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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