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)

$+16.8 KJ mol ^{-1}$

Answer 2

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