JEE Advance - Chemistry (2017 - Paper 2 Offline - No. 18)
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 $$ = {10^5}$$ $$Pa$$]
$${\Delta _f}{G^0}$$ [$$C$$(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 $$ = {10^5}$$ $$Pa$$]
$$14501$$ bar
$$58001$$ $$bar$$
$$1450$$ bar
$$29001$$ bar
Explanation
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
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