College Chemistry Quiz
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Using the data below, calculate ΔHf°\Delta H_f° for NH3(g)NH_3(g): N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) ΔH°=−92.2\Delta H° = -92.2 kJ ΔHf°[N2(g)]=0\Delta H_f°[N_2(g)] = 0 kJ/mol ΔHf°[H2(g)]=0\Delta H_f°[H_2(g)] = 0 kJ/mol
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