½N2+½O2 →NO,∆H=89KJmol^-1 If the entropy change for the reaction above at 25C is 11.8Jmol^-1 Calculate...
½N2+½O2 →NO,∆H=89KJmol^-1 If the entropy change for the reaction above at 25C is 11.8Jmol^-1 Calculate...
½N2+½O2 →NO,∆H=89KJmol^-1 If the entropy change for the reaction above at 25C is 11.8Jmol^-1 Calculate the change in free energy ,∆ci for the reaction at 25C?
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Asked by Khloe on 8th May, 2018
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