Question
Given that the bond enthalpies of the Xe-F and the F-F bonds are 144 and $155 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively, calculate the enthalpy of formation for $\mathrm{XeF}_{2}$.
Step 1
This is given by: \[ \mathrm{Xe} + \mathrm{F}_{2} \rightarrow \mathrm{XeF}_{2} \] Show more…
Show all steps
Your feedback will help us improve your experience
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The standard enthalpy of formation of XeF_is $-218 \mathrm{kJ} / \mathrm{mol} .$ Use this value and the enthalpy of dissociation of the $\mathrm{F}-\mathrm{F}$ bond to calculate the Xe-F bond dissociation enthalpy.
The standard enthalpy of formation of $\mathrm{XeF}_{4}$ is $-218 \mathrm{kJ} / \mathrm{mol} .$ Use this value and the enthalpy of dissociation of the $\mathrm{F}-\mathrm{F}$ bond to calculate the $\mathrm{Xe}-\mathrm{F}$ bond dissociation enthalpy.
Calculate the standard enthalpy of formation $\left(\Delta H_{\mathrm{f}}^{\circ}\right)$ of $\mathrm{ClO}$ from the following bond enthalpies: $\mathrm{Cl}_{2}$ :$242.7 \mathrm{kJ} / \mathrm{mol} ; \mathrm{O}_{2}: 498.7 \mathrm{kJ} / \mathrm{mol} ; \mathrm{ClO}: 206 \mathrm{kJ} / \mathrm{mol}$.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD