H = 90.8 kJ
Explanation
Reaction 1 : N2 (g) + 3 H2 (g) 2 NH3 (g) : H1 = -92.4 kJ
Reaction 2 : 4 NO (g) + 6 H2O (l) 4 NH3 (g) + 5 O2 (g) : H2 = 1167.1 kJ
Reaction 3 : 2 H2O (l) 2 H2 (g) + O2 (g) : H3 = 571.7 kJ
Desired reaction : 1/2 N2 (g) + 1/2 O2 (g) NO (g)
Consider the operation : (2 * R1 - R2 + 3 * R3) / 4
This operation gives the desired reaction.
H = (2 * H1 - H2 + 3 * H3) / 4
H = [2 * (-92.4 kJ) - (1167.1 kJ) + 3 * (571.7 kJ)] / 4
H = (-184.8 kJ - 1167.1 kJ + 1715.1 kJ) / 4
H = (363.2 kJ) / 4
H = 90.8 kJ
15. + -10.1 points 0/4 Submissions Used Calculate the standard enthalpy of formation of gaseous nitrogen...
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