Problem

Problem

Use the enthalpy change for the reaction of iodine with chlorine to complete the table. I2(g)+Cl2(g)→2IClΔH⊖=−70.2 kJ mol−1\text{I}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{ICl} \quad \Delta H^\ominus = -70.2 \text{ kJ mol}^{-1}
BondI–ICl–ClI–Cl
Bond enthalpy / kJ mol −1151242?

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