You visited us 1 times! Enjoying our articles? Unlock Full Access!

Physical Properties of Alkenes

Using the bon...

Question

Using the bond enthalpy data given below, calculate the enthalpy change for the reaction ((only magnitude in nearest integer in kj/mol),
$C_{2} H_{4} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g)$
Bond $C - C$ $C = C$ $C - H$ $H - H$
Bond Enthalpy $336.81 k J / m o l$ $606.68 k J / m o l$ $410.87 k J / m o l$ $431.79 k J / m o l$

Open in App

Solution

Energy is absorbed while breaking a bond and energy is released when a bond is formed.
Hence the net energy change for a reaction = Energy required to break various bonds in reactants $-$ Energy released by making bonds in products
Here
In reactants: $1 C = C$ bond, $4 C - H$ bonds, and $1 H - H$ bond
In products: $1 C - C$ bond, $6 C - H$ bonds

Therefore
$Δ H = 606.68 + (4 \times 410.87) + 431.79 - (336.81 + 6 \times 410.87) = - 120.08 k J / m o l^{- 1}$

Answer = 120

Suggest Corrections

Similar questions

C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	336.81 kJ / mol	606.68 kJ / mol	410.87 kJ / mol	431.79 kJ / mol

C - H = 413.4 k J {m o l}^{- 1}

C - C = 347.0 k J {m o l}^{- 1}

C = O = 728.0 k J {m o l}^{- 1}

O = O = 495.0 k J {m o l}^{- 1}

H - H = 435.8 k J {m o l}^{- 1}

Δ H_{s u b} C (s) = 718.4 k J {m o l}^{- 1}

C_{2} H_{6} \to C_{2} H_{4} + H_{2}

Δ_{r} H =

{kJ mol}^{- 1}

{kJ mol}^{- 1}

C_{2} H_{6} = - 83 k J / m o l

719 k J / m o l

H_{2} = 435 k J / m o l

414 k J / m o l

C_{2} H_{6} \to C_{2} H_{4} + H_{2}

Δ_{r} H

k J m o l^{- 1}

k J m o l^{- 1}

Join BYJU'S Learning Program