For the transformation, `H_(2)O(l,1 atm) "to H_(2)O ` `(g,1atm),` `DeltaH_(vap)=40.668kJmol^(-)` The change in entropy `(JK^(-)mol^(-)) is: `

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

For the equilibrium H_(2)O(l) iff H_(2)O(g) at 1 atm and 298 K

Steam undergoes decomposition at high temperature as per the reaction : H_(2)O(g)to H_(2)(g)+(1)/(2)O_(2)(g) Delta H^(@)=200kJmol^(-1)Delta S^(@)=40JK^(-1)mol^(-1) The temperature at which equilibrium constant is unity is :

For the process H_(2)O(l)hArrH_(2)O(g),DeltaH=40.8kJ*mol^(-1) at the boiling poinit of water. Calculate molar entropy change for vaporisation from liquid phase.

H_(2)O(l) rarr H_(2)O(g), DeltaH = +40.7 kJ DeltaH is the heat of ……………..of water.

H_(2)O(l) rarr H_(2)O(g), DeltaH = +40.7 kJ DeltaH is the heat of ……………..of water.