The enthalpy change at `298 K` of the reaction
`H_(2)O_(2)(l)toH_(2)O(l)+1//2O_(2)(g)` is `-23.5 kcal mol^(-1)` and enthalpy of formation of `H_(2)O_(2)(l)` is `-44.8 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O(l)` is

The enthalpy of the reaction : H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g) is -23.5 kcal mol^(-1) and the enthalpy of formation of H_(2)O(l) is -68.3 kcal mol^(-1) . The enthalpy of formation of H_(2)O_(2)(l) is

The enthaply at 298K of the reaction H_(2)O_(2)(l)rarr+(1)/(2)O_(2)(g) is -23.5Kcal mol^(-1) and the enthaply of formation of H_(2)O_(2)(l) is -44.8Kcal mol^(-1) . The enthaply of formatiom of H_(2)O(l) is

The enthalpy of the reaction H_(2)O_(2) ("?") to H_(2) O ("?") + 1//2 O_(2) (g) is -23.5 kcal mol^(-1) and the enthalpy of formation of H_(2) O (?) is -68.3 kcal mol^(-1) . The enthalpy of formation of H_(2) O_(2) (?) is -

The enthalpy of the reaction : H_(2)O_(2)(l) rarr H_(2)O(l) + (1)/(2)O_(2)(g) is -98.3 KJ mol^(-1) and the enthalpy of formation of H_(2)O(l) is -285.6 kJ mol^(-1) . The enthalpy of formation of H_(2)O_(2)(l) is :

What will be standard enthalpy of formation of H_(2)O(l) ?

The enthalpy change for the reaction, H_2 O_((s))to H_2 O_((l)),Delta H=+6.01 kJ mol^(-1) at 0^@C is called enthalpy of _______ .

From the data given below at 298 K for the reaction : CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) Calculate the enthalpy of formation of CH_(4)(g) at 298 K. Enthalpy of reaction is = -893.5 kJ Enthalpy of formation of CO_(2)(g) = 393. kJ mol^(-1) Enthalpy of formation of H_(2)O(l) = 286.0 kJ mol^(-1) .