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I How endothermic reactions happen?

  1. Mar 17, 2016 #1
    Ok, this is a large qiestion. Firstly, from second law of thermodynamics, thermal flow always happens between matters kept in different temparature. But while an endothermic reaction takes place, it may extract some heat from environment and consume it. Isnt it against second law?

    And for example, making solution can be an endothermmic or exothermic process. If it is an endothermic process, by La Chatelier's law, increasing temparature results in more compound being solufied. But if its an exothermic process opposite happens and the compound is less soluable in higher temparature. But if we physically look at it, thehigher the temparature and higher the molecular energy is, the better the compound should be solufied. How do you physically explain it?
  2. jcsd
  3. Mar 17, 2016 #2


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    All chemical reactions will obey the second law of thermodynamics and increase the entropy of the universe. There are two ways in which reactions can increase entropy: 1) increase the entropy of the system by becoming more disordered, or 2) increase the entropy of surroundings by converting chemical potential energy into heat. Of course, the opposite of these processes (creating order within the system or absorbing heat from the surroundings) will be thermodynamically disfavored.

    For dissolution reactions, you are taking solute molecules that are highly ordered within the crystal lattice of a solid, and freeing them to move throughout the volume of the liquid phase. Thus, dissolution increases the entropy of the system. Let's call the increase in entropy of the system associated with dissolution ΔSsys = ΔS > 0.

    Dissolution can either be an exothermic process (ΔH < 0) or an endothermic process (ΔH > 0). Assuming the reaction takes place at constant pressure, the ΔH of the reaction is equal to the amount of heat transferred to/from the surroundings. The change in entropy of the surroundings associated with transferring this amount of heat to the surroundings is ΔSsurr = -ΔH/T.

    The change in entropy of the universe is the sum of the changes in entropy of the system and surroundings. Furthermore, by the second law, we know that this value must be positive such that the entropy of the universe is always increasing:

    ΔSuniv = ΔSsys + ΔSsurr = ΔS - ΔH/T > 0

    Because T is always positive, we can re-arrange the equation to write: ΔH - TΔS < 0

    Since, ΔS > 0 for a dissolution reaction, any exothermic reaction (ΔH < 0) will always satisfy the relationship. For an exothermic reaction (ΔH > 0), whether the forward reaction is favorable (ΔH - TΔS < 0) or the reverse reaction is favorable (ΔH - TΔS > 0) depends on the temperature. For high temp, the forward reaction will be favored, and at low temp the reverse reaction will be favored. In other words, at low temperature, the system prefers to go to the lowest energy state (in this case the solid form) while at high temperatures, the system prefers to go to its highest entropy state (in this case the aqueous state).

    Because of its importance in determining the thermodynamic favorability of reactions, the quantity ΔH - TΔS is called the change in free energy (ΔG) of the reaction.

    TL;DR endothermic reactions can happen if the products are more disordered than the reactants
    Last edited: Mar 17, 2016
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