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,
$H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$ at $298 K$ will be:

- 125 k J

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+ 125 k J

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- 250 k J

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+ 250 k J

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Solution

The correct option is A $- 125 k J$
$H_{2} C = C H_{2} (g) + H_{2} (g) ⟶ H_{3} C - C H_{3} (g)$

According to Hess law,

$Δ H = Σ E_{R e a c t a n t s B o n d s B r o k e n} - Σ E_{P r o d u c t s B o n d s F o r m e d}$

$= [(C = C) + (H - H)] - [(C - C) + 2 (C H)]$

$= (615 + 435) - [347 + 2 (414)]$

$≃ - 125 K J$

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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

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:

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

kJ mol^{-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 $414$ , $347$ , $615$ 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:

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

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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 mol−1, the value of enthalpy change for the reaction,H2C=CH2(g)+H2(g)→H3C−CH3(g) at 298 K will be:

The correct option is A −125 kJH2C=CH2(g)+H2(g)⟶H3C−CH3(g)According to Hess law,ΔH=ΣEReactantsBondsBroken−ΣEProductsBondsFormed=[(C=C)+(H−H)]−[(C−C)+2(CH)]=(615+435)−[347+2(414)]≃−125KJ

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,
$H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$ at $298 K$ will be: