![]() If you feel less confident with fractions, just scale the reaction by #4# you will get integer coefficients, but you have to use #4# times as well when you convert to units of only #"kJ"#, so don't forget that. the enthalpy reaction for a combustion) of about = -"973.49 kJ/mol"#, you can easily say this as = -"973.49 kJ"#īut, only because the reaction involves #\mathbf(1)# equivalent of glycine. So, using the enthalpy of combustion for glycine (i.e. It is a bit convenient actually, because in this case, you do not have to consider whether or not you have to multiply the enthalpy of formation by the stoichiometric coefficient in front of glycine in the reaction-it's #1#! You should just note that when using fractions, you cannot have a fraction of an atom, so maybe consider it in terms of mass. The equation is still balanced, and you can still use it to correctly perform calculations (provided your math is sound). So, in some textbooks, you may see formation reactions that have fractional coefficients, and that is OK. In this case, the enthalpy of reaction is relative to #"1 mol"# of glycine (it is 'normalized on a per-mol basis'). Standard enthalpy of reaction, has units of #"kJ/mol"#, relative to the compound of interest being produced. So remember, you can use fractional coefficients to balance chemical equation, but make sure that the they make sense at the level of the atom.Īny fractional coefficient that gives you fractions of an atom is not used correctly.Īlthough Stefan has good reason to be concerned about the fractional stoichiometric coefficients, it is not unusual to see fractions in reactions that form a compound whose enthalpy of reaction is being referenced.ĮNTHALPY OF REACTION IS NORMALIZED ON A PER-MOL BASIS #9/2 xx "2 atoms of O" = "9 atoms of O" color(white)(a)color(green)(sqrt())# This time, the fractional coefficient makes sense because it gets you ![]() If you want to have a correct balanced chemical equation and still use fractional coefficients, you can divide all the chemical species by #2#. To do that, multiply all the chemical species by #4#. To get the correct balanced chemical equation, you'd have to get rid of the fractional coefficients. Represents a step in the process of balancing the equation that describes the combustion of glycine. So, the best-case scenario here is that this version of the equation ![]() #9/4 xx "2 atoms of O" = 9/2color(white)(a) "atoms of O"color(white)(a)color(red)(xx)# ![]() The same goes for #9/4"O"_2#, which is equivalent to You cannot have #2.5# atoms of oxygen because that would imply the splitting of atoms, which is not possible in ordinary chemical reactions. #5/2 xx "1 atom of O" = "2.5 atoms of O" color(white)(a)color(red)(xx)# #5/2 xx "2 atoms of H" = "5 atoms of H"color(white)(a)color(green)(sqrt())# #1/2 xx "2 atoms of N" = "1 atom of N" color(white)(a)color(green)(sqrt())#ĭoes not make sense because it implies that you're dealing with Makes sense because it describes half of one nitrogen molecule, #"N"_2#, which is equivalent to one atom of nitrogen. Now, the idea with using fractional coefficients to balance chemical equations is that you must make sure that you don't end up with fractions of atoms. You're dealing with the combustion of glycine, #"NH"_2"CH"_2"COOH"#, which produces carbon dioxide, #"CO"_2#, water, #"H"_2"O"#, and nitrogen gas, #"N"_2#. ![]()
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