The molar heat of formation of NH_4NO_{3(s)} | vedclass

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(B) The reaction is: $NH_4NO_{3(s)} 	o N_2O_{(g)} + 2H_2O_{(l)}$First,calculate the enthalpy change of the reaction $(\Delta H)$:$\Delta H = [\Delta

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$-125.04$

Vedclass · Answer

(B) The reaction is: $NH_4NO_{3(s)} 	o N_2O_{(g)} + 2H_2O_{(l)}$First,calculate the enthalpy change of the reaction $(\Delta H)$:$\Delta H = [\Delta H_f(N_2O_{(g)}) + 2 	imes \Delta H_f(H_2O_{(l)})] - [\Delta H_f(NH_4NO_{3(s)})]$$\Delta H = [81.46 + 2 	imes (-285.78)] - (-367.54)$$\Delta H = [81.46 - 571.56] + 367.54 = -490.1 + 367.54 = -122.56 \ kJ$Next,calculate the change in moles of gaseous species $(\Delta n_g)$:$\Delta n_g = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants}) = 1 - 0 = 1$Using the relation $\Delta H = \Delta U + \Delta n_g RT$:$\Delta U = \Delta H - \Delta n_g RT$Given $T = 298 \ K$ and $R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$:$\Delta U = -122.56 - (1 	imes 8.314 	imes 10^{-3} 	imes 298)$$\Delta U = -122.56 - 2.477 = -125.037 \ kJ \approx -125.04 \ kJ$

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