Question #de5b9

1 Answer
SCooke
Apr 24, 2016

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 ΔH_(f0) kJ/mol
CH_4 -74.81
C Cl_4 (g) -135.4
HCl (g) -92.31
Cl_2 (g) 0

ΔH_(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

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