Calculate the $Δ H$ for conversion of C (diamond) to C (graphite) when the following reactions are given:
$C (d i a m o n d) + O_{2} \to C O_{2} (g); Δ H = - 94.5 K c a l$
$C (g r a p h i t e) + O_{2} \to C O_{2} (g); Δ H = - 94 K c a l$

-188.5 K.cal

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+188.5 K.cal

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+0.5 K.cal

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-0.5 K.cal

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Solution

The correct option is D -0.5 K.cal
i) C (diamond) + $O_{2} (g) \to C O_{2} (g)$ , $Δ H = - 94.5$ kcal
C (graphite) + $O_{2} (g) \to C O_{2} (g)$ ), $Δ H = - 94$ kcal
ii) $C O_{2} (g) \to$ C (graphite) + $O_{2} (g)$ , $Δ H = 94$ kcal
Adding (i) & (ii)
C (diamond) $\to$ C (graphite) $Δ H = - 0.5$ kcal

Option D is correct.

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Similar questions

Calculate $Δ H$ for $C_{(g r a p h i t e)} \to C_{d i a m o n d}$ from the following data:

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

ii) $C (d i a m o n d) + O_{2} \to C O_{2} (g), Δ H = - 68 K c a l$

+ 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_{D i a m o n d} + O_{2} \to C O_{2} (g); Δ H = - 94.3 k c a l

C_{G r a p h i t e} + O_{2} \to C O_{2} (g); Δ H = - 97.6 k c a l

C_{d i a m o n d} \to C_{g r a p h i t e}

For the following reaction

$C_{d i a m o n d} + O_{2} \to C O_{2} (g)$ ; $Δ H = - 94.3 k c a l$

$C_{g r a p h i t e} + O_{2} \to C O_{2} (g)$ ; $Δ H = - 97.6 k c a l$

The heat required to change 1 g of $C_{d i a m o n d} \to C_{g r a p h i t e}$ is

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

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

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