Exothermic rxns being spontaneous

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Exothermic rxns being spontaneous

Postby 205192823 » Mon Feb 17, 2020 9:54 pm

Why are most exothermic reactions spontaneous?

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Re: Exothermic rxns being spontaneous

Postby chemboi » Mon Feb 17, 2020 9:58 pm

Spontaneity is determined by the sign of G, where G = H - T*S. For exothermic reactions., H is negative, so it is more common for G to be negative as well (S is positive, or negative T*S term is lower in absolute value than H)

Joowon Seo 3A
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Re: Exothermic rxns being spontaneous

Postby Joowon Seo 3A » Mon Feb 17, 2020 9:59 pm

Spontaneity means that the change in gibbs free energy is negative. Most exothermic reactions have a decrease in enthalpy meaning an increase in heat to the surroundings as well as an increase in entropy

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Re: Exothermic rxns being spontaneous

Postby saigorijavolu2k » Mon Feb 17, 2020 10:29 pm

Spontaneous reactions have negative gibbs free energy. G= H-TS (include the deltas). If H is negative (exo) then the likelihood of the overall reaction being negative increases as long as entropy is positive. if entropy is negative then H (-H in this case) must be greater than TS

Sanjana K - 2F
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Re: Exothermic rxns being spontaneous

Postby Sanjana K - 2F » Mon Feb 17, 2020 11:02 pm

Although reactions being exothermic/endothermic and spontaneous/nonspontaneous are closely related, you should also keep in mind that not all exothermic reactions will be spontaneous. When checking for spontaneity, look at the delta G sign (meaning you need the sign of delta H AND the sign of delta S and whether the temp is high/low) and when looking at whether or not heat is released, look at the delta H sign only. An example of heat released not corresponding to spontaneity is the condensation of steam.

Kevin Antony 2B
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Re: Exothermic rxns being spontaneous

Postby Kevin Antony 2B » Mon Feb 17, 2020 11:05 pm

Spontaneous reactions are characterized by having a deltaG. When a reaction gives off heat(exothermic) that means that H is negative. Now, unless S is positive, deltaG will be negative as deltaG = deltaH - T(deltaS).

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