Question

The enthalpy changes for the following processes are listed below $C{l}_2(g)2Cl(g),242.3kJmo{l}^{-1}$ $I_2(g)2I(g),151.0kJmo{l}^{-1}$ $ICl(g)I(g)+Cl(g),211.3kJmo{l}^{-1}$ $I_2(s)I_2(g),62.76kJmo{l}^{-1}$ Given that the standard states for iodine and chlorine are $I_2(s)$ and $C{l}_2(g)$ the standard enthalpy of formation of $ICI(g)$ 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)

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