The standard state Gibb's free energies of formation of C (graphite) and C (diamond) at T = 298   K are Δ f G o C   ( graphite = 0   kJ   mol - 1 Δ f G o C   diamond = 2 .9   kJ   mol - 1 The standard state means that the pressure should be 1 bar,…

Question

The standard state Gibb's free energies of formation of

C

(graphite) and

C

(diamond) at

T = 298 K

are

Δ_{f} G^{o} [C (graphite] = 0 kJ {mol}^{- 1}

Δ_{f} G^{o} [C (diamond)] = 2 .9 kJ {mol}^{- 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} {mol}^{- 1}

. If

C

(graphite) is converted to

C

(diamond) isothermally at

T = 298 K

, the pressure at which

C

(graphite) is in equilibrium with

C

(diamond), is
[Useful information:

1 J = 1 kg m^{2} s^{- 2}, 1 Pa = 1 kg m^{- 1} s^{- 2}; 1 bar = 10^{5} Pa

]

Aicharya · Accepted Answer

C graphite &#8594; C diamond ; &#160; &#916;G o = &#916; f G diamond o - &#916; f G graphite o = 2 .9 &#160; kJ / mol at 1 bar As dG T = V . dP &#8747; &#916;G 1 &#916;G 2 d &#916;G T = &#160; &#8747; P 1 P 2 &#916;V . dP &#916;G 2 - &#916;G 1 = &#916;V . &#160; P 2 - P 1 2 .9 &#215; 10 3 - 0 = - 2 &#215; 10 - 6 &#160; 1 - P 2 P 2 - 1 = 2 .9 &#215; 10 3 2 &#215; 10 - 6 Pa = 1 .45 &#215; 10 4 &#160; bar P 2 = 14501 &#160;&#160; bar .

The standard state Gibb's free energies of formation of C (graphite) and C (diamond) at T = 298   K…

Solution

The standard state Gibb&#39;s free energies of formation of C (graphite) and C (diamond) at T = 298 &#160; K…

Solution

The standard state Gibb's free energies of formation of C (graphite) and C (diamond) at T = 298 K…