Calculate the standard enthalpy of formation of gaseous nitrogen monoxide (NO) u

Question

Calculate the standard enthalpy of formation of gaseous nitrogen monoxide (NO) using the following thermochemical information:


N2(g) + 3 H2(g) 2 NH3(g) H = -92.4 kJ
4 NO(g) + 6 H2O(l) 4 NH3(g) + 5 O2(g) H = +1167.1 kJ
2 H2(g) + O2(g) 2 H2O(l) H = -571.7 kJ



H = kJ

Explanation / Answer

enthalpy (what is shown as Hrxn) being a thermodynamic state function - its current value depends on the current state of the system but not by the route that state was reached. the principle of conservation of energy: This means you can treat chemical equations and associated enthalpy changes arithmetically -- adding/subtracting, inverting, scaling. You can treat an overall reaction as being the result of intermediate steps, and you can exploit this to find thermodynamic data for intermediate reactions which, in practice, you may not be able to measure. 2 H2(g) + O2(g) ->2 H2O(l) >> H = -571.7 kJ 2x[N2(g) + 3 H2(g) ->2 NH3(g)] >>H = -92.4 kJ x2 4 NH3(g) + 5 O2(g) -> 4 NO(g) + 6 H2O(l) >>H = -1167.1 kJ ---------------------------------------------------------------------------------- 2N2 +6O2+ 8H2 -> 4NO + 8H2O >>H= 1832.2kJ

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