If at $298$ K the bond energies of $C - H, C - C, C = C$ and $H - H$ bonds are respectively $414, 347, 615$ and $435$ kJ mol^{-1}, the value of enthalpy change for the reaction:
$H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$ at $298 K$ will be:

- 125

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 125

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 250

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 250

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 125$ kJ
$H_{2} C = C H_{2} + H_{2} ⟶ H_{3} C - C H_{3}$
Bond break Energy Bond form Energy
$H - H$ $435 K J m o l^{- 1}$ $2 C - H$ $2 \times (- 414) K J m o l^{- 1}$
$C = C$ $615 K J m o l^{- 1}$ $C - C$ $- 347 K J m o l^{- 1}$
Total enthalpy change is
$\Rightarrow 435 + 615 - 2 \times 4141 - 347$
$= - 125 K J m o l^{- 1}$ .

Suggest Corrections

Similar questions

If at 298 K the bond energies of C - H, C - C, C=C and H - H bonds are respectively 414, 347, 615 and 435 kJ $m o l^{- 1}$ , the value of enthalpy change for the reaction

If at $298 K$ the bond energies of C - H, C - C, C - C and H - H bonds are respectively 414, 347, 615 and 435 $k J m o l^{- 1}$ , the value of enthalpy change for the reaction $H_{2} C = C H_{2} (g) \to H_{3} C - C H_{3} (g)$ at 298K will be

If at 298 K the bond energies of C - H, C - C, C=C and H - H bonds are respectively 414, 347, 615 and 435 kJ $m o l^{- 1}$ , the value of enthalpy change for the reaction
$C H_{2} = C H_{2} (g) + H_{2} (g) \to C H_{3} - C H_{3}$ is: