# The enthalpy changes for the following reactions can be measured: CH4(g) + 2O2(g)  -->  CO2(g) + 2H2O(g)    deltaH=-802.4KJ/mol-rxn CH3OH(g) + 3/2O2(g) -->  CO2(g) + 2H2O(g)   ...

The enthalpy changes for the following reactions can be measured:

CH4(g) + 2O2(g)  -->  CO2(g) + 2H2O(g)    deltaH=-802.4KJ/mol-rxn

CH3OH(g) + 3/2O2(g) -->  CO2(g) + 2H2O(g)    deltaH=-676KJ/mol-rxn

Use these values and Hess's law to determine the enthalpy change for the reaction:

CH4(g) + 1/2O2(g) --> CH3OH(g)

Draw an energy level diagram that shows the relationship between the energy quantities involved in this problem.

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CH4(g) + 2O2(g)  -->  CO2(g) + 2H2O(g)    deltaH=-802.4 kJ/mol

CH3OH(g) + 3/2O2(g) -->  CO2(g) + 2H2O(g)   deltaH=-676 kJ/mol

You had left out the CO2 term in the first reaction so I added it in as shown above.  First we need to rearrange the reactions to cancel out the terms we don't want and be left with the terms we do want.  As you can see, we should reverse the second equation and then add the two reactions algebraically.  This cancels out the CO2 terms and the 2H2O terms.  Remember that reversing a chemical reaction changes the sign of the deltaH (enthalpy) for the reaction.

CH4(g) + 2O2(g)  -->  CO2(g) + 2H2O(g)    deltaH=-802.4 kJ/mol

CO2(g) + 2H2O(g) --> CH3OH(g) + 3/2O2(g)     deltaH=676 kJ/mol

CH4(g) + 1/2O2(g) --> CH3OH(g)                 deltaH=-126.4 kJ/mol

So the enthalpy change for the desired reaction is -126.4 kJ/mol.

In terms of drawing a free energy diagram for the reaction, the negative sign for the deltaH shows that the reaction is an exothermic one.  This means that the energy levels of the products are lower than that of the reactants.  This means that there is a net release of energy in the form of heat.  In the image below, the second diagram is the one that corresponds to this chemical reaction.  The first diagram is for an endothermic reaction.

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