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.

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