The combustion of methane means that it is possible to burn it. Chemically, this combustion process consists of a reaction between methane and oxygen in the air. The combustion reaction is:
CH
4 + 2O
2 → CO
2 + 2H
2O + 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:
СН
4 + 2О
2 = СО
2 + 2Н
2О + 890 kJ
Negative value of enthalpy means that heat content reduced during reaction. In other word, the products of this reaction (СО
2 and 2Н
2О) contain on 891 kJ less than starting compounds (СН
4 and О
2).
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 = q
cal + q
wire + q
comb 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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