Consider the reaction 4NO_2(g) + O_2(g) right | vedclass

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(D) The given reaction is: $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH_1 = -111 \ kJ$. We want to find the enthalpy change for the reaction

Vedclass · Accepted Answer

$-219$

Vedclass · Answer

(D) The given reaction is: $4NO_2(g) + O_2(g) ightarrow 2N_2O_5(g)$, $\Delta_rH_1 = -111 \ kJ$. We want to find the enthalpy change for the reaction: $4NO_2(g) + O_2(g) ightarrow 2N_2O_5(s)$. The sublimation process is $N_2O_5(s) ightarrow N_2O_5(g)$, with $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$. For $2 \ mol$ of $N_2O_5$, the enthalpy change for condensation (reverse of sublimation) is $\Delta H_{cond} = -2 	imes 54 \ kJ = -108 \ kJ$. Adding the two reactions: $(4NO_2(g) + O_2(g) ightarrow 2N_2O_5(g)) + (2N_2O_5(g) ightarrow 2N_2O_5(s))$ $\Delta_rH_2 = \Delta_rH_1 + \Delta H_{cond} = -111 \ kJ + (-108 \ kJ) = -219 \ kJ$.

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)

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