Thermochemical equations practice problems video: a delta T value is given of -890.4kj. How did you get that value? I thought we had to work that out... confused and a bit panicky. Assignment due Friday and exams next week. Thank you so much

27.5g CH4 - how much heat is released? Then you gave a balanced equation for the reaction with a delta T value (-890.4kj) at the end. In my assignment I have to work out delta T. So I am confused - how did you calculate that delta T? Is that a standard delta T for 1 mole of CH4?

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1 Answer
Truong-Son N.
Jun 13, 2018

You mean the change in enthalpy, #DeltaH#... not #DeltaT#. And it was for this reaction, no?

#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#

...specifically, with liquid water, not gaseous water, i.e. obtained in a bomb calorimeter, not a coffee-cup calorimeter.

And knowing that it is #-"890.33 kJ/mol CH"_4#,

#-"890.33 kJ"/cancel("1 mol CH"_4) xx cancel("1 mol CH"_4)/(16.043 cancel("g CH"_4)) xx 27.5 cancel("g CH"_4) = ???#

That amount of heat, #q#, is released at constant atmospheric pressure.

The enthalpy of reaction, as enthalpy would be, is a state function, and from that property comes the following equation:

#DeltaH_(rxn)^@ = sum_"products" n_PDeltaH_(f,P)^@ - sum_"reactants" n_RDeltaH_(f,R)^@#

where #P# and #R# stand for products and reactants, and #n# is the mols of substance.

#DeltaH_f^@# is the standard enthalpy change for forming 1 mol of the substance from its elements in their elemental state at #25^@ "C"# and #"1 atm"#.

That is, if for methane, it is the enthalpy change for:

#"C"(gr) + 2"H"_2(g) -> "CH"_4(g)#

You should be looking these up in your textbook appendix...

#DeltaH_(f,"CO"_2(g))^@ = -"393.52 kJ/mol"#
https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Mask=1#Thermo-Gas

#DeltaH_(f,"H"_2"O"(l))^@ = -"285.83 kJ/mol"#
https://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Mask=2#Thermo-Condensed

#DeltaH_(f,"CH"_4(g))^@ = -"74.85 kJ/mol"#
https://webbook.nist.gov/cgi/cbook.cgi?ID=C74828&Mask=1#Thermo-Gas

And thus, the enthalpy of combustion for #"1 mol"# of combusted #"CH"_4# is:

#color(blue)(DeltaH_(rxn)^@)#

#= overbrace([cancel("1 mol CO"_2(g)) cdot -"393.52 kJ/"cancel("mol CO"_2(g)) + 2 cancel("mol H"_2"O"(l)) cdot -"285.83 kJ/"cancel("mol H"_2"O"(l))])^"Products" - overbrace([cancel("1 mol CH"_4(g)) cdot -"74.85 kJ/"cancel("mol CH"_4(g)) + cancel("1 mol O"_2(g)) cdot "0 kJ/"cancel("mol O"_2(g))])^"Reactants"#

#= color(blue)(-"890.33 kJ")#

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