This is a question about the number of moles of NO2 involved in the reaction. Since the chemical equation is already balanced and the enthalpy variation of the reaction is -72 kJ/mol for 3 moles reacted of NO2, in order to calculate the molar enthalpy of reaction for NO2, we have to do this calculation:
5 years ago
Answered By Leonardo F
This is a question about the number of moles of NO2 involved in the reaction. Since the chemical equation is already balanced and the enthalpy variation of the reaction is -72 kJ/mol for 3 moles reacted of NO2, in order to calculate the molar enthalpy of reaction for NO2, we have to do this calculation:
$\text{Δ}H_{molar}NO_{2\left(g\right)}=\frac{\text{Δ}_rH}{3}=\frac{-72}{3}=-24$ΔHmolarNO2(g)=ΔrH3 =−723 =−24 kJ/mol
This means that, for every mol of NO2(g) reacted, the reaction process releases 24 kJ of heat energy, since it is exothermic.