Question #de5b9

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Question #de5b9

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Answer 1 · 2016-04-24T02:31:33

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.
$C H_{4} + 4 C l_{2} \to C C l_{4} + 4 H C l$ $CH_4 + 4Cl_2 → C Cl_4 + 4HCl$

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

Compound Δ $H_{f 0}$ $H_(f0)$ kJ/mol
$C H_{4}$ $CH_4$ -74.81
$C C l_{4} (g)$ $C Cl_4 (g)$ -135.4
HCl (g) -92.31
$C l_{2} (g)$ $Cl_2 (g)$ 0

Δ $H_{f 0}$ $H_(f0)$ kJ/mol
$C C l_{4} + 4 H C l$ $C Cl_4 + 4HCl$ - ( $C H_{4} + 4 C l_{2}$ $CH_4 + 4Cl_2$ )
-135.4 + 4(-92.31) - (-74.81 + 4(0))
-135.4 + 4(-92.31) + 74.81 = -429.83 kJ/mol

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