Question

Estimate the ΔH (kcal/mol) for the reaction of hydrogen and oxygen to form water: 2 H2 + O2 2 H2O ΔH = ?

Answer 1

To estimate the deltaH for the reaction we can do it from the bond energies, that is, starting from the energy needed to form or break a bond.

The reaction energy will be equal to the binding energy of the reactants minus the binding energy of the products. To calculate each bond energy we must first identify which bonds are involved in the reaction. They give us a balanced reaction:

`2H_2+O_2\rightarrow2H_2O`

For the reaction, we have the following bonds

`\begin{gathered} 2\lbrack H-H\rbrack+1\lbrack O=O\rbrack\rightarrow2\lbrack H-O-H\rbrack \\ 2\lbrack H-H\rbrack+1\lbrack O=O\rbrack\rightarrow2*2\lbrack O-H\rbrack \end{gathered}`

Now we are going to calculate the energy taking into account the values that we can find in tables of the bond energy.

Bond Ee (kJ/mol)

H - H 436

O=O 499

O - H 460

Therefore the energy on each side of the reaction will be:

`\begin{gathered} 2\lbrack H-H\rbrack+1\lbrack O=O\rbrack\rightarrow2*2\lbrack O-H\rbrack \\ 2*436(kJ)/(mol)+1*499(kJ)/(mol)\rightarrow2*2*460(kJ)/(mol) \\ 1371(kJ)/(mol)\rightarrow1840(kJ)/(mol) \end{gathered}`

The reaction energy will be:

`\begin{gathered} \Delta H_r=\Delta H_(Reac\tan ts)-\Delta H_(Products) \\ \Delta H_r=1371(kJ)/(mol)-1840(kJ)/(mol)=-469(kJ)/(mol) \end{gathered}`

We have an exothermic reaction since the result is negative. This means that the energy of the products is greater than that of the reactants.

Now we will convert the energy units to kcal:

`\Delta H_r=-469(kJ)/(mol)*\frac{1\text{kcal}}{4.184kJ}=-112\frac{kcal}{\text{mol}}`

ΔH (kcal/mol) estimated of the reaction will be -112kcal/mol