Question

Calculate the enthalpy of reaction at 298 K for the reaction shown, given the average bond energies below. 1219) + Cl2(g) → 2 IC (9) Bond Bond Energy (kJ/mol) CI-CI 243 151 I-CI 208 *Make sure to indicate the sign in your answer.

Answer 1

Enthalpy of any reaction is overall heat released or absorbed in the reaction. When the bonds of a compound are broken, energy has to be supplied, and when new bonds are formed, than energy is released.

In the given reaction: bond between I2 and Cl2 is broken.

Therefore energy absorbed = bond energy of I2 + bond energy of Cl2

                                           = 1 mol x (151 + 243) kJ/mol         .....since 1 mole of each is there

                                           = 394 kJ

Hreactants
= + 394 kJ

The new bond formed is 2 moles of ICl.

Therefore energy released = 2 x bond energy of I-Cl

                                           = 2mol x 208 kJ/mol

                                           = 416 kJ

H products
= - 416 kJ

Note that sign of $\Delta H$ is taken as positive '+' when heat is absorbed and negative '-' when heat is released.

Hence, overall energy is released in the reaction;

- 416 kJ + 394 kJ= -22 kJ        .....Answer

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