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Standard XII

Chemistry

Introduction to Carbon Family

Using special...

Question

Using special techniques a high temperature and pressure with a suitable catalyst like tantalum, graphite can be converted into:

synthetic coal

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charcoal

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lead

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synthetic diamonds

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Solution

The correct option is D synthetic diamonds
A synthetic diamond (also known as a laboratory-grown diamond, cultured diamond, or cultivated diamond) is a diamond produced by a controlled process, as contrasted with a natural diamond created by geological processes or an imitation diamond made of non-diamond material that appears similar to a diamond.
They are prepared by using special techniques such as using high temperature and pressure with a suitable catalyst like tantalum, graphite can be converted into synthetic diamonds.

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Similar questions

We know that under high pressure, graphite can be converted to diamond. At room temperature, both are solids.

So,
A) Is the state same?
B) Is the phase same?

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)

at 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 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 (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

]

Q. Assertion :Graphite cannot be converted into diamond. Reason: Graphite is conducting in nature.

Q. Under suitable condition temperature,pressure and catalyst,methanol can be obtain by:

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)

at 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 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 (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

]

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