Question

Calculate the enthalpy change for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) $$ given the following enthalpies of reaction: $$ \begin{array}{ll} \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) & \Delta H=-1640.1 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2940.1 \mathrm{~kJ} \end{array} $$

Short answer

The enthalpy change for the reaction \(\mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)\) can be calculated using Hess's Law and the given enthalpy changes for the two other reactions. After reversing the first given reaction and adding it to the second given reaction to obtain the target reaction, the enthalpy change for the target reaction is found to be \(\Delta H_{target} = -1300 \mathrm{~kJ}\).

Step-by-step solution

Write down the target reaction

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

Analyze the given reactions and adjust them to match the target reaction

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) \( \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \)

Add up the adjusted enthalpy changes to find the enthalpy change for the target reaction

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.