From the following thermochemical equations- C2H4 - C2H6- H - -32-7 kcal C6H6-3H2 - C6H12- H- -49-2kcalCalculate resonance energy of benzene-

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Standard XII

Chemistry

Enthalpy

From the foll...

Question

From the following thermochemical equations:
$C_{2} H_{4} \to C_{2} H_{6}; △ H = - 32.7 k c a l$
$C_{6} H_{6} + 3 H_{2} \to C_{6} H_{12}; △ H = - 49.2 k c a l$
Calculate resonance energy of benzene.

- 48.9 k c a l

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- 49.2 k c a l

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- 16.5 k c a l

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- 98.1 k c a l

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Solution

The correct option is A $- 48.9 k c a l$
Enthalpy of hydrogenation of $C = C$ bond $= - 32.7 k c a l$
Calculated enthalpy of hydrogenation of benzene $= - 32.7 \times 3 = - 98.1 k c a l$
Actual enthalpy of hydrogenation of benzene $= - 45.2 k c a l$
Hence, Resonance energy of benzene $= - 98.1 - (- 45.2) = 52.9 k c a l$

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Similar questions

Q. Consider the following thermochemical equation:

C_{2} H_{4} + H_{2} \to C_{2} H_{6}; △ H_{1} = - 32.7 k c a l

1, 3 - b u t a d i e n e + 2 H_{2} \to C_{4} H_{10}; △ H_{2} = - 40.2 k c a l

Therefore the magnitude of resonance energy of 1,3-butadiene is:

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Q. Calculate the resonance energy of

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from the following data: (i)

Δ H^{\circ}

for

C_{6} H_{6} = - 358.5 k J m o l^{- 1}

(ii)

H e a t o f a t o m i s a t i o n o f C = 716.8 k J m o l^{- 1}

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and

H - - H

are 490, 340, 620 and

436.9 k J m o l^{- 1}

. Report your answer in kj (only value).

Following gaseous reaction is undergoing in a vessel $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}; Δ H = - 32.7 K c a l$ . Which will increase the equilibrium concentration of $C_{2} H_{6}$ ?

Q. For the Gas phase reaction

C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g) △ H = - 32.7 k c a l

caried out in a vessel, the equilibrium concentration of

C_{2} H_{4}

can be increased by:

Q. The enthalpy of hydrogenation of benzene is -207 kJ/mol and its resonance energy is -150 kJ/mol. Calculate the enthalpy of hydrogenation

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From the following thermochemical equations:C2H4→C2H6;△H=−32.7kcalC6H6+3H2→C6H12;△H=−49.2kcalCalculate resonance energy of benzene.

The correct option is A −48.9 kcalEnthalpy of hydrogenation of C=C bond =−32.7 kcalCalculated enthalpy of hydrogenation of benzene =−32.7×3=−98.1 kcalActual enthalpy of hydrogenation of benzene=−45.2 kcalHence, Resonance energy of benzene=−98.1−(−45.2)=52.9 kcal

From the following thermochemical equations:
$C_{2} H_{4} \to C_{2} H_{6}; △ H = - 32.7 k c a l$
$C_{6} H_{6} + 3 H_{2} \to C_{6} H_{12}; △ H = - 49.2 k c a l$
Calculate resonance energy of benzene.