STD 11 - 8.2 Organic chemistry isomerism — JEE STD 11 Science — Question

$A$ $\&$ $B$ are :

$\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = k\left[ {{N_2}{O_5}} \right]$ , 

$\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = k'\left[ {{N_2}{O_5}} \right]$,

$\frac{{d\left[ {{O_2}} \right]}}{{dt}} = k''\left[ {{N_2}{O_5}} \right]$

The relationship between $k$ and $k'$ and between $k$ and $k^{"}$ are

$\Delta_f G^0[\mathrm{C}(\text { graphite })]=0 \mathrm{kJmol}^{-1}$

$\Delta_f G^0[\mathrm{C}(\text { diamond })]=2.9 \mathrm{kJmol}^{-1}$

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 $2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mol}^{-1}$. If $\mathrm{C}$ (graphite) is converted to $\mathrm{C}$ (diamond) isothermally at $\mathrm{T}=298 \mathrm{~K}$, the pressure at which $C$ (graphite) is in equilibrium with C(diamond), is

[Useful information: $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^5 \mathrm{~Pa}$ ]