Given that
`H_2O(l)`` rarr` `H^+`(aq) + `OH^-` (aq), `DeltaH`= 57.32 kJ
`H_2`(g) + `1/2 O_2(g)` `rarr` `H_2O(l)`, `DeltaH` = -286.02 kJ
Then calculate the enthalpy of formation of `OH^-` at `25^@C`.

N_2(g) + 2O_2 (g) rarr 2NO_2 (g) + X kJ 2NO (g) + O_2(g) rarr 2NO_2 (g) + Y kJ the enthalpy of.formation of NO is

H_2O(l)rarr H_2O(g) ,DeltaH=+43.7 KJ H_2O(s) rarr H_2O(l) ,Delta h=+6.05KJ the value of Delta H_(vaporisation) of ice is

H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.8kJ . What will be the value of DeltaH for the reaction: 2H_(2)O(l)to2H_(2)(g) +O_(2)(g) ?

2H_(2)+O_(2)to2H_(2)O,DeltaH=-571.6kJ . Does this equation express the thermochemical equation for the formation of H_(2)O ?

H_(2)O(l)toH^(+)(aq)+OH^(-)(aq),DeltaH^(0)=57.32kJ*mol^(-1) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH^(0)=-285.8kJ*mol^(-1) at 25^(@)C . If DeltaH_(f)^(0)[H^(+)(aq)]=0 , then the standard heat of formation (kJ*mol^(-1)) for OH^(-)(aq) at 25^(@)C is-

Calculate the enthalpy of combustion of ethylene . Given : C_(2) H_(6) (g) + (7)/(2) O_(2) (g) to 2 CO_(2) + 3 H_(2)O (l) , Delta H = = 1562 kJ* mol^(-1) H_(2) (g) + (1)/(2) O_(2) (g) to H_(2) O (l) , Delta H = -286 kJ * mol^(-1) C_(2) H_(4) (g) + H_(2) (g) to C_(2) H_(6) (g) , Delta H = -32 kJ * mol^(-1)

Calculate Delta H-DeltaU at 25^@C for the reaction: H_2(g)+( 1/2)O_2(g) = H_2O(l) .

H_2(g)rarr2H(g), DeltaH=104.2 kcal . The bond energy of H-H bond is

The heat of reaction for the following reaction (DeltaH^(0)) at 25^(@)C temperatue is -1368 kJ: C_(2)H_(5)OH(l)+3O_(2)(g) to 2CO_(2)(g)+3H_(2)O(l) If the enthalpy of formation of CO_(2)(g) and H_(2)O(l) at 25^(@)C are -393.5 and -285.8 kJ*mol^(-1) respectively, then what will be the value of the standard enthalpy of formation of C_(2)H_(5)OH(l) ?