Question

The enthalpy changes for the following processes are listed below :
$C{l}_2(g)=2Cl(g),242.3kJmo{l}^{-1}$
$I_2(g)=21(g),151.kJmo{l}^{-1}$
$ICl(g)=I(g)+Cl(g),211.3kJmo{l}^{-1}$
$I_2(s)=I_2(g).62.76kJmo{l}^{-1}$
Given that the standard states for iodine and chlorine are $I_2$ (s) and $C{l}_2$ (g), the standard enthalpy of formation of $ICl$ (g) is :

• A. $-14.6kJmo{l}^{-1}$
• B. $-16.8kJmo{l}^{-1}$
• C. $+16.8kJmo{l}^{-1}$
• D. $+244.8kJmo{l}^{-1}$

Answer 1

Answer: (C)

$\frac{1}{2}{I}_2\left(S\right)+\frac{1}{2}C{l}_2\left(g\right)\to ICl\left(g\right)$
$\Delta H=\left[\frac{1}{2}\Delta H_{s\to g}+\frac{1}{2}\Delta H_{diss,}\left(C{l}_2\right)+\frac{1}{2}\Delta H_{diss,}\left(I_2\right)\right]-\Delta H_{ICl}$
$=\left(\frac{1}{2}\times 62.76+\frac{1}{2}\times 242.3+\frac{1}{2}\times 151.0\right)-211.3$
$=228.03-211.3$
$\Delta H=16.73$