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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

The densities...

Question

The densities of graphite and diamond at $298 K$ are $2.25$ and $3.31$ $g {c m}^{- 3}$ respectively. If the standard free energy difference is $1895 J {m o l}^{- 1}$ , the pressure at which graphite will be transformed into diamond is:

9.92 \times 10^{8} P a

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9.92 \times 10^{7} P a

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9.92 \times 10^{6} P a

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None of these

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Solution

The correct option is A None of these
$Δ G = W_{c o m p r e s s} = - p Δ V$

$1895 = - p [(\frac{12}{3.31} - \frac{12}{2.25}) \times 10^{- 6} m^{3}]$

$1895 = - p (3.62 - 5.33) \times 10^{- 6}$

$p = \frac{1895 \times 10^{6}}{1.71} P a = 11.02 \times 10^{8} P a$

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Similar questions

Q. The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^-3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is

A. 9.92 × 10⁵

B. 9.92 × 10^{8

C. 9.92 × 10^7}

D. 9.92 × 10^{6

All options' unit is Pascal
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What should be the answer of this question according to you and why?
Please explain in detail.}

Q. The densities of graphite and diamond at

298 K

are

2.25

and

3.31 g c m^{- 3}

, respectively, If the standard free energy difference

(Δ G^{0})

is equal to

1898 J m o l^{- 1}

, the pressure at which graphite will be transformed diamond at

298 K

is:

Q. The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm

^{- 3}

, respectively. If the standard free energy difference

(Δ G^{o})

is equal to 1895 J mol

^{- 1}

, the pressure at which graphite will be transformed into diamond at 298 K is:

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. Densities of diamond and graphite are

3.5

and

2.3

g / m L

respectively. Increase of pressure on the equilibrium:

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

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The densities of graphite and diamond at 298K are 2.25 and 3.31 g cm−3 respectively. If the standard free energy difference is 1895 J mol−1, the pressure at which graphite will be transformed into diamond is:

The correct option is A None of theseΔG=Wcompress=−pΔV1895=−p[(123.31−122.25)×10−6m3]1895=−p(3.62−5.33)×10−6p=1895×1061.71Pa=11.02×108Pa

The densities of graphite and diamond at $298 K$ are $2.25$ and $3.31$ $g {c m}^{- 3}$ respectively. If the standard free energy difference is $1895 J {m o l}^{- 1}$ , the pressure at which graphite will be transformed into diamond is:

The correct option is A None of these
$Δ G = W_{c o m p r e s s} = - p Δ V$

$1895 = - p [(\frac{12}{3.31} - \frac{12}{2.25}) \times 10^{- 6} m^{3}]$

$1895 = - p (3.62 - 5.33) \times 10^{- 6}$

$p = \frac{1895 \times 10^{6}}{1.71} P a = 11.02 \times 10^{8} P a$