A scientist measures the standard enthalpy change for the following reaction to be -910.7 kJ:2NH3(g) + 3 N2O(g)4N2(g) + 3 H2O(g)Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(g) is ??kJ/mol.

Standard enthalpy (∆H°f) of formation of a substance is furnished as the enthalpy change associated with the formation of the substance (one mole) from its constituent elements in their standard state. The numerical values of ∆H°f is expressed in units of kJ/mol. The enthalpy of formation associated with a substance in its standard state has zero value. In a reaction, the magnitude of ∆H°f is furnished the enthalpy change associated with the formation of products and the reactants in their standard state. The standard enthalpy of the reaction of ammonia with nitrous oxide is expressed as follows: ∆H=∑∆H°f (products)-∑∆H°f (reactants)-910.7 kJ=3(∆H°f H2O(g))+4(∆H°f N2(g))-[2(∆H°f NH3(g))+3(∆H°f N2O(g)]∆H°f H2O(g)=x kJ/mol∆H°f N2(g)=0 kJ/mol∆H°f NH3(g)=-46.3 kJ/mol∆H°f N2O(g)=81.56 kJ/mol Substitute the enthalpy of formation values of each substance in the above equation to determine the enthalpy of formation of water. -910.7 kJ/mol=3(∆H°f H2O(g))+4(∆H°f N2(g))-[2(∆H°f NH3(g))+3(∆H°f N2O(g)]-910.7 kJ/mol=[3x+4(0)] kJ/mol-[2(-46.3)+3(81.56)] kJ/mol-910.7 kJ/mol=3x-152.08 kJ/mol3x=-758.62 kJ/molx=-252.8 kJ/mol Hence the standard enthalpy of formation of H2O(g) is determined as -252.8 kJ.