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Chemistry
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Calculate ∆H(rxn)for the reaction:

5N2O4(l) +4N2H3CH3(l) = 12H2O(g) + 9N2(g) + 4CO2(g)
Given:
N2H3CH3(l)     ∆Hf(rxn) = 54.0 kJ/mole
N2O4(l)            ∆Hf(rxn) = -20.0kJ/mole
Apr 8th, 2015


dH rxn = sum(dHf product*coeff) - sum(dHf reactant*coeff) 

dHrxn = [12*(-241.95) + 9*0 + 4*(-393.51)] - [5*(-29)+ 4*(54)] = -4548.44

Apr 8th, 2015

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