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

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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.

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Updated on August 01, 2022

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  • chemquestion
    chemquestion 13 days

    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 $181\times 2 = 362$?

    • MollyCooL
      MollyCooL over 4 years
      Be sure to take this tour to know about the community. Also use MathJax to format questions and answers.
  • ralk912
    ralk912 over 4 years
    It would be worth saying that the units for entalpy are just $\text{kJ}$, whereas those of enthalpy of formation are $\pu{kJ mol^-1}$
  • ralk912
    ralk912 over 4 years
    Also, standard conditions also include $\pu{298 K}$.
  • MollyCooL
    MollyCooL over 4 years
    No there is no condition on temperature as per IUPAC but commonly known is 298K. @ralk912 Refer standard states
  • ralk912
    ralk912 over 4 years
    Do you have a link for that?
  • MollyCooL
    MollyCooL over 4 years
    I gave one in my answer as well as comment. Will update regarding units @ralk912
  • ralk912
    ralk912 over 4 years
    Interesting. I wonder why the IUPAC would decide to do that instead of just fixing a value. In any case, then the temperature should be given when speaking of standard states I think.
  • MollyCooL
    MollyCooL over 4 years
    Yes, it is not very intuitive. But I was taught like that.
  • ralk912
    ralk912 over 4 years
    The IUPAC has proven themselves they're not usually intuitive. You were taught right.
  • MollyCooL
    MollyCooL over 4 years
  • chemquestion
    chemquestion over 4 years
    Thanks for the help! I'm also wondering what the units are for standard enthalpy of reaction. So far I've seen kJ and kJ/mol. Can they be used interchangeably? Does the /mol part add anything?
  • MollyCooL
    MollyCooL over 4 years
    Mol part indicates the enthalpy change per mol of the reaction. (If that makes sense) @chemquestion Unit for standard enthalpy change of a reaction is kJ/mol.(usual) Although we can use any combination of energy/mass or amount like kcal/mol