Question

The enthalpy changes for the following processes are listed below :
$Cl_2 (g) = 2Cl (g), 242.3 \,kJ \,mol^{-1}$
$I_2 (g) = 21 (g),\, 151. \, kJ\, mol^{-1}$
$ICl (g) = I (g) + Cl (g),\, 211.3 \,kJ \,mol^{-1}$
$I_2(s) = I_2 (g).\, 62.76 \,kJ \,mol^{-1}$
Given that the standard states for iodine and chlorine are $I_2$ (s) and $Cl_2$ (g), the standard enthalpy of formation of $ICl$ (g) is :

• A. $- 14.6 \,kJ \,mol^{-1}$
• B. $- 16.8 \,kJ \,mol^{-1}$
• C. $+ 16.8 \,kJ \,mol^{-1}$
• D. $+ 244.8 \,kJ \,mol^{-1}$

Answer 1

Answer: (C)

$\frac{1}{2}I_{2}\left(S\right)+\frac{1}{2}Cl_{2}\left(g\right) \to ICl\left(g\right)$
$\Delta H = \left[\frac{1}{2}\Delta H_{s\to g} + \frac{1}{2}\Delta H_{diss,} \left(Cl_{2}\right) +\frac{1}{2}\Delta H_{diss,} \left(I_{2}\right)\right]-\Delta H_{ICl}$
$= \left(\frac{1}{2}\times62.76+\frac{1}{2}\times242.3+\frac{1}{2}\times151.0\right)- 211.3$
$= 228.03-211.3$
$\Delta H = 16.73$