91Ó°ÊÓ

First, let's write down the reaction we want to find the enthalpy change for: \( \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \)

We have the following two given reactions: 1) \( \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) \quad \Delta H=-1640.1 \mathrm{~kJ} \) 2) \( \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \quad \Delta H=-2940.1 \mathrm{~kJ} \) Looking at the target reaction, we can see that P4O6 is a reactant, but in the first given reaction, it is a product. So, we'll reverse reaction 1: Reversed Reaction 1) \( \mathrm{P}_{4} \mathrm{O}_{6}(s) \longrightarrow \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=1640.1 \mathrm{~kJ} \) Now let's add the reversed reaction 1 with reaction 2: Reversed Reaction 1 + Reaction 2) \( \mathrm{P}_{4} \mathrm{O}_{6}(s) + \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) + \mathrm{P}_{4} \mathrm{O}_{10}(s) \) We can cancel out the P4(s) from both sides of the equation, and we will be left with the target reaction: Target Reaction) \( \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \)

Since we reversed the first reaction and added it to the second reaction to obtain the target reaction, we have to add their enthalpy changes according to Hess's Law: \( \Delta H_{target} = 1640.1 \mathrm{~kJ} - 2940.1 \mathrm{~kJ} \) Now, let's calculate the enthalpy change for the target reaction: \( \Delta H_{target} = -1300 \mathrm{~kJ} \) The enthalpy change for the target reaction is -1300 kJ.