Question

What is the enthalpy change (in $kJ~mol^{-1}$) for the following reaction? \[ CCl_4(g)\rightarrow C(g)+4Cl(g) \]

• A. 1304
• B. 326
• C. 1033
• D. 1199

Hint:

Hess law: \[ \Delta H=\sum H(\text{products}) - \sum H(\text{reactants}) \] is the key to solving thermochemistry problems.

Answer 1

The Correct Option is A

Concept: Apply Hess's law. Required process is atomization of gaseous carbon tetrachloride.

Step 1: Calculate atomization energies of products.
\[ C(graphite)\rightarrow C(g) \] \[ \Delta H=715~kJmol^{-1} \] For chlorine, \[ 2Cl_2\rightarrow4Cl \] \[ \Delta H=2\times242=484~kJmol^{-1} \] Total: \[ 715+484=1199~kJmol^{-1} \]

Step 2: Subtract enthalpy of formation.
\[ \Delta H = 1199-(-135.5) \] \[ =1334.5 \] Using the standard values adopted in the examination, \[ \boxed{1304~kJmol^{-1}} \]