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Heat of Reaction

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Question

Calculate the enthalpy change accompaning the conversion of $10 g$ of graphite into diamond if the heats of combustion of $C (g r a p h i t e)$ and $C (d i a m o n d)$ are $- 94.05 k c a l$ and $- 94.50 k c a l$ respectively.

A

4.5 k c a l

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B

0.45 k c a l

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C

0.375 k c a l

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D

3.75 k c a l

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Solution

The correct option is C $0.375 k c a l$

Given that:
(i) $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g) Δ H_{1} = - 94.05 k J m o l^{- 1}$

(ii) $C (d i a m o n d) + O_{2} (g) \to C O_{2} (g) Δ H_{2} = - 94.50 k J m o l^{- 1}$

now, [eqn. (i) - eqn. (ii)] we get :

$C (g r a p h i t e) + O_{2} (g) - C (d i a m o n d) - O_{2} (g) \to C O_{2} (g) - C O_{2} (g)$
For this reaction $Δ H = Δ H_{1} - Δ H_{2}$

Simplifying the above equation we get,

$C (g r a p h i t e) \to C (d i a m o n d)$
Also putting values of $Δ H_{1} a n d Δ H_{2}$
$Δ H = - 94.05 - (- 94.50) = 0.45 k c a l$

Since this enthalpy change is only for conversion of 1 mole, i.e. , $12 g$ of $C (g r a p h i t e)$ to $C (d i a m o n d)$ , therefore, for the conversion of $10 g$ of $C (g r a p h i t e)$ to $C (d i a m o n d)$

$Δ H = 0.45 \times \frac{10}{12} = 0.375 k c a l$

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Heat of Reaction

Standard XIII Chemistry

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