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(B) We need to find the enthalpy of formation of $H_2O_2(l)$,which corresponds to the reaction: $H_2(g) + O_2(g) \rightarrow H_2O_2(l)$.From the given

Vedclass · Accepted Answer

$-187$

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(B) We need to find the enthalpy of formation of $H_2O_2(l)$,which corresponds to the reaction: $H_2(g) + O_2(g) \rightarrow H_2O_2(l)$.From the given equations:$(i) N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l); \Delta_r H_1^\circ = -818 \, kJ/mol$$(ii) N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l); \Delta_r H_2^\circ = -622 \, kJ/mol$$(iii) H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l); \Delta_r H_3^\circ = -285 \, kJ/mol$Perform the operation: $\frac{1}{2} [ (ii) + 2(iii) - (i) ]$:$\frac{1}{2} [ (N_2H_4 + O_2 - N_2H_4 - 2H_2O_2 + 2H_2 + O_2) \rightarrow (N_2 + 2H_2O - N_2 - 4H_2O + 2H_2O) ]$$\frac{1}{2} [ 2H_2 + 2O_2 - 2H_2O_2 \rightarrow 0 ]$$H_2 + O_2 \rightarrow H_2O_2$$\Delta_f H^circ (H_2O_2) = \frac{1}{2} [ \Delta_r H_2^\circ + 2 \Delta_r H_3^\circ - \Delta_r H_1^\circ ]$$= \frac{1}{2} [ -622 + 2(-285) - (-818) ]$$= \frac{1}{2} [ -622 - 570 + 818 ]$$= \frac{1}{2} [ -374 ] = -187 \, kJ/mol$.

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