What is the change in enthalpy for the following reaction?
4NH3 (g) + 302 (g) 2N2 (g) + 6H2O (l)
Given:
NH3: ∆H = - 46.2 kJ
Given: H2O: ∆H = - 286 kJ
A. - 1530.4 kJ
B. - 1020.3 kJ
C. 1530.4 kJ
D. 1020.3 kJ
E. - 510.1 kJ
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