You visited us 2 times! Enjoying our articles? Unlock Full Access!

Gibbs Free Energy & Spontaneity

If the enthal...

Question

If the enthalpy of combustion of diamond and graphite are $- 395.4 k J m o l^{- 1}$ and $- 393.6 k J m o l^{- 1}$ ,
The enthalpy change for the $C (g r a p h i t e) ⟶ C (d i a m o n d)$ is :

Δ H = 1.8 k J m o l^{- 1}

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

Δ H = 2.8 k J m o l^{- 1}

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

Δ H = 5.8 k J m o l^{- 1}

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

Δ H = 6.8 k J m o l^{- 1}

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

Open in App

Solution

The correct option is A $Δ H = 1.8 k J m o l^{- 1}$
$C (d i a m o n d) + O_{2} (g) ⟶ C O_{2} (g)$ ..... $Δ H = - 395.4 k J m o l^{- 1} . . . (i)$
$C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g)$ $Δ H = - 393.6 k J m o l^{- 1} . . . (i i)$

Subtracting equation (i) from equation (ii), we get-

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

$Δ H = - 393.6 k J - (- 395.4 k J)$

$Δ H = + 1.8 k J m o l^{- 1}$ .

Suggest Corrections

Similar questions

C (g r a p h i t e) \to C (d i a m o n d), δ H = 1.9 k J .

C O_{2},

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

$Δ H = - 393.5 k . J$

$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 395.4 k . J$

Reason(R): graphite is more stable than diamond, so heat of combustion is less.

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}

m o l^{- 1}

B_{2} H_{6 (g)} + 3 O_{2 (g)} \to B_{2} O_{3 (s)} + 3 H_{2} O (g)

2 B (s) + (3 / 2) O_{2 (g)} \to B_{2} O_{3 (s)}

Δ H = - 1273 k J m o l^{- 1}

H_{2 (g)} + (1 / 2) O_{2 (g)} \to H_{2} O (1)

Δ H = - 286 k J m o l^{- 1}

H_{2} O_{(} 1) \to H_{2} O_{(} g)

Δ H = 44 k J m o l^{- 1}

2 B (s) + 3 H_{2 (g)} \to B_{2} H_{6 (g)}

Δ H = 36 K J m o l^{- 1}

Join BYJU'S Learning Program