At $$57$$°C, gaseous dinitrogen tetroxide is $$50$$ % dissociated. Calculate the standard free energy change per mole of $$N2O4$$ (g) at this temperature and at $$1$$ atm.(R=8.3JK-1$$ $$mol-1$$ , In $$10=2.3$$,$$log2=0.3$$,log $$3=0.48)

Question

At $$57$$°C, gaseous dinitrogen tetroxide is $$50$$ % dissociated. Calculate the standard free energy change per mole of $$N2O4$$ (g) at this temperature and at $$1$$ atm.(R=8.3JK-1$$ $$mol-1$$ , In $$10=2.3$$,$$log2=0.3$$,log $$3=0.48)

Infinity Learn · Accepted Answer

Given Temperature $$T1=570C+2730C=330KGaseous$$ dinitrogen $$tetroxide=50$$ % $$dissociatedR=8.3JK$$−$$1mol$$−$$1$$ ln $$10=2.3$$, $$log2=0.3$$, $$log3=0.48Standard$$ free energy, Δ$$G0=$$?The Reaction is:                                 $$N2O4g$$⇔$$2NO2(g)Initial$$  (t$$ $$=$$ $$1)                     $$1$$                    $$0$$ At equilibrium (t$$ $$=$$ $$eq)    $$1-a$$                $$2aTotal$$ moles   $$=1$$−α$$+2$$α$$=1+$$αa $$=$$ $$0.5$$  and P $$=$$ $$1$$ atm∴P  $$N2O4=1$$−α$$1+$$α P $$T=1$$−$$0.51+0.5$$ ×$$1=0.51.5$$ atm∴P $$NO2=2$$α$$1+$$α P $$T=2$$×$$0.51+0.5$$ ×$$1=11.5$$ atm∴$$KP=P$$ $$NO22P$$ $$N2O4=11.520.51.5=43Gibbs$$ energy formula, ∴ Δ$$G0=$$−$$2.303$$ nRT log $$KP=$$−$$2.303$$×$$1$$×$$8.3$$×$$330$$×log $$43=$$−$$2.303$$×$$8.3$$×$$330$$ $$0.6$$−$$0.48=$$−$$756$$ J mol−$$1$$

At 57°C, gaseous dinitrogen tetroxide is 50 % dissociated. Calculate the standard free energy change per mole of N2O4 (g) at this temperature and at 1 atm.(R=8.3JK-1 mol-1 , In 10=2.3,log2=0.3,log 3=0.48)

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