For the reactionC2H6-C2H4-H2the reaction enthalpy- -rH k J mol-1-Round off to the Nearest Integer-Given - Bond enthalpies in k J mol-1C - C - 347- C - C - 611-C - H - 414- H - H - 436-

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

Byju's Answer

Standard XII

Chemistry

Properties of Enthalpy

For the react...

Question

For the reaction
$C_{2} H_{6} \to C_{2} H_{4} + H_{2}$
the reaction enthalpy, $Δ_{r} H$ _____ $k J m o l^{- 1}$ .
[Round off to the Nearest Integer]
[Given : Bond enthalpies in $k J m o l^{- 1}$
C - C : 347, C = C : 611;
C - H : 414, H - H : 436]

Open in App

Solution

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

$Δ_{r} H = Sum of bond dissociation energy - Sum of bond formation energy$
$Δ_{r} H = {(B . E)}_{C - C} + 6 \times {(B . E)}_{C - H} - [{(B . E)}_{C = C} + 4 \times {(B . E)}_{C - H} + {(B . E)}_{H - H}]$

$Δ_{r} H = 347 + 6 \times 414 - (611 + 4 \times 414 + 436)$
$Δ_{r} H = 2831 - 2703$
$Δ_{r} H = 128 k J m o l^{- 1}$

Suggest Corrections

Similar questions

Q. 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]

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

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:

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

Calculate the enthalpy ,

Δ H

, for the given reaction using the the given bond energies ?

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

Bond energies	kJ/ mol
$C - C$	$347$
$C = C$	$612$
$C - C l$	$341$
$C - H$	$414$
$C l - C l$	$243$

Join BYJU'S Learning Program

For the reaction C2H6→C2H4+H2 the reaction enthalpy, ΔrH _____k J mol−1. [Round off to the Nearest Integer] [Given : Bond enthalpies in kJmol−1 C - C : 347, C = C : 611; C - H : 414, H - H : 436]

C2H6→C2H4+H2 ΔrH=Sum of bond dissociation energy - Sum of bond formation energy ΔrH=(B.E)C−C+6×(B.E)C−H−[(B.E)C=C+4×(B.E)C−H+(B.E)H−H] ΔrH=347+6×414−(611+4×414+436) ΔrH=2831−2703 ΔrH=128 kJ mol−1

For the reaction
$C_{2} H_{6} \to C_{2} H_{4} + H_{2}$
the reaction enthalpy, $Δ_{r} H$ _____ $k J m o l^{- 1}$ .
[Round off to the Nearest Integer]
[Given : Bond enthalpies in $k J m o l^{- 1}$
C - C : 347, C = C : 611;
C - H : 414, H - H : 436]