Why is there a minus in the answer? The answer should be -102 kJ/mol, but I get 102 kJ/mol

Using the set of enthalpy bonds calculate the standard reaction enthalpy of methane with chlorine and write the thermochemical equation.
# ΔH_b^°#of #Cl-Cl##=#243 kJ/mol
# ΔH_b^°# of #CH_3# =438 kJ/mol
# ΔH_b^°#of #H-Cl#= 432 kJ/mol
# ΔH_b^°# of #CH_3-Cl#= 351 kJ/mol

1 Answer
anor277
Jan 1, 2018

Well, there is a sign convention associated with bond energy....

Explanation:

Bond-breaking is always considered an ENDOTHERMIC process. Bond-making is conceived to be an EXOTHERMIC process...and so the reported bond enthalpies are always quoted as endothermic, i.e. bond-breaking, quantities....

We want....

#CH_4(g) + Cl_2(g) rarr CH_3Cl(g) + HCl(g)#

And thus we want to interrogate the thermochemical process...

#DeltaH_"rxn"^@=Sigma_"bonds broken"-Sigma_"bonds formed"#

In the given equation.... we BREAK #1xxC-H#, and #1xxCl-Cl# bonds...AND we make #1xxC-Cl#, and #1xxH-Cl# bond...

And so we make the summation on that basis...

#"Bonds broken"={438+243}*kJ*mol^-1=681*kJ*mol^-1#

#"Bonds formed"=-{351+432}*kJ*mol^-1=-783*kJ*mol^-1#

And so.......

#DeltaH_"rxn"^@={681-783}*kJ*mol^-1=-102*kJ*mol^-1#.

...which gives you the required answer with units of kilojoules per mole of REACTION as written....

If this does not address your question I am willing to try again. Certainly you should be mindful of the given conventions.

Related questions
Impact of this question
1368 views around the world
You can reuse this answer
Creative Commons License