At 227^{circ} C,dinitrogen tetraoxide is 60 % | vedclass

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(D) The dissociation of dinitrogen tetraoxide is given by: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ Given: Temperature $T = 227 + 273 = 500 \ K$ De

Vedclass · Accepted Answer

$-806 \ cal / mol$

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(D) The dissociation of dinitrogen tetraoxide is given by: $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$ Given: Temperature $T = 227 + 273 = 500 \ K$ Degree of dissociation $\alpha = 60 \% = 0.6$ Total pressure $P = 1 \ atm$ The equilibrium constant $K_p$ for this reaction is: $K_p = \frac{4\alpha^2 P}{1-\alpha^2}$ $K_p = \frac{4 	imes (0.6)^2 	imes 1}{1 - (0.6)^2} = \frac{4 	imes 0.36}{1 - 0.36} = \frac{1.44}{0.64} = 2.25$ The standard free energy change $\Delta G^{\circ}$ is: $\Delta G^{\circ} = -RT \ln K_p = -2.303 RT \log K_p$ $\Delta G^{\circ} = -2.303 	imes 1.987 	imes 500 	imes \log(2.25)$ $\Delta G^{\circ} = -2.303 	imes 1.987 	imes 500 	imes 0.3522 \approx -806 \ cal / mol$.

At $227^{\circ} C$,dinitrogen tetraoxide is $60 \%$ dissociated. What is the standard free energy change at this temperature and at $1 \ atm$ pressure?

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