The following reaction takes place at 298 K and 1 atm: CO(g) + CH_3OH(l) rightar
ID: 897815 • Letter: T
Question
The following reaction takes place at 298 K and 1 atm: CO(g) + CH_3OH(l) rightarrow CH_3COOH(l) Use the thermodynamic data table posted on D2L to calculate Delta H Degree_r for the reaction. Use the gas law to calculate Delta U degree_r assuming all gas are perfect. Use the D2L table to calculate Delta Sdegree_r, for the reaction. Use the results of question 1 and 3 to calculate Delta G degree_r for the reaction. Use the D2L table to confirm the result of question 4. Is the reaction spontaneous at 298 K? Would the reaction be more spontaneous or less spontaneous if the temperature was raised? Explain your answer. What temperature should be reached to prevent this reaction from proceeding?Explanation / Answer
1. dHorxn = dHo(product) - dHo(reactants)
= dHo(CH3COOH) - [dHo(CO) + dHo(CH3OH)]
= -276.981 - (-110.525 - 238.66)
= 72.204 kJ/mol
2. dUorxn = dHorxn - (dn)RT
= dHorxn - (1-2)RT
= 72.204 - (-1) x 8.314 x 298
= 2549.78 kJ/mol
3. dSorxn = dSo(product) - dSo(reactants)
= dSo(CH3COOH) - [dSo(CO) + dSo(CH3OH)]
= 160.666 - (197.674 + 126.8)
= -163.808 J/K.mol
4. dGorxn = dHorxn - TdSorxn
= 72.204 - 298 x (-0.163808)
= 121.02 kJ/mol
5. dGorxn = dGo(product) - dGo(reactants)
= dGo(CH3COOH) - [dGo(CO) + dGo(CH3OH)]
= -173.991 - (-137.168 - 166.27)
= 129.45 kJ/mol
The reaction is non-spntaneous at 298 K. The sign of dGorxn is +ve.
6. For an endothermic reaction heat is absorbed dueing the reaction, so If more heat is supplied in the form of temperature, the reaction would be favoured for the product formation. Thus, here as the reaction is endothermic higher temperature would favor the product formation and the reaction would be more spontaneous.
7. dGorxn = 0 = dHorxn - TdSorxn
= 72.204 - T x (-0.163808)
T = -440.8 K
For the reaction to stop, it must be at equilibrium so dGorxn must be equal to zero. The remperature should be -440.8 K.
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