`Delta_(vap)H = 30 kJ mol^(-1)` and `Delta_(vap)S = 75 mol^(-1)K^(-1)`. Find the temperature of vapour, at `1` atm.

Delta_(vap)H = 30 kJ mol^(-1) and Delta_(vap)S = 75Jmol^(-1)K^(-1) . Find the temperature of vapour, at 1 atm.

A liquid is in equilibrium with its vapour at a certain temperature. If deltaH_(vap) = 60.24 kJ mol^(-1) and deltaS_(vap) = 150.6 J mol^(-1) then the boiling point of the liquid will be

A liquid is in equilibrium with its vapour at a certain temperature. If DeltaH_(vap) = 60.24 kJ mol^-1 and DeltaS_(vap) = 150.6 J mol^-1 then the boiling point of the liquid will be

Delta_(vap)H^(@) for water is 40.73 kJ mol^(-1) and Delta_(vap)S^(@) is 109 J K^(-1) "mol"^(-1) . Calculate the temperature at which liquid water wil be in the equilibrium with water vapour.

For the reactions at 298 K, 2A + B to C, Delta H = 400 kj mol^(-1) and Delta S = 0.2 kj K^(-1) mol^(-1) . At what temperature will the reaction become spontaneous considering Delta H and Delta S to be constant over the temperature range ?

For the reaction at 298 k,2A +B rarr C, Delta H= 400 KJ mol^(-1) and Delta S=0.2 KJ k^(-1) mol^(-1). At what temperature will the reaction become spontaneous ?

DeltaH=30kJmol^(-1), DeltaS=75J//K//mol. "find boiling temperature at" 1atm: