One gram mole of graphite and diamond were burnt to form CO2 gas- C graphite - O 2 g- C O 2 g- H o -399-5kJ C diamond - O 2 g- C O 2 g- H o -395-4kJthen- which of the following is the correct about diamond and graphite-

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Allotropes of Carbon

One gram mole...

Question

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?

Graphite is more stable than diamond

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Diamond is more stable than graphite

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Graphite has greater affinity with oxygen

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Diamond has greater affinity with oxygen

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Solution

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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.

Δ H

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

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

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

$Δ H = - 94.05 k c a l m o l^{- 1}$

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

Δ H_{t}

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

25^{o} C

{F e}_{2} O_{3} (s) + 3 C_{g r a p h i t e} ⟶ 2 F e (s) + 3 C O (g); Δ H^{o} = 492 k J / m o l

F e O (s) + C_{g r a p h i t e} ⟶ F e (s) + C O (g); Δ H^{o} = 155.8 k J / m o l

C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 393.51 k J / m o l

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 282.98 k J / m o l

F e O (s)

{F e}_{2} O_{3} (s)

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