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Standard XII

Physics

Calorimetry

H2C = CH2g+ H...

Question

$H_{2} C = C H_{2} (g) + H_{2} (g) ⟶ H_{3} C - C H_{3} (g)$
The bond energy of $C - H, C - C, C = C, H - H$ are $414, 347, 615, 435 k J m o l^{- 1}$ respectively.
Enthalpy change of the above reaction is:

125

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-125

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150

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None of these

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Solution

The correct option is B -125
$Δ H^{o} = Σ Δ H^{o} ( reactant bonds) - Σ Δ H^{o} ( product bonds)$
$Δ H^{o} = Δ H^{o} (C = C) + 4 Δ H^{o} (C - H) + Δ H^{o} (H - H) - [Δ H^{o} (C - C) + 6 Δ H^{o} (C - H)]$
$Δ H^{o} = 615 + 4 \times 414 + 435 - [347 + 6 \times 414]$
$Δ H^{o} = - 125 k J / m o l$

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Similar questions

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}

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:

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:

H_{2} C = = C H_{2} (g) + H_{2} (g) ⟶ H_{3} C - - C H_{3} (g)

T h e b o n d e n e r g y o f C - - H, C - - C, C = = C a n d H - - H a r e 414, 347, 615 a n d 435 k J m o l^{- 1} r e s p e c t i v e l y .

If the enthalpy change of the above reaction is

x

k J

, then

- x

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H2C=CH2(g)+H2(g)⟶H3C−CH3(g)The bond energy of C−H,C−C,C=C,H−H are 414,347,615,435kJmol−1 respectively.Enthalpy change of the above reaction is:

The correct option is B -125ΔHo=ΣΔHo ( reactant bonds) −ΣΔHo ( product bonds) ΔHo=ΔHo(C=C)+4ΔHo(C−H)+ΔHo(H−H)−[ΔHo(C−C)+6ΔHo(C−H)] ΔHo=615+4×414+435−[347+6×414] ΔHo=−125 kJ/mol

$H_{2} C = C H_{2} (g) + H_{2} (g) ⟶ H_{3} C - C H_{3} (g)$
The bond energy of $C - H, C - C, C = C, H - H$ are $414, 347, 615, 435 k J m o l^{- 1}$ respectively.
Enthalpy change of the above reaction is: