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Standard XII

Chemistry

Hess' Law

If at 298 K t...

Question

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:

-250 kJ

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+125 kJ

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-125 kJ

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+250 kJ

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Solution

The correct option is C
-125 kJ

$C H_{2} = C H_{2} (g) + H_{2} (g) \to C H_{3} - C H_{3} Enthalpy change = Bond energy of raeactants -Bond energy of products . Δ H = 1 (C = C) + 4 (C - H) + 1 (H - H) - 1 (C - C) - 6 (C - H) = 1 (C = C) + 1 (H - H) - 1 (C - C) - 2 (C - H) = 615 + 435 - 347 - 2 \times 414 = 1050 - 1175 = - 125 k J .$

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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

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:

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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 CH2=CH2(g)+H2(g)→CH3−CH3 is:

The correct option is C -125 kJ CH2=CH2(g)+H2(g)→CH3−CH3Enthalpy change = Bond energy of raeactants -Bond energy of products.ΔH=1(C=C)+4(C−H)+1(H−H)−1(C−C)−6(C−H)=1(C=C)+1(H−H)−1(C−C)−2(C−H)=615+435−347−2×414=1050−1175=−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 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: