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The enthalpy changes for the following processes are listed below: Cl2(g) = 2Cl(g), 242.3 kJ mol^-1 I2(g) = 2I(g), 151.0 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 for ICl(g) is:

  1. +16.8 kJ mol^-1
  2. +244.8 kJ mol^-1
  3. −14.6 kJ mol^-1
  4. −16.8 kJ mol^-1

Correct answer: −16.8 kJ mol^-1

Solution

The standard enthalpy of formation for ICl(g) is calculated using Hess's law by combining the enthalpy changes of the relevant reactions. The formation of ICl from its elements involves the sublimation of iodine, the dissociation of chlorine, and the formation of ICl, leading to a net enthalpy change of -16.8 kJ mol^-1.

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