The standard enthalpy of reaction, $\Delta H_{\text{rxn}}^\circ$, can be determined using the enthalpy of formation values:

$\Delta H_{\text{rxn}}^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}$

Given data:

• $\Delta H_f^\circ [CO(g)] = -110.5$ kJ/mol
• $\Delta H_f^\circ [COCl_2(g)] = -219.1$ kJ/mol
• $\Delta H_f^\circ [Cl_2(g)] = 0$ kJ/mol (since it is in its elemental form)

Balanced reaction:

$\text{CO}(g) + \text{Cl}_2(g) \rightarrow \text{COCl}_2(g)$

Applying the equation:

[ \Delta H_{\text{rxn}}^\circ = \left[ \Delta H_f^\circ \text{(COCl}_2\text{)} \right] - \left[ \Delta H_f^\circ \text{(CO)} + \Delta H_f^\circ \text{(Cl}_2\text{)} \right] ]

$= (-219.1) - (-110.5 + 0)$

$= -219.1 + 110.5$

$= -108.6 \approx -109 \text{ kJ}$

Answer:

The correct answer is –109 kJ.

---

Would you like further explanation on enthalpy calculations?

Here are some related questions for practice:

1. How do you calculate the enthalpy of reaction when multiple moles of reactants are involved?
2. Why is the enthalpy of formation of elemental substances in their standard state zero?
3. What is Hess’s Law, and how does it help in determining reaction enthalpies?
4. How does bond enthalpy relate to enthalpy of reaction?
5. How does the enthalpy of reaction relate to spontaneity in chemical reactions?

Tip:

The enthalpy of formation values always refer to the formation of one mole of a compound from its elements in their standard states.