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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 $336.81 k J / m o l$ $606.68 k J / m o l$ $410.87 k J / m o l$ $431.79 k J / m o l$

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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

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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} = - 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 =

{kJ mol}^{- 1}

{kJ mol}^{- 1}

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}

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