C diamond - C graphite- - S300-10 cal mol-1 C diamond -O2- CO2- - H-91 kcal mol-1 at 300K C graphite -O2- CO2- - H-x at 300K- x is-

The correct option is D 88 kcal C (diamond) → C (graphite), Δ S_300=10 cal mol^ 1 C(d)+ O_2→ CO_2–(1) Δ H = 91 kcal mol ...

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Bond Order and Enthalpy

C diamond →...

Question

$C$ (diamond) $\to$ $C$ (graphite), $Δ S_{300} = 10$ cal $m o l^{- 1}$
$C$ (diamond) $+ O_{2} \to C O_{2}$ ; $Δ H = - 91$ kcal $m o l^{- 1}$ at $300$ K
$C$ (graphite) $+ O_{2} \to C O_{2}$ ; $Δ H = x$ at $300$ K.
$x$ is:

- 81

kcal

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101

kcal

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- 94

kcal

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88

kcal

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Solution

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Similar questions

Δ H

C (d i a m o n d) + O_{2} \to C O_{2} (g); Δ H = - 94.5 K c a l

C (g r a p h i t e) + O_{2} \to C O_{2} (g); Δ H = - 94 K c a l

+ O_{2} \to C O_{2} (g); Δ H = - 94.3 k c a l

+ O_{2} \to C O_{2} (g); Δ H = - 97.6 k c a l

\to

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

$Δ H = - 94.05 k c a l m o l^{- 1}$

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

10 g

C (g r a p h i t e)

C (d i a m o n d)

- 94.05 k c a l

- 94.50 k c a l

C (s) + O_{2} (g) \to C O_{2}, (g);

Δ H = - 94.3

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g);

Δ H = - 67.4

O_{2} (g) \to 2 O (g);

Δ H = 117.4

C O (g) \to C (g) + O (g);

Δ H = 230.6

Δ H

C (s) \to C (g)

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