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Internal Energy

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Question

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

A

58001 bar

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B

1450 bar

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C

14501 bar

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D

29001 bar

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Solution

The correct option is C 14501 bar
$G = H - T S = U + p V - T S$
$\Rightarrow d G = d U + p d V + V d p - T d S - S d T = V d p - S d T$
$[∵ d U + p d V = d q = T d S]$
$\Rightarrow d G = V d p$ if isothermal process (dT = 0)
$\Rightarrow Δ G = v Δ p$
Now taking initial state as standard state
$G_{g r} - G_{g r}^{\circ} = V_{g r} Δ p$ . . . . .(i)
$G_{d} - G_{d}^{\circ} = V_{d} Δ P$ . . . . . .(ii)
Now (iii) – (i) gives,
$(V_{d} - V_{g r}) Δ p = G_{d} - G_{g r} + (G_{g r}^{\circ} - g_{d}^{\circ})$
At equilibrium, $G_{d} - G_{g r}$
$\Rightarrow (V_{g r} - V_{d}) Δ p = G_{d}^{\circ} - G_{g r}^{\circ} = 2.9 \times 10^{3} J$
$\Rightarrow Δ p = \frac{2.9 \times 10^{3}}{2 \times 10^{- 6}} P a = \frac{29}{2} \times 10^{8} P a = \frac{29000}{2} b a r$
$p = p_{0} + \frac{29000}{2} = 1 + \frac{29000}{2} = 14501 b a r$

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Similar questions

Q. The standard state Gibbs free energies of formation of

C (g r a p h i t e)

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}

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. The standard state Gibbs free energies of formation of

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

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

T = 298 K

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)}

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

reduces its volume by

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

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)}

[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. The standard state Gibbs free energies of formation of

C (g r a p h i t e)

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}

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

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)}

Q. Calculate the enthalpy change accompaying the conversion of 10 grams of graphite into diamond if the standard enthalpy of combustion of

C_{g r a p h i t e}

C_{d i a m o n d}

are -94.05 and -94.50 kcal respectively.

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