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A process will be spontaneous when the change in Gibbs free energy is negative.

The change in Gibbs free energy can be calculated from the equation:

G2 - G1 = H2 - H1 - T(S2 - S1)

where

G is Gibbs free energy

H is Enthalpy

T is absolute temperature (when T is given in Kelvin or Rankine it is an absolute temperature)

S is Entropy

In this case

H2 - H1 = 125 kJ

T = 293 K

S2 - S1 = 35 J/K = 0.035 kJ/K

so doing the math you get

G2 - G1 = 125 - 293(0.35) = 22.45 > 0 so the process is not spontaneous in the direction where enthalpy change and entropy change are being measured. The reverse process would be spontaneous.

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2d ago

The spontaneity of a reaction is determined by Gibbs free energy (ΔG = ΔH - TΔS). In this case, the reaction would be spontaneous at 293 K if the calculated ΔG value is negative. If ΔG is negative, it means that the reaction can occur spontaneously because the increase in entropy outweighs the enthalpy change.

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Q: How If a reaction occurs at 293 K that has a change in enthalpy of plus 125 kJ and a change in entropy of plus 35 J K then is the reaction spontaneous If so why If not why not?
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