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In Cambridge Chemistry Coursebook [1, p. 94] it’s written that

A rise in temperature is given a positive sign. So the value of $\Delta H$ is negative for an exothermic reaction. A fall in temperature is given a negative sign. So the value of $\Delta H$ is positive for an endothermic reaction.

Why is the sign of enthalpy for exothermic reaction negative? Doesn’t exothermic reaction rise the temperature of surrounding?

Why is the sign of enthalpy for endothermic reaction positive? Doesn’t endothermic reaction take in energy and cool down the surrounding?

Reference

  1. Ryan, L.; Norris, R. Cambridge International AS and A Level Chemistry Coursebook, 2nd ed.; Cambridge International Examinations; Cambridge University Press: Cambridge, 2014. ISBN 978-1-107-63845-7.
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  • $\begingroup$ Is the rising or falling of the T that is not required here. This is often see at introductory level in spire of being unnecessary. See answer $\endgroup$
    – Alchimista
    Nov 24 '18 at 15:44
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Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give that heat to surroundings. If you do so, than since our system lost heat its enthalpy decreases (first law of thermodynamics) and because of that enthalpy of exothermic reaction is taken as negative. The opposite reasoning holds for endothermic reactions. You need to bring heat from surroundings into system to keep it at the same temperature and that heat increases enthalpy of the system.

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The reaction is said to be exothermic or endothermic according to the sign of the Enthalpy change. To obtain Enthalpy you subtract the enthalpy of formation for reactants from that of products. So, if the final enthalpy of the system is less than the initial one the result is negative and this difference manifests it self in the system as heat and temperature increases. The opposite is true for endothermic reactions.

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    $\begingroup$ This answer is incorrect. Whether a reaction is exothermic or endothermic is determined by the sign of the enthalpy, not that of the Gibbs free energy. Indeed, for many reactions, depending upon the conditions, the enthalpy and the Gibbs free energy have can have the opposite sign. Thus, unless the entropy change is negligible, they cannot even be approximately equated to each other. $\endgroup$
    – theorist
    Jan 10 at 2:36
  • $\begingroup$ @theorist, You're right. the right term for free energy is endergonic or exergonic. Thank you. $\endgroup$
    – M.ghorab
    May 7 at 13:00
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    $\begingroup$ In your correction, you have the signs reversed. You wrote: "To obtain Enthalpy you subtract the heat of formation for products (enthalpy) from that of reactants. So, if the final enthalpy (heat) of the system is more than the initial one the result is negative". In fact, enthalpy change = (enthalpy products) – (enthalpy reactants). [This is generally true for all state functions, not just enthalpy: change = (final state) – (initial state).] Thus if the final enthalpy is higher, the enthalpy change is positive. $\endgroup$
    – theorist
    May 7 at 17:09
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    $\begingroup$ Also: enthalpy is not heat. enthalpy is a state function, while heat is a path function. I don't mean to be discouraging, but I would strongly recommend you learn more about thermodynamics before attempting to answer questions on the subject. $\endgroup$
    – theorist
    May 7 at 17:10
  • $\begingroup$ Thank you for the advice. $\endgroup$
    – M.ghorab
    May 10 at 8:54

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