question_answer 13)

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

Answer:

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) \[{{K}_{p}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{{{p}_{{{N}_{2}}{{O}_{5}}}}}=\frac{{{\left( \frac{1}{1.5} \right)}^{2}}}{\left( \frac{0.5}{1.5} \right)}\] \[=\frac{1}{0.5\times 1.5}=1.33\] \[\Delta {{G}^{{}^\circ }}=-2.303RT{{\log }_{10}}{{K}_{p}}\] \[=-2.303\times 8.314\times 333\log 1.33\] \[=-789.67kJ\,mo{{l}^{-1}}\]