Question
You wish to know the enthalpy change for the formation of liquid $\mathrm{PCl}_{3}$ from the elements.$$\mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{r} H^{\circ}=?$$The enthalpy change for the formation of $\mathrm{PCl}_{5}$ from the elements can be determined experimentally, as can the enthalpy change for the reaction of $\mathrm{PCl}_{3}(\ell)$ with more chlorine to give $\mathrm{PCl}_{5}(\mathrm{s}):$ $\mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s})$$\Delta_{i} H^{\circ}=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}$$\mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s})$$$\Delta_{r} H^{\circ}=-123.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{Dxn}$$Use these data to calculate the enthalpy change for the formation of 1.00 mol of $\mathrm{PCl}_{3}(\ell)$ from phosphorus and chlorine.
Step 1
This gives us: $$ \mathrm{PCl}_{5}(\mathrm{s}) \rightarrow \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) $$ Since we reversed the reaction, we also need to change the sign of $\Delta_{r} H^{\circ}$, so it becomes $123.8 \mathrm{kJ} / Show more…
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You wish to know the enthalpy change for the formation of liquid $\mathrm{PCl}_{3}$ from the elements. $$\mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{\mathrm{r}} H^{\circ}=?$$The enthalpy change for the formation of $\mathrm{PCl}_{5}$ from the elements can be determined experimentally, as can the enthalpy change for the reaction of $\mathrm{PCl}_{3}(\ell)$ with more chlorine to give $\mathrm{PCl}_{5}(\mathrm{s}):$ $\mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s})$ $$\Delta_{r} H^{\circ}=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{r} \mathrm{xn}$$ $\mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s})$ $$\Delta_{\mathrm{r}} H^{\circ}=-123.8 \mathrm{k} \mathrm{J} / \mathrm{mol}-\mathrm{r} \mathrm{xn}$$ Use these data to calculate the enthalpy change for the formation of 1.00 mol of $\mathrm{PCl}_{3}(\ell)$ from phosphorus and chlorine.
You wish to know the enthalpy change for the formation of liquid $\mathrm{PCl}_{3}$ from the elements. $\mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell)$ $\Delta H^{\circ}=?$ The enthalpy change for the formation of $\mathrm{PCl}_{5}$ from the elements can be determined experimentally, as can the enthalpy change for the reaction of $\mathrm{PCl}_{3}(\ell)$ with more chlorine to give $\mathrm{PCl}_{5}(\mathrm{~s})$ : $\mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s}) \quad \Delta H^{\circ}=-1774.0 \mathrm{~kJ}$ $\mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) \quad \Delta H^{\circ}=-123.8 \mathrm{~kJ}$ Use these data to calculate the enthalpy change for the formation of $1.00 \mathrm{~mol}$ of $\mathrm{PCl}_{3}(\ell)$ from phosphorus and chlorine.
You wish to know the enthalpy change for the formation of liquid PCl grom the elements. $$\mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta H^{\circ}=?$$ The enthalpy change for the formation of $\mathrm{PCl}_{5}$ from the elements can be determined experimentally, as can the enthalpy change for the reaction of $\mathrm{PCl}_{3}(\ell)$ with more chlorine to give $\mathrm{PCl}_{5}(\mathrm{s}):$ $$\begin{array}{cc}\mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \Delta H^{\circ}=-1774.0 \mathrm{kJ} \\\mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \Delta H^{\circ}=-123.8 \mathrm{kJ}\end{array}$$ Use these data to calculate the enthalpy change for the formation of 1.00 mol of $\mathrm{PCl}_{3}(\ell)$ from phosphorus and chlorine.
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