Question

calculate the standard reaction enthalpy for the reaction between clf₃ and o₂. 2clf₃(g) + 2o₂(g) → cl₂o(g) + 3of₂(g) \

$$\begin{tabular}{|c|c|} \\hline compound & $\\delta h^\\circ_f$ (kj/mol) \\\\ \\hline clf₃ & -163.2 \\\\ \\hline cl₂o & 80.3 \\\\ \\hline of₂ & 24.5 \\\\ \\hline \\end{tabular}$$

$\delta h_{rxn} = ?$ kj enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for standard reaction enthalpy

The formula for the standard reaction enthalpy ($\Delta H_{\text{rxn}}$) is $\Delta H_{\text{rxn}} = \sum n\Delta H^{\circ}_f(\text{products}) - \sum m\Delta H^{\circ}_f(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of the products and reactants respectively.

Step2: Identify the enthalpies of formation for each compound

• For $\text{Cl}_2\text{O}(g)$: $\Delta H^{\circ}_f = 80.3\ \text{kJ/mol}$, stoichiometric coefficient $n = 1$.
• For $\text{OF}_2(g)$: $\Delta H^{\circ}_f = 24.5\ \text{kJ/mol}$, stoichiometric coefficient $n = 3$.
• For $\text{ClF}_3(g)$: $\Delta H^{\circ}_f = -163.2\ \text{kJ/mol}$, stoichiometric coefficient $m = 2$.
• For $\text{O}_2(g)$: The standard enthalpy of formation of an element in its standard state is $0\ \text{kJ/mol}$, so $\Delta H^{\circ}_f(\text{O}_2) = 0\ \text{kJ/mol}$, stoichiometric coefficient $m = 2$.

Step3: Calculate the sum of enthalpies of formation for products

$\sum n\Delta H^{\circ}_f(\text{products}) = (1\times\Delta H^{\circ}_f(\text{Cl}_2\text{O})) + (3\times\Delta H^{\circ}_f(\text{OF}_2))$
Substitute the values:
$= (1\times80.3) + (3\times24.5)$
$= 80.3 + 73.5$
$= 153.8\ \text{kJ}$

Step4: Calculate the sum of enthalpies of formation for reactants

$\sum m\Delta H^{\circ}_f(\text{reactants}) = (2\times\Delta H^{\circ}_f(\text{ClF}_3)) + (2\times\Delta H^{\circ}_f(\text{O}_2))$
Substitute the values:
$= (2\times(-163.2)) + (2\times0)$
$= -326.4 + 0$
$= -326.4\ \text{kJ}$

Step5: Calculate the reaction enthalpy

$\Delta H_{\text{rxn}} = \sum n\Delta H^{\circ}_f(\text{products}) - \sum m\Delta H^{\circ}_f(\text{reactants})$
Substitute the values from Step3 and Step4:
$\Delta H_{\text{rxn}} = 153.8 - (-326.4)$
$= 153.8 + 326.4$
$= 480.2\ \text{kJ}$

Answer:

+480.2