In the equation
`2H_(2)(g)+O_(2)(g)to2H_(2)O(L), DeltaH=-571.6KJ"mol"^(-1)`
What is the enthalpy of formation of a water molecule.

For the reaction, 2H_(2)O(g) rarr 2H_(2)(g) + O_(2)(g), Delta H = 571.6 kJ Delta_(f) H^(theta) of water 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 :

The following thermochemical equations represent combustion of ammonia and hydrogen. 4NH_(3)(g) + 3O_(2)(g) rarr 6H_(2)O(l) + 2N_(2)(g), Delta H = -1516 kJ 2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) , Delta H = -572 kJ Calculate enthelpy of formation of ammonia.

For the reaction : H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l), Delta H = - 68 kcalmol^(-1) The heat change for the decomposition of 7.2 g of water is

For the thermochemical equation, 2H_(2(g)) + O_(2(g)) rarr 2H_(2)O_(l) , Delta H = - 571.6 kJ Heat of decomposition of water is :

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) .

Delta H for the reaction, SO_(2) (g) + (1)/(2) O_(2)(g) hArr SO_(3)(g) " " Delta H = -98.3kJ If the enthalpy of formation of SO_(3)(g) is -395.4kJ, then enthalpy of formation of SO_(2)(g) is: