Determine the enthalpy of formation for H_2O_ | vedclass

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(B) We need to find the enthalpy of formation for $H_2O_{2(l)}$,which corresponds to the reaction: $H_{2(g)} + O_{2(g)} 	o H_2O_{2(l)}$.Let the given

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$-187 \ kJ/mol$

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(B) We need to find the enthalpy of formation for $H_2O_{2(l)}$,which corresponds to the reaction: $H_{2(g)} + O_{2(g)} 	o H_2O_{2(l)}$.Let the given reactions be:$(1) N_2H_{4(l)} + 2H_2O_{2(l)} 	o N_{2(g)} + 4H_2O_{(l)}$; $\Delta_r H_1^o = -818 \ kJ/mol$$(2) N_2H_{4(l)} + O_{2(g)} 	o N_{2(g)} + 2H_2O_{(l)}$; $\Delta_r H_2^o = -622 \ kJ/mol$$(3) H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(l)}$; $\Delta_r H_3^o = -285 \ kJ/mol$To obtain the target reaction,we perform: $(2) + 2 	imes (3) - (1)$ divided by $2$:$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [\Delta_r H_2^o + 2 	imes \Delta_r H_3^o - \Delta_r H_1^o]$$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [-622 + 2(-285) - (-818)]$$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [-622 - 570 + 818]$$\Delta_f H^o(H_2O_{2(l)}) = \frac{1}{2} [-374] = -187 \ kJ/mol$.

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