Find the value of "a" if the phase transition temperature in Kelvin for the conversion of grey tin to white tin is found to be "100 a ".`Delta H_("transition")=21kJmol^(-1)``Delta S_("transition")=70Jmol^(-1)K^(-1)`

Calculate the entrpoy change for the following reversible process: alpha-Tin hArr beta-Tin at 1 mol at 1 atm 1 mol at 1 atm 300 K (Delta_(trans)H = 2090 J mol^(-1))

What is the entropy that takes place in conversion of 1 mol of alpha- tin to 1 mol beta - tin at 17^(@)C , 1 atm pressure if enthalpy of transition is 2.90 kJ mol^(-1) ?

The temperature at which the given reaction is at equilibrium Ag_(2)O_(s) rightarrow 2Ag(s) + 1/2 O_(2)(g) Delta H = 40.5 kJ mol^(-1) and DeltaS= 0.086 k J mol^(-1) K^(-1)

The entropy change for the conversion of 1 mole of alpha-" tin" (at 13^(@)C, 1 atm) to 1 mole of beta-" tin "(13^(@)C, 1 atm) if the enthalpy of transition is 2.095 kJ mol^(-1)

Calculate the temperature at which liquid water will be in equilibrium with water vapour. Delta_(vap)H = 40.00 kJ mol^(-1) and Delta_(vap)S = 0.100 kJ mol^(-1)K^(-1)

As O_2 (I) is cooled at 1 atm pressure , it freezes to form solid I at 54.5 K. At a lower temperature , solid rearrange to solid II, which has a different crystal that for the phase transition solid to slid II , DeltaH=-743.1 Jmol^(-1) and DeltaS=-17.0JK^(-1) mol^(-1) . At what temperature are solids I and II in equilibrium ?

Calculate the value of molal elevation constant for water if Delta S_("vaporisation") is 26.33 cal K^(-1) mol^(-1) .

Delta_(f) H of graphite is 0.23 kJ mol^(-1) and Delta_(f) H of diamond is 1.896 kh mol^(-1) . Delta H_(transition) from graphite to diamond is