Please Answer ALL. Thanks. -->Methanol is manufactured by the partial combustion
ID: 532135 • Letter: P
Question
Please Answer ALL.
Thanks.
-->Methanol is manufactured by the partial combustion of methane under pressure, using a copper catalyst:
CH4(g) + ½ O2(g) CH3OH(l)
Enthalpies of combustion are (in kJ mol-1):
CH4(g), -883
CH3OH(l), -730
Calculate the standard enthalpy change for the manufacture of 1 mole methanol by the reaction
CH4(g) + ½ O2(g) CH3OH(l)
Follow the procedures based on Hess’s Law:
First write down the reactions corresponding to the enthalpies of combustion you have been given, reverse one of the equations remembering to change the sign of Ho, then combine these equations to give the required process, and evaluate the associated Ho
--> Standard enthalpies of formation are obtained from thermodynamic tables as:
C2H5OH(l) -225 kJ/mol
CO2(g) -391 kJ/mol
H2O(l) -283 kJ/mol.
Calculate the enthalpy change of the reaction
C2H5OH(l) + 3 O2 2 CO2(g) + 3 H2O
Follow the procedures based on Hess's Law:
First write down the reactions corresponding to the enthalpies of formation you have been given, reverse the equations if necessary, remembering to change the sign of Ho, then combine the equations to give the required process, and evaluate the associated Ho
Explanation / Answer
The enthalpy change for a chemical reaction is the difference
H = Hproducts – Hreactants
If the reaction in question represents the formation of one mole of the compound from its elements in their standard states, as in
H2(g) + ½ O2(g) H2O(l) H = –286 kJ
then we can arbitrarily set the enthalpy of the elements to zero and write
Hf ° = Hf °products – Hf °reactants= –286 kJ – 0 = –268 kJ mol–1
which defines the standard enthalpy of formation of water at 298K.
The value Hf ° = –268 kJ tells us that when hydrogen and oxygen, each at a pressure of 1 atm and at 298 K (25° C) react to form 1 mole of liquid water also at 25°C and 1 atm pressure, 268 kJ will have passed from the sytstem (the reaction mixture) into the surroundings. The negative sign indicates that the reaction is exothermic: the enthalpy of the product is smaller than that of the reactants.