Calculate the enthalpy of the following reaction- H2C-CH2g-H2g-CH3-CH3gThe bond energies of C-H-C-C-C-C and H-H are 99-83-147 and 104 kcal respectively-

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Standard XII

Chemistry

Heat of Reaction

Calculate the...

Question

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 $C - H, C - C, C = C$ and $H - H$ are $99, 83, 147$ and $104$ $k c a l$ respectively.

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Solution

The reaction is given as (figure)
$Δ H =$ Sum of bond energies of reactants $-$ Sum of bond energies of products
$[Δ H_{C = C} + 4 \times Δ H_{C - H} + Δ H_{H - H}] - [Δ H_{C - C} + 6 \times Δ H_{C - H}]$
$= (147 + 4 \times 99 + 104) - (83 + 6 \times 99)$
$= - 30 k c a l$

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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.

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 $414$ , $347$ , $615$ 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:

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:

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Calculate the enthalpy of the following reaction:H2C=CH2(g)+H2(g)→CH3−CH3(g)The bond energies of C−H,C−C,C=C and H−H are 99,83,147 and 104 kcal respectively.

The reaction is given as (figure)ΔH= Sum of bond energies of reactants − Sum of bond energies of products[ΔHC=C+4×ΔHC−H+ΔHH−H]−[ΔHC−C+6×ΔHC−H]=(147+4×99+104)−(83+6×99)=−30kcal

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 $C - H, C - C, C = C$ and $H - H$ are $99, 83, 147$ and $104$ $k c a l$ respectively.

The reaction is given as (figure)
$Δ H =$ Sum of bond energies of reactants $-$ Sum of bond energies of products
$[Δ H_{C = C} + 4 \times Δ H_{C - H} + Δ H_{H - H}] - [Δ H_{C - C} + 6 \times Δ H_{C - H}]$
$= (147 + 4 \times 99 + 104) - (83 + 6 \times 99)$
$= - 30 k c a l$