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)

+ 1.9 kJ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 1.9 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.8 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 3.8 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$C (d i a m o n d) \to C (g r a p h i t e); Δ H = + 1.9 k J$

Suggest Corrections

Similar questions

m o l^{- 1}

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

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

- 393.5

- 395.4 k J / m o l

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

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

k J / m o l

m o l^{- 1}

m o l^{- 1}

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

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

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

Join BYJU'S Learning Program