Question

Question #de5b9

Answer 1

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