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Enthalpy

At 500 kiloba...

Question

At $500$ kilobar pressure density of diamond and graphite are $3 g / c c$ and $2 g / c c$ respectively, at certain temperature 'T'. Find the value of $| Δ H - Δ U | (k J / m o l e)$ for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T'.

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Solution

C (graphite) $\to$ C (diamond)
$Δ H = Δ U + P_{2} V_{2} - P_{1} V_{1}$
$Δ H - Δ U = (500 \times 10^{3} \times 10^{5} N / m^{2}) (\frac{12}{2} - \frac{12}{3}) \times 10^{- 6}$ $= 500 \times 2 \times 10^{3} \times 10^{5} \times 10^{- 6}$
$= 100 k J / m o l e$

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500

kilobar pressure density of diamond and graphite are

3 g / c c

2 g / c c

respectively, at certain temperature 'T'. Find the value of

| Δ H - Δ U | (k J / m o l e)

for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T'.

At 500 Kbar pressure density of diamond and graphite 3 gm/cc and 2gm/cc respectively at a certain temperature T .{delta H-deltaU} in KJ/ mole for the conversion of one mole of graphite to one mole of diamond at 500 Kbar pressure is 10^X .The value of X is ?

[given 1 bar =10⁵N/m²]

ans =2 how ?

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

]

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