Question

$X_{2}O_{\left(\right. s \left.\right)}\rightleftharpoons2X_{\left(\right. s \left.\right)}+\frac{1}{2}O_{2}\Delta \left(G^{o}\right)_{\left(\right. X_{2} O \left.\right)}=-1.72\,kJ\left(mol\right)^{- 1}$
For the above reaction, what is the value of equilibrium pressure of $O_{2}$ at $25^{o}C$ in $Nm^{- 2}$ ?

Answer 1

$\left(ΔG\right)^{o}=-2.303 \, RTlog P_{O_{2} \left(at \, equil\right)}^{\frac{1}{2}}$
$-1720J=-2.303\times 8.314J\times 298log P_{O_{2}}^{\frac{1}{2}}$
$log P_{O_{2}}^{\frac{1}{2}}=0.3\Rightarrow \, P^{\frac{1}{2}}=2Nm^{- 2}$
$\therefore \, P=4 \, Nm^{- 2}$