What is the difference between enthalpy of reaction and standard enthalpy?

Question

What is the difference between enthalpy of reaction ($\Delta H$) and standard enthalpy ($\Delta H^\circ$)? I was told that the standard enthalpy of reaction is the change in heat when one mole of matter is transformed. I don't quite get what this one mole refers to in a chemical equation.

For example, I came across this equation in my textbook: $$\ce{N2 + O2 -> 2 NO \tag{ΔH = +181kJ/mol}}$$

Does the $\Delta H$ here mean the amount of energy absorbed during the formation of 1 mole of $\ce{NO}$? But since 2 moles of $\ce{NO}$ are produced, shouldn't it be $181\times 2 = 362$?

Answer 1

Yes, the standard enthalpy of reaction ($\Delta_\mathrm{r}H^\circ$) is the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states. The only condition is that the participants have to be in their standard states, ie. usually at 1 bar pressure.

On the other hand, enthalpy of reaction ($\Delta_\mathrm{r}H$) is amount of energy absorbed or released at any condition. Considering your example, $$ \ce{N2 + O2 -> 2NO} \qquad \tag{$H = \pu{+181kJ}$}$$ Here, $\Delta$H represents the amount of heat absorbed in the formation of 2 moles of $\ce{NO}$. For the formation of one mole of a substance from constituent elements, it is called enthalpy of formation($\Delta_\mathrm{f}H$)and if in standard states, standard enthalpy of formation. $$ \ce{ 1/2 N2 + 1/2 O2 -> NO} \tag{$\Delta_\mathrm{f}H$}$$ As we can see, $\Delta_\mathrm{f}H$ = $\frac{1}{2}\Delta H$. Hence it should be $\Delta_\mathrm{f}H$= +90.5 kJ/mol.

Note: Unit of enthalpy is that of energy but unit of enthalpy for a reaction is of energy per unit mass or amount.