Given Delta _rS^o = -266 J K^{-1} and the l | vedclass

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(C) The reaction is: $4 Fe_3O_{4(s)} + O_{2(g)} 	o 6 Fe_2O_{3(s)}$Given $\Delta_rS^o = -266 \ J \ K^{-1}$$S_m^o(O_2) = 205 \ J \ K^{-1} \ mol^{-1}$$S

Vedclass · Accepted Answer

$145.75$

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(C) The reaction is: $4 Fe_3O_{4(s)} + O_{2(g)} 	o 6 Fe_2O_{3(s)}$Given $\Delta_rS^o = -266 \ J \ K^{-1}$$S_m^o(O_2) = 205 \ J \ K^{-1} \ mol^{-1}$$S_m^o(Fe_2O_3) = 87 \ J \ K^{-1} \ mol^{-1}$Using the formula: $\Delta_rS^o = \sum 
u_p S_m^o(products) - \sum 
u_r S_m^o(reactants)$$-266 = [6 	imes S_m^o(Fe_2O_3)] - [4 	imes S_m^o(Fe_3O_4) + 1 	imes S_m^o(O_2)]$$-266 = (6 	imes 87) - (4 	imes S_m^o(Fe_3O_4) + 205)$$-266 = 522 - 4 	imes S_m^o(Fe_3O_4) - 205$$4 	imes S_m^o(Fe_3O_4) = 522 - 205 + 266$$4 	imes S_m^o(Fe_3O_4) = 583$$S_m^o(Fe_3O_4) = \frac{583}{4} = 145.75 \ J \ K^{-1} \ mol^{-1}$

Given $\Delta _rS^o = -266 \ J \ K^{-1}$ and the listed $S_m^o$ values,calculate $S^o$ for $Fe_3O_{4(s)}$ for the reaction: $4 Fe_3O_{4(s)} + O_{2(g)} \to 6 Fe_2O_{3(s)}$. Given $S_m^o(Fe_2O_3) = 87 \ J \ K^{-1} \ mol^{-1}$ and $S_m^o(O_2) = 205 \ J \ K^{-1} \ mol^{-1}$.

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