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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

Form ICl from elements: 1/2 Cl2(g)->Cl = 121.15, 1/2 I2(g)->I = 75.5, 1/2 I2(s)->1/2 I2(g) = 31.38, and I+Cl->ICl = -211.3. Sum = +16.8 kJ/mol.

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