If ΔfH∘C2H6=−84kJmol−1,Δsub H∘C, graphite )=720kJmol−1,ΔatH2=435kJmol−1 and the bond enthalpy εC−H=414kJmol−1 then the bond enthalpy εC−C will be about

# If  and the bond enthalpy ${\epsilon }_{\mathrm{C}-\mathrm{H}}=414\mathrm{kJ}{\mathrm{mol}}^{-1}$ then the bond enthalpy ${\epsilon }_{\mathrm{C}-\mathrm{C}}$ will be about

1. A

2. B

3. C

4. D

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

According to Hess's law $\begin{array}{r}{\epsilon }_{\mathrm{C}-\mathrm{C}}=2{\mathrm{\Delta }}_{\mathrm{sub}}{H}^{\circ }\left(\mathrm{C}\right)+3{\mathrm{\Delta }}_{\mathrm{at}}{H}^{\circ }\left({\mathrm{H}}_{2}\right)-6{\epsilon }_{\mathrm{C}-\mathrm{H}}-{\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }=\left[2×720+3×435-6×414+84\right]\mathrm{kJ}{\mathrm{mol}}^{-1}\\ =345\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

## Related content

 Distance Speed Time Formula Refractive Index Formula Mass Formula Electric Current Formula Ohm’s Law Formula Wavelength Formula Electric Power Formula Resistivity Formula Weight Formula Linear Momentum Formula