How do you determine the sign of heat flow #q# as compared to #DeltaH#? How do their units relate and what is the difference between #"kJ"# and #"kJ/mol"#?

1 Answer
Feb 17, 2017

It always depends on context, but it is never the case that #-q = DeltaH#.

By definition, #q = nDeltaH# at constant pressure, and both #q# and #DeltaH# are typically in units of #"J"# or #"kJ"#, as long as they agree.

#q# and #DeltaH# are ALWAYS the same sign, for the SAME process. i.e. #q_"rxn" > 0# implies that #DeltaH_"rxn" > 0#, #q_"cal" > 0# implies #DeltaH_"cal" > 0#, because the number of #"mol"#s is always positive.

If you see #DeltaH# in #"kJ/mol"#, then it's an entirely different context than #q# if #q# is in #"J"#. The units are necessarily different if we are talking about #n# #"mol"#s of gas (#"J"# or #"kJ"#, an extensive unit), compared to some nonspecific quantity of gas (#"J/mol"# or #"kJ/mol"#, an intensive unit).

This can be seen by simply using the units. If #q = nDeltaH#, then:

#"J" = "mol" xx "J"/"mol"#

Thus, if #q# is units of #"J"#, we must have from #q = nDeltaH# that #DeltaH# is in #"J/mol"#. If it's anything else, and #q# is chosen arbitrarily to be in #"J"#, it's simply incorrect use of units on #DeltaH#, or there was a conversion to different units that was not stated, i.e.

#"2500 J"/"mol" xx "1 kJ"/"1000 J" = "2.5 kJ"/"mol"#