QJEEAdv17Ch8075: JEE Advanced 2017 Chemistry - Class 11 JEEAdv17Ch8075 Solution: The standard state Gibbs free energies of formation of…

Question

Verified Solution • 2025–26 Session moderate

The standard state Gibbs free energies of formation of

C

(graphite) and

C

(diamond) at

T=298

K

are

{\Delta _f}{G^0}

[

C

(graphite)]

= 0kJmo{l^{ - 1}}

{\Delta _f}{G^0}

[

C

(diamond)]

= 2.9kJmo{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}\,mo{l^{ - 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

]

Accepted Answer

We have

C(graphite) $\to$ C(diamond)

For solid at constant temperature, Gibbs energy can be calculated as

$dG = \Delta Vdp$

$\int\limits_{\Delta {G_1}}^{\Delta {G_2}} {d(\Delta {G_T}) = \int\limits_{{p_1}}^{{p_2}} {\Delta Vdp} }$

$\Delta {G_2} - \Delta {G_1} = \Delta V({p_2} - {p_1})$

$(2.9 \times {10^3} - 0) = 2 \times {10^{ - 6}}({p_2} - 1)$

$({p_2} - 1) = {{2.9 \times {{10}^3}} \over {2 \times {{10}^{ - 6}}}}$ Pa

$({p_2} - 1) = {{2.9 \times {{10}^3}} \over {2 \times {{10}^{ - 6}} \times {{10}^5}}}$ bar

p₂ = 14501 bar

Step-by-Step Verified Solution