Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

The target reaction is given as: \[ \mathrm{P}_{4}\mathrm{O}_{10}(s) + 6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3}\mathrm{PO}(g). \]

In order to get to the desired reaction, we must find a way to combine the given reactions. To do this, we will manipulate the given reactions accordingly: 1. Reverse the second reaction so that P4O10 is on the left side of the reaction. After reversing, it becomes: \[ \mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g), \quad \Delta H = +2967.3 \mathrm{~kJ}. \] Note that the enthalpy change also changes its sign since the reaction is reversed. 2. Multiply the third reaction by 6 so that 6 PCl5 will be on the left side: \[ 6\mathrm{PCl}_{3}(g) + 6\mathrm{Cl}_{2}(g) \longrightarrow 6\mathrm{PCl}_{5}(g), \quad \Delta H = -84.2 \cdot 6 \mathrm{~kJ}. \] 3. Multiply the fourth reaction by 10 so that there will be 10 Cl3PO on the right side of the reaction: \[ 10\mathrm{PCl}_{3}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g), \quad \Delta H = -285.7 \cdot 10 \mathrm{~kJ}. \]

Add the manipulated reactions together to form the target reaction: \[ \begin{aligned} \cancel{\mathrm{P}_{4}\mathrm{O}_{10}(s)} &\longrightarrow \cancel{\mathrm{P}_{4}(s)} + \cancel{5\mathrm{O}_{2}(g)} \\ \cancel{\mathrm{P}_{4}(s)} + 6\cancel{\mathrm{Cl}_{2}(g)} &\longrightarrow 4\cancel{\mathrm{PCl}_{3}(g)} \\ 6\cancel{\mathrm{PCl}_{3}(g)} + 6\cancel{\mathrm{Cl}_{2}(g)} &\longrightarrow 6\mathrm{PCl}_{5}(g) \\ 10\cancel{\mathrm{PCl}_{3}(g)} + \cancel{5\mathrm{O}_{2}(g)} &\longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g) \\ \hline \mathrm{P}_{4}\mathrm{O}_{10}(s) + 6\mathrm{PCl}_{5}(g) &\longrightarrow 10\mathrm{Cl}_{3}\mathrm{PO}(g) \end{aligned} \]

Add up the enthalpy changes of the manipulated reactions to get the enthalpy change for the target reaction: \[ \Delta H_\text{target} = (+2967.3) + (-1225.6) + (-84.2 \cdot 6) + (-285.7 \cdot 10) \mathrm{~kJ}. \] Calculate the numerical value: \[ \Delta H_\text{target} = 2967.3 - 1225.6 - 505.2 - 2857 \mathrm{~kJ} = -1620.5 \mathrm{~kJ}. \] So the enthalpy change for the target reaction is \(\Delta H=-1620.5 \mathrm{~kJ}\).