For vaporization of water at 1 atmospheric pressure, the values of Delta H and Delta S are 40.63 KJ/mol""^(1) and 108.8 JK^(-1)/mol^(-1), respectively. The temperature when Gibbs energy change(Delta G) for this transformation will be zero, is :

For vaporization of water at 1 atmospheric pressure, the values of Del

1.	For vaporization of water at 1 atmospheric pressure, the values of Delta H and Delta S are 40.63 KJ/mol""^(1) and 108.8 JK^(-1)/mol^(-1), respectively. The temperature when Gibbs energy change(Delta G) for this transformation will be zero, is :
Answer» 293.4 K 273.4 K 393.4 K 373.4 K Solution :`H_(2) O_((L)) overset(1 "atm")(hArrr) H_(2) O_((g))` `Delta H = 40630 "J mol"^(-1)` `Delta S = 108.8 "JK mol"^(-1)` `Delta G = Delta H - T Delta S`. (When `Delta G = 0, Delta H - T Delta S= 0`) `T= (Delta H)/( Delta S)= (40630 "J mol"^(-1) )/( 108.8 "J mol"^(-1) )` `= 373.4` K `THEREFORE` Correct answer is (D)