Calculate the standard reaction enthalpy for the following reaction:
$\begin{array}{ll}2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) +\mathrm{O}_{2}(\mathrm{~g}) \\\text { From the standard reaction enthalpies of }\\\mathrm{N}_{2}(\mathrm{~g}) +\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}(\mathrm{g}) & \Delta H^{\circ}=+180.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}) +2 \mathrm{O}_{2}(\mathrm{~g}) & \Delta H^{\circ}=-66.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\end{array}$

Question

Calculate the standard reaction enthalpy for the following reaction:

\begin{array}{ll}2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) +\mathrm{O}_{2}(\mathrm{~g}) \\\text { From the standard reaction enthalpies of }\\\mathrm{N}_{2}(\mathrm{~g}) +\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}(\mathrm{g}) & \Delta H^{\circ}=+180.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}) +2 \mathrm{O}_{2}(\mathrm{~g}) & \Delta H^{\circ}=-66.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\end{array}

Quizplus · Accepted Answer

To find the standard reaction enthalpy for the given reaction, we use Hess's Law, which allows us to add or subtract given reactions to get the desired reaction. First, we reverse the reaction $2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g})$ to get $\mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$, which changes the sign of $\Delta H^{\circ}$ to $+66.4 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$. Then, we add this to the reaction $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}(\mathrm{g})$ with $\Delta H^{\circ}=+180.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$. The sum of these enthalpies gives the standard reaction enthalpy for the desired reaction: $+66.4 \mathrm{~kJ} + 180.5 \mathrm{~kJ} = +246.9 \mathrm{~kJ}$. However, since the question asks for the reaction enthalpy of $2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})$, and we've effectively calculated the reverse, we need to subtract the given reactions correctly. The correct interpretation of Hess's Law actually leads to the realization that the calculation described initially was incorrect in its final step of explanation. The correct calculation should directly use the given reactions without the need for reversing and reinterpreting the enthalpy changes as described. The direct addition of the given reactions' enthalpies does not apply here as initially explained. The correct approach involves understanding that the given reactions should be manipulated algebraically to match the target reaction, and the enthalpy change for the target reaction is directly given by the algebraic sum of the manipulated reactions' enthalpy changes.

Calculate the Standard Reaction Enthalpy for the Following Reaction A) -1141 KJ