Determine enthalpy of formation for H2O2(l),, using the listed enthalpies of reaction :N2H4(l)+2H2O2(l)⟶N2(g)+4H2O(l) ΔrH1∘=−818 kJ/molN2H4(l)+O2(g)⟶N2(g)+2H2O(l); ΔrH2∘=−622 kJ/molH2(g)+12O2(g)⟶H2O(l); ΔrH3∘=−285 kJ/mol

Change in enthalpyΔH=−12×(−818)+12(−622)+−(285)=-187 kJ/mol

Determine enthalpy of formation for H2O2(l),, using the listed enthalpies of reaction :N2H4(l)+2H2O2(l)⟶N2(g)+4H2O(l) ΔrH1∘=−818 kJ/molN2H4(l)+O2(g)⟶N2(g)+2H2O(l); ΔrH2∘=−622 kJ/molH2(g)+12O2(g)⟶H2O(l); ΔrH3∘=−285 kJ/mol

Change in enthalpyΔH=−12×(−818)+12(−622)+−(285)=-187 kJ/mol

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Determine enthalpy of formation for $H_{2} O_{2} (l),$ , using the listed enthalpies of reaction :
$\begin{aligned} N_{2} H_{4} (l) + 2 H_{2} O_{2} (l) ⟶ N_{2} (g) + 4 H_{2} O (l) \\ Δ_{r} H_{1}^{\circ} = - 818 kJ / mol \\ N_{2} H_{4} (l) + O_{2} (g) ⟶ N_{2} (g) + 2 H_{2} O (l); \\ Δ_{r} H_{2}^{\circ} = - 622 kJ / mol \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ_{r} H_{3}^{\circ} = - 285 kJ / mol \end{aligned}$

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-383 kJ/mol

-187 kJ/mol

-498 kJ/mol

None of these

answer is B.

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Change in enthalpy
$\begin{array}{l} ΔH = - \frac{1}{2} \times (- 818) + \frac{1}{2} (- 622) + - (285) \\ = - 187 kJ / mol \end{array}$

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Determine enthalpy of formation for H2O2(l),, using the listed enthalpies of reaction :N2H4(l)+2H2O2(l)⟶N2(g)+4H2O(l) ΔrH1∘=−818 kJ/molN2H4(l)+O2(g)⟶N2(g)+2H2O(l); ΔrH2∘=−622 kJ/molH2(g)+12O2(g)⟶H2O(l); ΔrH3∘=−285 kJ/mol

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