Determine enthalpy of formation for H2g+O2g→H2Ol using listed enthalpies of reactionN2H41+2H2O2→N2g+4H2O1: ΔrH1∘=−818KJ/mol N2H41+O2g→N2g+2H2O1: ΔrH2∘=−622KJ/mol N2H2g+1/O2g→H2O1: ΔrH3∘=−285KJ/mol

For H2g+O2g→H2O21 ΔfH0H2O2,1=ΔrH30+ΔrH202−ΔrH102 =−285+−6222−−8182 =−187kJ/mol

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Determine enthalpy of formation for $H_{2 (g)} + O_{2 (g)} \to H_{2} O (l)$ using listed enthalpies of reaction
$N_{2} H_{4} (1) + 2 H_{2} O_{2} \to N_{2} (g) + 4 H_{2} O (1) : Δ_{r} H_{1}^{\circ} = - 818 K J / m o l N_{2} H_{4} (1) + O_{2} (g) \to N_{2} (g) + 2 H_{2} O (1) : Δ_{r} H_{2}^{\circ} = - 622 K J / m o l N_{2} H_{2} (g) + 1 / O_{2} (g) \to H_{2} O (1) : Δ_{r} H_{3}^{\circ} = - 285 K J / m o l$

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answer is 187.

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Detailed Solution

$ F o r H_{2} (g) + O_{2} (g) \to H_{2} O_{2} (1) Δ_{f} H^{0} (H_{2} O_{2}, 1) = Δ_{r} H_{3}^{0} + \frac{Δ_{r} H_{2}^{0}}{2} - \frac{Δ_{r} H_{1}^{0}}{2} = (- 285) + (\frac{- 622}{2}) - (\frac{- 818}{2}) = - 187 k J / m o l$

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