For the reaction- C2H6- C2H4 - H2 - The reaction enthalpy -rH - kJ mol-1- Round off to the Nearest Integer-Given- Bond enthalpies in kJ mol-1- C - C- 347- C - C - 611- C - H - 414- H - H - 436-

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

Internal Energy

For the react...

Question

For the reaction: $C_{2} H_{6} \to C_{2} H_{4} + H_{2}$ : The reaction enthalpy $Δ_{r} H =$ ______ ${kJ mol}^{- 1}$ . (Round off to the Nearest Integer).

[Given: Bond enthalpies in ${kJ mol}^{- 1}$ ; C – C: 347; C = C : 611; C – H : 414; H – H : 436]

Open in App

Solution

Suggest Corrections

Similar questions

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}

△ H_{r}^{o}

C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .

414 k J m o l^{- 1}

330 k J m o l^{- 1} .

Δ H

C_{2} H_{4} + C l_{2} \to C l H_{2} C - C H_{2} C l

(δ H)

2 C_{2} H_{2 (g)} + 5 O_{2 (g)} \to 4 C O_{2 (g)} + 2 H_{2} O_{(g)}

C - H, C \equiv C, O = O, C = O, O - H

414, 814, 499, 724

640 k J m o l^{- 1}

C_{2} H_{4} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g)

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	$336.81 k J / m o l$	$606.68 k J / m o l$	$410.87 k J / m o l$	$431.79 k J / m o l$

Join BYJU'S Learning Program