Given the following data $$ \begin{aligned} 2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) & \Delta H &=167.4 \mathrm{~kJ} \\ 2 \mathrm{ClF}_{3}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+3 \mathrm{~F}_{2} \mathrm{O}(g) & \Delta H &=341.4 \mathrm{~kJ} \\\ 2 \mathrm{~F}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{~F}_{2} \mathrm{O}(g) & \Delta H &=-43.4 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g) $$

First, we need to determine which reactions we need to combine to get the desired reaction: $$ \mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g) $$ From the given reactions, we know that the above reaction is a combination of the first and third reaction.

We will manipulate the first and third reactions so that they can be added together to create the desired reaction equation. In this case, we will reverse the first reaction and divide the third reaction by 2. This will give us the correct stoichiometric coefficients for our target reaction. Reversed first reaction: $$ \mathrm{Cl}_{2}\mathrm{O}(g)+\mathrm{F}_{2}\mathrm{O}(g) \longrightarrow 2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) $$ and the new enthalpy is $$ \Delta H_{1}' = -167.4 \mathrm{~kJ} $$ Divide the third reaction by 2: $$ \mathrm{F}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{F}_{2}\mathrm{O}(g) $$ and the new enthalpy is $$ \Delta H_{3}' = -\frac{43.4}{2} \mathrm{~kJ} = -21.7 \mathrm{~kJ} $$

Now we will add the adjusted reactions from Step 2 to form the desired reaction, and add the corresponding enthalpies to find the enthalpy for the desired reaction. Add the adjusted reactions: $$ (\mathrm{Cl}_{2}\mathrm{O}(g)+\mathrm{F}_{2}\mathrm{O}(g) \longrightarrow 2\mathrm{ClF}(g)+\mathrm{O}_{2}(g)) + (\mathrm{F}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{F}_{2}\mathrm{O}(g)) $$ This simplifies to $$ \mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g) $$ Add the adjusted enthalpy changes: $$ \Delta H = \Delta H_{1}' + \Delta H_{3}' = -167.4 \mathrm{~kJ} - 21.7 \mathrm{~kJ} = -189.1 \mathrm{~kJ} $$

The enthalpy change, \(\Delta H\), for the reaction $$ \mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g) $$ is -189.1 kJ.