• question_answer 13) At 60C, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

The given reaction is:   ${{N}_{2}}{{O}_{4}}(g)$ $\rightleftharpoons$ $2N{{O}_{2}}(g)$ ${{t}_{initial}}$ $1$   $0$ ${{t}_{equilibrium}}$ $1-0.5$   $2\times 0.5$     Total mole = $0.5+1=\,1.5$ Mole fraction $\frac{0.5}{1.5}$   $\frac{1}{1.5}$ Partial pressure $\frac{0.5}{1.5}\times 1$   $\frac{1}{1.5}\times 1$   (Partial pressure of gas = Mole fraction of gas $\times$Total pressure) ${{K}_{p}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{{{p}_{{{N}_{2}}{{O}_{5}}}}}=\frac{{{\left( \frac{1}{1.5} \right)}^{2}}}{\left( \frac{0.5}{1.5} \right)}$ $=\frac{1}{0.5\times 1.5}=1.33$ $\Delta {{G}^{{}^\circ }}=-2.303RT{{\log }_{10}}{{K}_{p}}$ $=-2.303\times 8.314\times 333\log 1.33$ $=-789.67kJ\,mo{{l}^{-1}}$