C graphite - O 2 g - CO 2 g - H -94-05 kcal mol -1C diamond - O 2 g - CO 2 g - - H -94-50 kcal mol -1 thereforeA- Cdiamonal - Cgraphite - H298 K-450 cal mol -1B- Cgraphite - Cdianond - - H298 K-450 cal mol -1C- Graphite is the stabler allotropeD- Diamond is harder than graphite

The correct option is C Graphite is the stabler allotrope nbsp; Heat energy is also involved when one allotropic form of an element is converted in to anothe ...

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Byju's Answer

Standard XII

Chemistry

Introduction

C graphite + ...

Question

$C_{g r a p h i t e} + O_{2} (g) \to C O_{2} (g)$

$Δ H = - 94.05 k c a l m o l^{- 1}$

$C_{d i a m o n d} + O_{2} (g) \to C O_{2} (g); Δ H = - 94.50 k c a l m o l^{- 1}$ therefore

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Graphite is the stabler allotrope

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Diamond is harder than graphite

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Solution

The correct option is C
Graphite is the stabler allotrope

Heat energy is also involved when one allotropic form of an element is converted in to another graphite is the stabler is the stabler allotrope because the heat of transformation of $C_{d i a m o n d} \to C_{g r a p h i t e}$

(i) $C_{(d i a)} + O_{2} (g) = C O_{2} (g) Δ H = - 94.5 K c a l$

(ii) $C_{(g r a p h i t e)} + O_{2} (g) = C O_{2} (g) Δ H = - 94.0 K c a l$

$Δ H_{(t r a n s f o r m a t i o n)} = - 94.5 - (- 94.0)$

= $0.54 K c a l$ .

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Similar questions

Assertion- $C (g r a p h i t e) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 393.5 k . J$

$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 395.4 k . J$

Reason(R): graphite is more stable than diamond, so heat of combustion is less.

Q. The heat of transition (

Δ H_{t}

) of graphite into diamond would be,
where,

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)

Δ H = x k J

C (d i a m o n d) + O_{2} (g) \to C O_{2} (g)

Δ H = y k J

Q. One gram mole of graphite and diamond were burnt to form

{C O}_{2}

gas.

C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 399.5 k J

C_{d i a m o n d} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 395.4 k J

then, which of the following is the correct about diamond and graphite?

Q. Given that:

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 393.7 k J

Calculate the quantity of graphite (in gm) that must be burnt to evolve

5000 k J

of heat.

Q. Calculate the

Δ H

in joules for:

C (g r a p h i t e) \to C (d i a m o n d)

from the data:

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 393.5 k J

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 395.4 k J

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Cgraphite + O2(g) → CO2(g) ΔH = −94.05kcal mol−1 Cdiamond + O2(g) → CO2(g); ΔH = −94.50 kcal mol−1 therefore

$C_{g r a p h i t e} + O_{2} (g) \to C O_{2} (g)$

$Δ H = - 94.05 k c a l m o l^{- 1}$

$C_{d i a m o n d} + O_{2} (g) \to C O_{2} (g); Δ H = - 94.50 k c a l m o l^{- 1}$ therefore