Given the following data
3 H2(g) + N2(g) 2 NH3(g) Hrxn° = -92.4 kJ/molrxn
2 H2(g) + O2(g) 2 H2O(l) Hrxn° = -571.7 kJ/molrxn
Calculate Hrxn° for the following reaction:
4 NH3(g) + 3 O2(g) 2 N2(g) + 6 H2O(l)
A) -1899.9 kJ/molrxn
B) -1715.1 kJ/molrxn
C) -1530.3 kJ/molrxn
D) -479.3 kJ/molrxn
E) 1530.3 kJ/molrxn
Correct Answer:
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