Question
The enthalpy changes for the following processes are listed below :
$$\eqalign{
& C{l_2}\left( g \right) \to 2Cl\left( g \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,242.3\,kJ\,mo{l^{ - 1}} \cr
& {I_2}\left( g \right) \to 2I\left( g \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,151.0\,kJ\,mo{l^{ - 1}} \cr
& ICl\left( g \right) \to I\left( g \right) + Cl\left( g \right),\,\,\,\,211.3\,kJ\,mo{l^{ - 1}} \cr
& {I_2}\left( s \right) \to {I_2}\left( g \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,62.76\,kJ\,mo{l^{ - 1}} \cr} $$
Given that the standard states for iodine and chlorine are $${I_2}\left( s \right)$$ and $$C{l_2}\left( g \right),$$ the standard enthalpy of formation for $$ICl\left( g \right)$$ is :
A.
$$ + 16.8\,kJ\,mo{l^{ - 1}}$$
B.
$$ + 244.8\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 14.6\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 16.8\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ + 16.8\,kJ\,mo{l^{ - 1}}$$
Solution :
$${I_2}\left( s \right) + C{l_2}\left( g \right) \to 2ICl\left( g \right)$$
$$\Delta A = \left[ {\Delta {I_2}\left( s \right) \to {I_2}\left( g \right) + \Delta {H_{I - I}} + \Delta {H_{Cl - Cl}}} \right]$$ $$ - 2\left[ {\Delta {H_{I - Cl}}} \right]$$
$$\eqalign{
& = 151.0 + 242.3 + 62.76 - 2 \times 211.3 \cr
& = 33.46 \cr} $$
$$\Delta H_f^ \circ \left( {ICl} \right) = \frac{{33.46}}{2} = 16.73\,kJ/mol$$