Correct order of standard enthalpy of formation of diamond graphite, and fullerene is

Δ_{f} H_{fullerene}^{\circ} > Δ_{f} H_{diamond}^{\circ} > Δ_{f} H_{graphite}^{\circ}

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Δ_{f} H_{diamond}^{\circ} > Δ_{f} H_{graphite}^{\circ} > Δ_{f} H_{fullerene}^{\circ}

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Δ_{f} H_{fullerene}^{\circ} > Δ_{f} H_{graphite}^{\circ} > Δ_{f} H_{diamond}^{\circ}

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Δ_{f} H_{graphite}^{\circ} > Δ_{f} H_{diamond}^{\circ} > Δ_{f} H_{fullerene}^{\circ}

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Solution

The correct option is A $Δ_{f} H_{fullerene}^{\circ} > Δ_{f} H_{diamond}^{\circ} > Δ_{f} H_{graphite}^{\circ}$
Graphite is thermodynamically the most stable allotrope of carbon and therefore $Δ_{f} H^{\circ}$ of graphite is taken as zero.
$Δ_{f} H^{\circ} of diamond: 1.90 k J m o l^{-} 1$
$Δ_{f} H^{\circ} of fullerene: 38.1 k J m o l^{-} 1$

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Similar questions

Choose the odd one out :

Diamond; soot; graphite; fullerene.

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(a) Why is graphite a good conductor of electricity but not diamond ?

(b) Why is diamond very hard ?

(d) What impurity is present in black diamond ?

How the formation enthalpy of a compound is related to its stability ?For eg- enthalpy of formation of graphite is zero whereas enthalpy of formation of fullerness and diamond is 38.1 and 1.90 kj mol^-1 respectively. Here graphite is more stable than diamond and fullerness. How ?

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Q. Calculate the enthalpy change accompaying the conversion of 10 grams of graphite into diamond if the standard enthalpy of combustion of

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Correct order of standard enthalpy of formation of diamond graphite, and fullerene is

The correct option is A ΔfH∘fullerene>ΔfH∘diamond>ΔfH∘graphiteGraphite is thermodynamically the most stable allotrope of carbon and therefore ΔfH∘ of graphite is taken as zero. ΔfH∘ of diamond: 1.90 kJ mol−1 ΔfH∘ of fullerene: 38.1 kJ mol−1