The enthalpy changes for the following processes are listed below:
\[\mathrm{Cl}_2(\mathrm{g})\rightarrow 2\mathrm{Cl}(\mathrm{g}), \quad 242.3\mathrm{kJmol}^{-1}\]
\[\mathrm{I}_2(\mathrm{g})\rightarrow 2\mathrm{I}(\mathrm{g}), \quad 151.0\mathrm{kJmol}^{-1}\]
\[\mathrm{ICl}(\mathrm{g})\rightarrow \mathrm{I}(\mathrm{g}) + \mathrm{Cl}(\mathrm{g}), \quad 211.3\mathrm{kJmol}^{-1}\]
\[\mathrm{I}_2(\mathrm{s})\rightarrow \mathrm{I}_2(\mathrm{g}), \quad 62.76\mathrm{kJmol}^{-1}\]
Given that the standard states for iodine and chlorine are \(\mathrm{I}_2(\mathrm{s})\) and \(\mathrm{Cl}_2(\mathrm{g})\), the standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is:
Choose Your Option
A. \(+16.8\mathrm{kJmol}^{-1}\)
B. \(+244.8\mathrm{kJmol}^{-1}\)
C. \(-14.6\mathrm{kJmol}^{-1}\)
D. \(-16.8\mathrm{kJmol}^{-1}\)
Solution & Explanation
No details explanation available for this question. Option A is the verified correct answer.