1. Tardigrade
  2. Question
  3. Chemistry
  4. The enthalpy changes for the following processes are listed below: Cl2 (g) = 2Cl (g), 242.3 kJ mol-1 I2 (g) = 21 (g), 151. kJ mol-1 ICl (g) = I (g) + Cl (g), 211.3 kJ mol-1 I2(s) = I2 (g). 62.76 kJ mol-1 Given that the standard states for iodine and chlorine are I2 (s) and Cl2 (g), the standard enthalpy of formation of ICl (g) is :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :

AIEEEAIEEE 2008

Solution:

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