To calculate the standard reaction enthalpy (ΔH°) for the reaction:

$2\text{HNO}_2(g) + \text{O}_2(g) \rightarrow 2\text{HNO}_3(g)$

we'll use the enthalpy of formation (ΔH_f°) values for the substances involved. The reaction enthalpy can be calculated using the following formula:

$\Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants})$

1. Write the expression for the reaction enthalpy: $\Delta H^\circ_{\text{reaction}} = 2\Delta H^\circ_f(\text{HNO}_3(g)) - \left[2\Delta H^\circ_f(\text{HNO}_2(g)) + \Delta H^\circ_f(\text{O}_2(g))\right]$

   Note that for elemental substances in their standard states, such as $\text{O}_2(g)$, the enthalpy of formation is zero.

2. Obtain the standard enthalpy of formation (ΔH_f°) values from the table (units are kJ/mol):

   • $\Delta H^\circ_f(\text{HNO}_3(g)) = -134.3 \, \text{kJ/mol}$
   • $\Delta H^\circ_f(\text{HNO}_2(g)) = +33.2 \, \text{kJ/mol}$
   • $\Delta H^\circ_f(\text{O}_2(g)) = 0 \, \text{kJ/mol}$

3. Plug the values into the equation:

   $\Delta H^\circ_{\text{reaction}} = 2(-134.3) - \left[2(33.2) + 0\right]$

4. Simplify the expression:

   $\Delta H^\circ_{\text{reaction}} = -268.6 - (66.4)$

   $\Delta H^\circ_{\text{reaction}} = -335.0 \, \text{kJ}$

So, the reaction enthalpy under standard conditions is −335 kJ.

Would you like more details or clarifications on this?

Here are five related questions you might find useful:

1. How does the enthalpy of formation relate to bond energy?
2. What is the significance of a negative reaction enthalpy?
3. How do changes in temperature affect reaction enthalpy?
4. What is the difference between enthalpy of formation and enthalpy of combustion?
5. How do catalysts affect reaction enthalpy?

Tip: Always ensure that you use the correct units when calculating enthalpy changes, as incorrect units can lead to errors in your results.