Search any question & find its solution
Question:
Answered & Verified by Expert
Average bond enthalpy of water is $464 \cdot 5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If the energy required to break first $0-\mathrm{H}$ bond is $502 \mathrm{~kJ} \mathrm{~mol}^{-1}$, how much energy per mol is required to break second 0 -H bond?
Options:
Solution:
2377 Upvotes
Verified Answer
The correct answer is:
$427 \mathrm{~kJ}$
Average bond enthalpy of $\mathrm{H}_{2} \mathrm{O}$ is $464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\begin{array}{ll}
\mathrm{H}_{2} \mathrm{O}_{(z)} \longrightarrow \mathrm{H}_{(z)}+\mathrm{OH}_{(z)} & \Delta \mathrm{H}_{1}=502 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\mathrm{OH}_{(z)} \longrightarrow \mathrm{H}_{(z)}+\mathrm{O}_{(g)} & \Delta \mathrm{H}_{2}=?
\end{array}$
Average bond enthalpy $=\frac{\Delta \mathrm{H}_{1}+\Delta \mathrm{H}_{2}}{2}$
$\therefore \quad 464.5 \times 2=502+\Delta \mathrm{H}_{2}$
$\therefore \Delta \mathrm{H}_{2}=929-502=427 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\begin{array}{ll}
\mathrm{H}_{2} \mathrm{O}_{(z)} \longrightarrow \mathrm{H}_{(z)}+\mathrm{OH}_{(z)} & \Delta \mathrm{H}_{1}=502 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\mathrm{OH}_{(z)} \longrightarrow \mathrm{H}_{(z)}+\mathrm{O}_{(g)} & \Delta \mathrm{H}_{2}=?
\end{array}$
Average bond enthalpy $=\frac{\Delta \mathrm{H}_{1}+\Delta \mathrm{H}_{2}}{2}$
$\therefore \quad 464.5 \times 2=502+\Delta \mathrm{H}_{2}$
$\therefore \Delta \mathrm{H}_{2}=929-502=427 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.