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ExamsJEE MainChemistry

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

Correct answer: +16.8 kJ mol−1

Solution

The standard enthalpy of formation for ICl(g) is calculated by considering the enthalpy changes of the constituent elements in their standard states. The formation reaction can be derived from the provided enthalpy values, leading to a net positive value of +16.8 kJ mol−1, indicating that the formation of ICl from its elements is endothermic.

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