If at $298 K$ , the bond energies of $C - H, C - C, C = C$ and $H - H$ bonds are $414$ , $347$ , $615$ and $435 k J / m o l$ respectively, then the value of enthalpy change for the reaction $H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$ at $298 K$ will be:

- 250 k J

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+ 125 k J

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- 125 k J

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+ 250 k J

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Solution

The correct option is C $- 125 k J$
Given that the bond energies of $C - H, C - C, C = C$ and $H - H$ bonds are $414, 347, 615$ and $435$ kJ/mol respectively at $298$ K.

We need to find the enthalpy change at $298$ K for the following reaction: $H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$

$Δ H =$ Bond energy for reactants - Bond energy for products

$Δ H = [4 \times (C - H) + C = C + (H - H)] - [6 \times (C - H) + (C - C)]$

$Δ H = [4 \times 414 + 615 + 435] - [(6 \times 414) + 347]$

$Δ H = 2706 - 2831$

$Δ H = - 125 k J$

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Similar questions

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

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_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)

The bond energy of

C - H, C - C, C = C

and

H - H

are 414, 347, 615 and 435 kJ mol

^{- 1}

respectively.
If the enthalpy change of the above reaction is

x

kJ, then

\frac{- x}{25}

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If at 298 K, the bond energies of C−H, C−C, C=C and H−H bonds are 414, 347, 615 and 435 kJ/mol respectively, then the value of enthalpy change for the reaction H2C=CH2(g)+H2(g)→H3C−CH3(g) at 298 K will be:

If at $298 K$ , the bond energies of $C - H, C - C, C = C$ and $H - H$ bonds are $414$ , $347$ , $615$ and $435 k J / m o l$ respectively, then the value of enthalpy change for the reaction $H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$ at $298 K$ will be: