How can a reaction be spontaneous when ∆S^o rxn is negative?
anyoneee?
anyoneee?
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It would have to be an exothermic reaction, and the temperature would have to be low enough. The Second Law of Thermodynamics tells us that, for any spontaneous occurrence, the entropy of the universe increases. The change in entropy of the universe = the change in entropy of the systen + the change in entropy of the surroundings. Therefore, if the change in entropy of the surroundings is a positive amount greater than the negative change in entropy of the system, the reaction will be spontaneous. delta S surr = -delta H/T where delta H is the change in enthalpy and T is the absolute temperature. For an exothermic reaction, delta H is negative, so delta S surr is positive. At lower temperatures, -delta H/T will have a larger value.
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Whether a reaction will be spontaneous or not depends on the equation:
∆G=H-T∆S
G is Gibb's free energy. H is enthalpy, T is temperature and S is entropy. If ∆G is negative, it means that the reaction is spontaneous because it creates free energy. In this case, by looking at the equation you can see that if S is negative, the equation would look like ∆G=H+T∆S and if enthalpy of the reaction is negative and larger than T∆S, ∆G would be negative and the reaction would be spontaneous. A negative enthalpy means that the reaction is exothermic (it releases heat as it proceeds). (T is measured in Kelvins so it can never be negative).
the general rule is that universe only favors more disorder, hence more "positive entropy". A reaction will only proceed if:
a. It increases the entropy of the particles that it involves. In order to do this it can either:
i. Cause a phase change from solid to liquid or liquid to gas.
ii. Increase moles of gas.
iii. Increase amount of solute in a solvent.
b. It increases the entropy of the environment. In order to do this, it should release heat energy. (Hence be an exothermic reaction.)
So if the entropy of the environment can be increased enough to overcome the entropy reduction in the system (composing of the particles that are in the reaction), the reaction is spontaneous. The opposite is true for endothermic reactions which absorb heat energy from the environment hence decrease entropy of the environment; if the reaction decreases the entropy of the environment less than it increases the entropy of the system, it is spontaneous. THe reaction I gave above is a way to calculate this balance.
So, the short answer to your question is that if ∆S is negative, the entropy of the system decreases. But if the reaction is exothermic enough so that the entropy of the environment increases more than the reduction in the entropy of the system, the reaction can be spontaneous.
Hope this helps!
∆G=H-T∆S
G is Gibb's free energy. H is enthalpy, T is temperature and S is entropy. If ∆G is negative, it means that the reaction is spontaneous because it creates free energy. In this case, by looking at the equation you can see that if S is negative, the equation would look like ∆G=H+T∆S and if enthalpy of the reaction is negative and larger than T∆S, ∆G would be negative and the reaction would be spontaneous. A negative enthalpy means that the reaction is exothermic (it releases heat as it proceeds). (T is measured in Kelvins so it can never be negative).
the general rule is that universe only favors more disorder, hence more "positive entropy". A reaction will only proceed if:
a. It increases the entropy of the particles that it involves. In order to do this it can either:
i. Cause a phase change from solid to liquid or liquid to gas.
ii. Increase moles of gas.
iii. Increase amount of solute in a solvent.
b. It increases the entropy of the environment. In order to do this, it should release heat energy. (Hence be an exothermic reaction.)
So if the entropy of the environment can be increased enough to overcome the entropy reduction in the system (composing of the particles that are in the reaction), the reaction is spontaneous. The opposite is true for endothermic reactions which absorb heat energy from the environment hence decrease entropy of the environment; if the reaction decreases the entropy of the environment less than it increases the entropy of the system, it is spontaneous. THe reaction I gave above is a way to calculate this balance.
So, the short answer to your question is that if ∆S is negative, the entropy of the system decreases. But if the reaction is exothermic enough so that the entropy of the environment increases more than the reduction in the entropy of the system, the reaction can be spontaneous.
Hope this helps!
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When the ambient temperature is equal to the "flash -point" temperature.