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1. Use the data in the table to calculate the enthalpy change of formation of ethanol Substance & Delta_(mathrm(C)) H^circ / mathrm(kJ) mathrm(mol)^-1 mathrm(C)(mathrm(s)) & -394 mathrm(H)_(2)(mathrm(~g)) & -286 mathrm(C)_(4) mathrm(H)_(10)(mathrm(~g)) & -2877 mathrm(C)_(2) mathrm(H)_(5) mathrm(OH)(mathrm(l)) & -1367

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1. Use the data in the table to calculate the enthalpy change of formation of ethanol

 Substance & Delta_(mathrm(C)) H^circ / mathrm(kJ) mathrm(mol)^-1 
 mathrm(C)(mathrm(s)) & -394 
 mathrm(H)_(2)(mathrm(~g)) & -286 
 mathrm(C)_(4) mathrm(H)_(10)(mathrm(~g)) & -2877 
 mathrm(C)_(2) mathrm(H)_(5) mathrm(OH)(mathrm(l)) & -1367

1. Use the data in the table to calculate the enthalpy change of formation of ethanol Substance & Delta_(mathrm(C)) H^circ / mathrm(kJ) mathrm(mol)^-1 mathrm(C)(mathrm(s)) & -394 mathrm(H)_(2)(mathrm(~g)) & -286 mathrm(C)_(4) mathrm(H)_(10)(mathrm(~g)) & -2877 mathrm(C)_(2) mathrm(H)_(5) mathrm(OH)(mathrm(l)) & -1367

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IrvingElite · Tutor for 8 years

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Entropy change for formation of ethanol is -716058 KJ/mol.

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This problem requires calculating the ΔHº isn’t provided for ethanol (C2H5OH). Base on enthalpy stability rule, the ethanol formation reaction would be within following equation (<br />reagents 0 -> products):<br /><br />C(s) + H2(g) -> C2H5OH(l)<br /><br />Due to the presences of respective quantity in equation, we will need to create a balanced Eq: <br /><br />2C(s) + 3H2(g) -> C2H5OH(l)<br /><br />Now as the enthalpy change during given equation is:<br /><br />ΔHº = Total enthalpy of products - Total enthalpy of reactants <br /><br />Therefore:<br /><br />ΔHº = [1 * Hº (C2H5OH(l)] - [ 2*Hº (C(s)) + 3*Hº (H2(g))]<br /><br />Put given values of Hº:<br />{\displaystyle \Delta = \Delta_e^\circ[\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l})]-[\Delta_c^\circ\mathitC({s})+2\Delta_h^\circ[\mathitH_{\mathrm2(\mathrm{g})}<br />ΔHº = [-1367 KJ/mol] - [(2*-394 KJ/mol)+(3*-286KJ/mol)]= -716058 KJ/mol
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