2O_{3(g)} rightleftharpoons 3O_{2(g)}At 300 | vedclass

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(C) For the reaction $2O_{3(g)} \rightleftharpoons 3O_{2(g)}$,let the initial moles of $O_3$ be $1$.At equilibrium,moles of $O_3 = 1 - 0.5 = 0.5$ and

Vedclass · Accepted Answer

$747$

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(C) For the reaction $2O_{3(g)} ightleftharpoons 3O_{2(g)}$,let the initial moles of $O_3$ be $1$.At equilibrium,moles of $O_3 = 1 - 0.5 = 0.5$ and moles of $O_2 = \frac{3}{2} 	imes 0.5 = 0.75$.Total moles = $0.5 + 0.75 = 1.25$.Mole fraction of $O_3 = \frac{0.5}{1.25} = 0.4$ and $O_2 = \frac{0.75}{1.25} = 0.6$.Partial pressures at $P = 1 \ atm$: $P_{O_3} = 0.4 \ atm$,$P_{O_2} = 0.6 \ atm$.$K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} = \frac{(0.6)^3}{(0.4)^2} = \frac{0.216}{0.16} = 1.35$.$\Delta G^{\circ} = -RT \ln K_p$.$\Delta G^{\circ} = -8.3 	imes 300 	imes \ln 1.35$.$\Delta G^{\circ} = -8.3 	imes 300 	imes 0.3 = -747 \ J \ mol^{-1}$.

$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).
[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

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$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]