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Enthalpy of Neutralisation

The enthalpy ...

Question

The enthalpy of transition of C(diamond) to C(graphite) is (Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ $m o l^{- 1}$ respectively)

A

+ 1.9 kJ

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B

- 1.9 kJ

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C

3.8 kJ

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D

- 3.8 kJ

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Solution

The correct option is A + 1.9 kJ
Our aim is:
$C (d i a m o n d) \to C (g r a p h i t e); Δ H = ?$
The available data is :
(i) $C (d i a m o n d) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J$
(ii) $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ H = - 395.4 k J$
Subtracting equation (ii) from equation (i)
$S u b t r a c t C (d i a m o n d) - C (g r a p h i t e) \to O; Δ H = (- 393.5 + 395.4)$
Or
$C (d i a m o n d) \to C (g r a p h i t e); Δ H = + 1.9 k J$

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Similar questions

Q. The enthalpy of transition of C(diamond) to C(graphite) is?
Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ

m o l^{- 1}

Q. The enthalpy of transition of C(diamond) to C(graphite) is?
Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ

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Q. The enthalpies of combustion of

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

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- 395.4 k J / m o l

respectively. The enthalpy of conversion of

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

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- 393.6 k J m o l^{- 1}

,
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Q. Calculate the enthalpy change accompaying the conversion of 10 grams of graphite into diamond if the standard enthalpy of combustion of

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