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

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 $181\times 2 = 362$?

• Be sure to take this tour to know about the community. Also use MathJax to format questions and answers. Mar 15, 2018 at 13:53

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.
• It would be worth saying that the units for entalpy are just $\text{kJ}$, whereas those of enthalpy of formation are $\pu{kJ mol^-1}$ Mar 15, 2018 at 8:59
• Also, standard conditions also include $\pu{298 K}$. Mar 15, 2018 at 9:01