Question #de5b9

1 Answer
Apr 24, 2016

Answer:

Use other available reaction ΔH values to “build up” the desired reaction. Then combine them into a final value of ΔH for this reaction.

Explanation:

Product Enthalpies minus Reactant Enthalpies equals reaction enthalpy.
#CH_4 + 4Cl_2 → C Cl_4 + 4HCl #

http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm

Compound Δ#H_(f0)# kJ/mol
#CH_4 # -74.81
#C Cl_4 (g) # -135.4
HCl (g) -92.31
#Cl_2 (g) # 0

Δ#H_(f0)# kJ/mol
#C Cl_4 + 4HCl # - (#CH_4 + 4Cl_2 #)
-135.4 + 4(-92.31) - (-74.81 + 4(0))
-135.4 + 4(-92.31) + 74.81 = -429.83 kJ/mol