Answer to Question #4913 in Physical Chemistry for Khatlyn
CH4 + 2O2 → CO2 + 2H2O + energy
One molecule of methane, combined with two oxygen molecules, react to form a carbon dioxide molecule, and two water molecules usually given off as steam or water vapor during the reaction and energy. The amount of heat evolved in the combustion of methane at standard conditions is a well-known fact, that was determined empirically years ago. It is ≈ 890-891 kiloJoules/mol. So, the reaction can be rewritten as:
СН4 + 2О2 = СО2 + 2Н2О + 890 kJ
Negative value of enthalpy means that heat content reduced during reaction. In other word, the products of this reaction (СО2 and 2Н2О) contain on 891 kJ less than starting compounds (СН4 and О2).
Right answer: e - 890 kiloJoules.
p.s. How was this value obtained? The enthalpy change for this reaction is measured by pressurizing a strong metal reaction vessel (called a bomb) with a mixture of methane and oxygen gas. The bomb is immersed in a calorimeter filled with water. An electrical current is passed through ignition wire (a fine iron wire), which ignites the wire and the gas mixture. The heat balance for this calorimetry experiment is:
0 = qcal + qwire + qcomb
The heat for the calorimeter, qcal, is determined from the heat capacity of the calorimeter and the temperature change for the calorimetry experiment. Typically the amount of water in the calorimeter is always the same; therefore Ccal includes the heat capacities of the calorimeter, the water, and the bomb itself. The burning of the ignition wire releases heat, qwire, and this heat must be included in the calculations. (This heat is treated separately, because the amount of ignition wire used varies from one measurement to the next.)
The heat released by the combustion reaction is qcomb, which is related to the molar enthalpy of combustion by:
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