negative since heat energy is evolved, rather than absorbed.) n0.115 g C2H5OH1 mol C2H5OH46.0682 g C2H5OH0.00250 mol C2H5OH
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