Find delta H^0 for N2H4(l) + O2(g) -->N2(g)+2H2O(l) given equations: (1) 2NH3(g)

Question

Find delta H^0 for N2H4(l) + O2(g) -->N2(g)+2H2O(l) given equations: (1) 2NH3(g) + 3N2O(g) -->4N2(g) + 3H2O(l) delta H1 = -1010 (2) N2O(g) + 3H2(g) --> N2H4(l) + H2O(l) delta H2 = -317 (3) 2NH(g) + 1/2Og(g) --> N2H4(l) + H2O(l) delta H3 = -143 (4) H2(g) + 1/2O2 --> H2O(l) delta H4 = -286.

Explanation / Answer

2NH3(g) + 3N2O(g) --> 4N2(g) + 3H2O(l) (delta)H = -1010. kJ N2O(g) + 3H2(g) --> N2H4(l) + H2O(l) (delta)H = -317 kJ 2NH3(g) + 1/2O2(g) --> N2H4(l) + H2O(l) (delta)H = -143 kJ H2(g) + 1/2O2(g) --> H2O(l) (delta)H = -286 kJ according to hess's law, the sum of the deltaH products - sum of the deltaH reactants = deltaH for the reaction. N2H4(l) + O2(g) --> N2(g) + 2H2O(l) is the reaction N2H4 formation = -143kJ O2 formation = 0 since this is a pure element N2 formation = 0 since this is a pure element H2O formation = -286kJ products - reactants = reaction -286kJ - (-143kJ) = -143kJ

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