$H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$
The bond energy of $C - H, C - C, C = C$ and $H - H$ are 414, 347, 615 and 435 kJ mol $^{- 1}$ respectively.
If the enthalpy change of the above reaction is $x$ kJ, then $\frac{- x}{25}$ is

Open in App

Solution

$∵ C H_{2} = C H_{2} (g) + H_{2} (g) \to C H_{3} - C H_{3} (g); Δ H = ?$
$∴ Δ H_{R e a c t i o n} =$ Bond energy data used for formation of bond + Bond energy data used for dissociation of bond
$∴ Δ H_{R e a c t i o n} = - [e_{(C - - C)} + 6 \times e_{(C - - H)}] + [e_{(C = = C)} + 4 \times e_{(C - - H)}] + e_{(H - - H)}$
$= - e_{(C - - C)} - 2 \times e_{(C - - H)} + e_{(C = = C)} + e_{(H - - H)}$
$= - 347 - 2 \times 414 + 615 + 435$
$Δ H_{R e a c t i o n} = - 125 k J$

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 $k J m o l^{- 1}$ , 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 $414$ , $347$ , $615$ and $435 k J / m o l$ respectively, then 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:

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: