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Standard XII

Physics

Calorimetry

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 $k J m o l^{- 1}$ , 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

+250 kJ

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-250 kJ

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+125 kJ

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-125 Kj

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Solution

The correct option is D
-125 Kj

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

$414 \times 4 = 1656414 \times 6 = 2484$

$615 \times 1 = 615347 \times 1 = 347$

$435 \times 1 = 435 - -- - 347 \times 1 = 347 - -- -$

$2706 - --- - 2831 - --- -$

$Δ H = 2706 - 2831 = - 125 k J$

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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 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 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 respectively 414, 347, 615 and 435 kJ $m o l^{- 1}$ , the value of enthalpy change for the reaction

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 H2C = CH2(g) → H3C − CH3(g) at 298K will be

The correct option is D -125 Kj CH2 = CH2(g) + H2(g) → H3C − CH3(g) 414 × 4 = 1656 414 × 6 = 2484 615 × 1 = 615 347 × 1 = 347 435 × 1 = 435–––– 347 × 1 = 347–––– 2706––––– 2831––––– ΔH = 2706 − 2831 = − 125 kJ

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

The correct option is D
-125 Kj

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

$414 \times 4 = 1656414 \times 6 = 2484$

$615 \times 1 = 615347 \times 1 = 347$

$435 \times 1 = 435 - -- - 347 \times 1 = 347 - -- -$

$2706 - --- - 2831 - --- -$

$Δ H = 2706 - 2831 = - 125 k J$