Question
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
Answer
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(271 Votes)
Irving
Elite · Tutor for 8 years
Answer
Entropy change for formation of ethanol is -716058 KJ/mol.
Explanation
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 (reagents 0 -> products):C(s) + H2(g) -> C2H5OH(l)Due to the presences of respective quantity in equation, we will need to create a balanced Eq: 2C(s) + 3H2(g) -> C2H5OH(l)Now as the enthalpy change during given equation is:ΔHº = Total enthalpy of products - Total enthalpy of reactants Therefore:ΔHº = [1 * Hº (C2H5OH(l)] - [ 2*Hº (C(s)) + 3*Hº (H2(g))]Put given values of Hº:{\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})}ΔHº = [-1367 KJ/mol] - [(2*-394 KJ/mol)+(3*-286KJ/mol)]= -716058 KJ/mol