The value of - Htransition of C graphite- C diamond is 1-9 kJ-mol at 25-C- Entropy of graphite is higher than entropy of diamond- This implies that-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Enthalpy

The value of ...

Question

The value of $Δ H_{t r a n s i t i o n}$ of $C (g r a p h i t e) ⟶ C (d i a m o n d)$ is $1.9 k J / m o l$ at $25^{\circ} C$ . Entropy of graphite is higher than entropy of diamond. This implies that:

C (d i a m o n d)

is more thermodynamically stable than

C (g r a p h i t e)

25^{\circ} C

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

C (g r a p h i t e)

is more thermodynamically stable than

C (d i a m o n d)

25^{\circ} C

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

Diamond will provide more heat on complete combustion at

25^{\circ} C

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

Δ G_{t r a n s i t i o n}

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

- v e

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

Open in App

Solution

The correct options are
B $C (g r a p h i t e)$ is more thermodynamically stable than $C (d i a m o n d)$ at $25^{\circ} C$
C Diamond will provide more heat on complete combustion at $25^{\circ} C$
D $Δ G_{t r a n s i t i o n}$ of $C (d i a m o n d) ⟶ C (g r a p h i t e)$ is $- v e$
C (graphite) is more thermodynamically stable than C (graphite).
For the reverse process, $Δ H < 0$ , Also $Δ S$ changes very negligible due to this transition.
Hence, $Δ G$ is <0 for the transition from C(diamond) to C (graphite)
More heat is evolved due to combustion of less stable form.
Hence, options B, C, D are the appropriate answers.

Suggest Corrections

Similar questions

Q. The value of

{Δ H}_{t r a n s i t i o n}

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

1.9 k J / m o l

25^{o} C

and entropy of graphite is higher than entropy of diamond. This implies that:

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.

Q. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 K J m o l^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

. If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information:

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a

Q. If

C_{d i a m o n d} \to C_{g r a p h i t e}; Δ H = - v e .

This implies:

Q. The standard state Gibbs free energies of formation of

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

and

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

T = 298 K

are

Δ_{f} G^{\circ} C_{(g r a p h i t e)} = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} C_{(d i a m o n d)} = 2.9 k J m o l^{- 1}

The standard state means that the pressure should be

1 b a r

, and substance should be pure at a given temperature. The conversion of graphite

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

to diamond

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

reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

. If

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

is converted to

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

isothermally at

T = 298 K

, the pressure at which

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

is in equilibrium with

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

, is?

[Useful information:

1 J = 1 k g m^{2} s^{- 2}; 1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a]

Join BYJU'S Learning Program

The value of ΔHtransition of C(graphite)⟶C(diamond) is 1.9kJ/mol at 25∘C. Entropy of graphite is higher than entropy of diamond. This implies that:

The value of $Δ H_{t r a n s i t i o n}$ of $C (g r a p h i t e) ⟶ C (d i a m o n d)$ is $1.9 k J / m o l$ at $25^{\circ} C$ . Entropy of graphite is higher than entropy of diamond. This implies that: