The enthalpy change for the reaction H2g - C2H4g - C2H6g is- The bond energies are- H - H - 103- C - H - 99- C -C - 80 and C -C - 145 K cal mol-1

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Bond Order and Enthalpy

The enthalpy ...

Question

The enthalpy change for the reaction $H_{2} (g) + C_{2} H_{4} (g) \to C_{2} H_{6} (g)$ is:
The bond energies are, $H - H = 103, C - H = 99, C - C = 80$ and $C = C = 145$ $K c a l m o l^{- 1}$

- 10 K c a l m o l^{- 1}

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+ 10 K c a l m o l^{- 1}

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- 30 K c a l m o l^{- 1}

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+ 30 K c a l m o l^{- 1}

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Solution

The correct option is D $+ 30 K c a l m o l^{- 1}$
$H_{2}$ has a sigle $H - H$ bond. So $Δ {_{f} H}_{H_{2}} = 103 k c a l m o l^{- 1}$
$C_{2} H_{4}$ has structure $H - C | H = C | H - H$ So, $Δ_{f} H = 4 \times B E_{C - H} + B E_{C = C}$
$Δ_{f} H_{C_{2} H_{4}} = 4 \times 99 + 145 = 541 k c a l m o l^{- 1}$
$C_{2} H_{6}$ has 6 $C - H$ bonds and one $C - C$ bond. So, $Δ_{f} H_{C_{2} H_{6}} = 6 \times 99 + 80 = 674 k c a l m o l^{- 1}$
Enthalpy change for the given reaction $Δ H = 674 - [541 + 103]$
$Δ H = + 30 k c a l m o l^{- 1}$

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Similar questions

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

Using the bond enthalpy data given below, calculate the enthalpy change for the reaction.

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

Q. 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) + H_{2} (g) \to H_{3} C - C H_{3} (g)

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:

H | C | H = H | C | H + H - H \to H - H | C | H - H | C | H - H

From the following bond energies:

H - H

bond energy :

431.37 k J m o l^{- 1}

C = C

bond energy :

606.10 k J m o l^{- 1}

C - C

bond energy :

336.49 k J m o l^{- 1}

C - H

bond energy :

410.50 k J m o l^{- 1}

Enthalpy for the reactions, will be?

Calculate the enthalpy of the following reaction. $H_{2} C = C H_{2} (g) + H_{2} (g) \to C H_{3} - C H_{3} (g)$

The bond energies of C - H, C - C, C = C & H - H are 99, 83, 147 & 104 Kcal respectively.

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The enthalpy change for the reaction H2(g)+C2H4(g)→C2H6(g) is: The bond energies are, H−H=103, C−H=99, C−C=80 and C=C=145 K cal mol−1

The enthalpy change for the reaction $H_{2} (g) + C_{2} H_{4} (g) \to C_{2} H_{6} (g)$ is:
The bond energies are, $H - H = 103, C - H = 99, C - C = 80$ and $C = C = 145$ $K c a l m o l^{- 1}$