Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(g) & \Delta H=-1225.6 \mathrm{kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2967.3 \mathrm{kJ} \end{aligned} $$ $$ \begin{array}{cc}{\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g)} & {\Delta H=-84.2 \mathrm{kJ}} \\\ {\mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g)} & {\Delta H=-285.7 \mathrm{kJ}}\end{array} $$ 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 we want to find the enthalpy change for is: \( \mathrm{P}_{4} \mathrm{O}_{10}(s) + 6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) \)

First, let's work with the second given reaction (\( \mathrm{P}_{4}(s) + 5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \)), as it is the only one with the P4O10 compound. Since the target reaction begins with 1 P4O10, we keep the second reaction the same and not need to manipulate it: \( \mathrm{P}_{4}(s) + 5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \, (\Delta H_2 = -2967.3 \,\mathrm{kJ}) \) Now, let's look at the first given reaction (\( \mathrm{P}_{4}(s) + 6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(g) \)). To get 6 PCl5 in our target reaction, we will need to reverse this first given reaction and multiply it by 1.5. This gives: \( 6 \mathrm{PCl}_{3}(g) \longrightarrow \mathrm{P}_{4}(s) + 9 \mathrm{Cl}_{2}(g) \, (-1.5\cdot \Delta H_1 = 1.5 \cdot 1225.6 \,\mathrm{kJ}) \) Next, let's work with the third given reaction (\( \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) \)). Since we reversed the first given reaction and obtained 6 PCl3 in it, we'll reverse this reaction also and multiply it by 6: \( 6 \mathrm{PCl}_{5}(g) \longrightarrow 6 \mathrm{PCl}_{3}(g) + 6 \mathrm{Cl}_{2}(g) \, (-6 \cdot \Delta H_3 = 6 \cdot 84.2 \, \mathrm{kJ}) \) Lastly, for the fourth given reaction (\( \mathrm{PCl}_{3}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) \)), we need 10 Cl3PO in our target reaction. So, we'll multiply it by 10: \( 10 \mathrm{PCl}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) \, (10 \cdot \Delta H_4 = 10 \cdot-285.7 \, \mathrm{kJ}) \)

Now, let's add all the manipulated reactions together: \( 6 \mathrm{PCl}_{3}(g) \longrightarrow \mathrm{P}_{4}(s) + 9 \mathrm{Cl}_{2}(g) \, (1.5 \cdot 1225.6 \,\mathrm{kJ}) \) \( \mathrm{P}_{4}(s) + 5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \, (-2967.3 \,\mathrm{kJ}) \) \( 6 \mathrm{PCl}_{5}(g) \longrightarrow 6 \mathrm{PCl}_{3}(g) + 6 \mathrm{Cl}_{2}(g) \, (6 \cdot 84.2 \,\mathrm{kJ}) \) \( 10 \mathrm{PCl}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) \, (10 \cdot -285.7 \, \mathrm{kJ}) \) -------------------------------------------------------------------------------------------- \( \mathrm{P}_{4} \mathrm{O}_{10}(s) + 6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) \, (\Delta H_{target}) \)

Now, we can sum up the enthalpy changes from the manipulated reactions to find the enthalpy change for the target reaction: \( \Delta H_{target} = 1.5 \cdot 1225.6 - 2967.3 + 6 \cdot 84.2 + 10 \cdot (-285.7) \) \( \Delta H_{target} = 1838.4 - 2967.3 + 505.2 - 2857 \) \( \Delta H_{target} = -4470.7 \, \mathrm{kJ} \) The enthalpy change for the target reaction is -4470.7 kJ.