Calculate the heat change for the reaction 4N | vedclass

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(C) Given:$\Delta H_f(NH_3) = -46.0 \ kJ \ mol^{-1}$$\Delta H_f(H_2O) = -286.0 \ kJ \ mol^{-1}$$\Delta H_f(N_2) = 0 \ kJ \ mol^{-1}$ (element in stand

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$-1532$

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(C) Given:$\Delta H_f(NH_3) = -46.0 \ kJ \ mol^{-1}$$\Delta H_f(H_2O) = -286.0 \ kJ \ mol^{-1}$$\Delta H_f(N_2) = 0 \ kJ \ mol^{-1}$ (element in standard state)$\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$ (element in standard state)The enthalpy change for the reaction is given by:$\Delta H_{rxn} = \sum \Delta H_f(	ext{products}) - \sum \Delta H_f(	ext{reactants})$$\Delta H_{rxn} = [2 	imes \Delta H_f(N_2) + 6 	imes \Delta H_f(H_2O)] - [4 	imes \Delta H_f(NH_3) + 3 	imes \Delta H_f(O_2)]$Substituting the values:$\Delta H_{rxn} = [2(0) + 6(-286.0)] - [4(-46.0) + 3(0)]$$\Delta H_{rxn} = [-1716] - [-184]$$\Delta H_{rxn} = -1716 + 184 = -1532 \ kJ$

Calculate the heat change for the reaction $4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(l)$ in $kJ$. The enthalpies of formation of $NH_3(g)$ and $H_2O(l)$ at $298 \ K$ are $-46.0 \ kJ \ mol^{-1}$ and $-286.0 \ kJ \ mol^{-1}$ respectively.

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