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Calculate the free energy change of following reaction at298 kelvin H2+I2=2HI if the enthalpy change is 51.8816kj and the entropy change is 166.15j is this reaction thermodyanamically possible?
Solution) delta g=51.8816kj-298*166.15/1000kjdelta g=2.3689since delta g is +ve therefore this reaction is not themodynamically possible
Interactive periodic table In this interactive periodic table animation you can view details of all components through clicking on it. You can select the colored through option
The equilibrium cell potential of the galvanic cell Pt | H 2 (g, f =1 bar) HCl (aq, 0.500mol kg -1 ) | Cl 2 (g, f =1 bar) | Pt is found to be E cell, eq = 1.410V at 298.15
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Free energy change=enthalpy change-298(entropy change) reaction is thermodynamically possible if free energy change is negative.
Free energy change=enthalpy change-298(entropy change)
reaction is thermodynamically possible if free energy change is negative.
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