Question

$\frac{3}{2}{O}_2\left(g\right)⟶O_3\left(g\right)$ $K_p$ for this reaction is $2.47\times {10}^{-29}$. At 298 K, ${\Delta }_r{G}^o$ for conversion of oxygen to ozone will be:

• A. 100 kJ mol${}^{-1}$
• B. 150 kJ mol${}^{-1}$
• C. 163 kJ mol${}^{-1}$
• D. 2303 kJ mol${}^{-1}$

Answer 1

The correct option is C 163 kJ mol${}^{-1}$

For the reaction,

$\frac{3}{2}{O}_2\left(g\right)⟶O_3\left(g\right)$

The free energy is calculated as

${\Delta }_r{G}^0=-RT\ln K_p$

Where T=298K

R=8.314

$K_p=$ equilibrium constant = $2.47\times {10}^{-29}$

${\Delta }_r{G}^0=-8.314\times 298\times \ln (2.47\times {10}^{-29})⟹{\Delta }_r{G}^0=163228.93J=163.2KJ$