Question

calculate the standard reaction enthalpy for the reaction between no and o₃.
no(g) + o₃(g) → no₂(g) + o₂(g)
compound δh°_f (kj/mol)
no(g) 90.
o₃(g) 143
no₂(g) 34
δh_rxn = ? kj
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for reaction enthalpy

The formula for standard reaction enthalpy ($\Delta H_{rxn}$) using standard enthalpies of formation ($\Delta H_f^\circ$) is:
$\Delta H_{rxn} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

Step2: Identify products and reactants

Products: $\text{NO}_2(g)$ and $\text{O}_2(g)$.
Reactants: $\text{NO}(g)$ and $\text{O}_3(g)$.

Step3: Note $\Delta H_f^\circ$ for $\text{O}_2(g)$

For elemental oxygen ($\text{O}_2$), $\Delta H_f^\circ = 0$ kJ/mol (by definition, as it is in its standard state).

Step4: Calculate sum of $\Delta H_f^\circ$ for products

$\sum \Delta H_f^\circ(\text{products}) = \Delta H_f^\circ(\text{NO}_2) + \Delta H_f^\circ(\text{O}_2)$
$= 34\ \text{kJ/mol} + 0\ \text{kJ/mol} = 34\ \text{kJ/mol}$

Step5: Calculate sum of $\Delta H_f^\circ$ for reactants

$\sum \Delta H_f^\circ(\text{reactants}) = \Delta H_f^\circ(\text{NO}) + \Delta H_f^\circ(\text{O}_3)$
$= 90\ \text{kJ/mol} + 143\ \text{kJ/mol} = 233\ \text{kJ/mol}$

Step6: Compute $\Delta H_{rxn}$

$\Delta H_{rxn} = 34 - 233 = -199\ \text{kJ}$

Answer:

-199