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 $$ = {10^5}$$ $$Pa$$]

For the following cell,

$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$

when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]

The reaction of compound $$P$$ with $$C{H_3}MgBr$$ (excess) in $${\left( {{C_2}{H_5}} \right)_2}O$$ followed by addition of $${H_2}O$$ gives $$Q.$$ The compound $$Q$$ on treatment with $${H_2}S{O_4}$$ at $${0^ \circ }C$$ gives $$R.$$ The reaction of $$R$$ with $$C{H_3}COCl$$ in the presence of anhydrous $$AlC{l_3}$$ in $$C{H_2}C{l_2}$$ followed by treatment with $${H_2}O$$ producess compound $$S.$$ [$$Et$$ in compound $$P$$ is ethyl group]

The reactions, $$Q$$ to $$R$$ and $$R$$ to $$S,$$ are

If f : R $$ \to $$ R is a twice differentiable function such that f"(x) > 0 for all x$$ \in $$R, and $$f\left( {{1 \over 2}} \right) = {1 \over 2}$$, f(1) = 1, then