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
C1530.4 kJ
D1020.3 kJ
E-510.1 kJ
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
C1530.4 kJ
D1020.3 kJ
E-510.1 kJ