$$\ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \hskip{5em} \Delta H^\circ = \pu{-196 kJ/mol_{rxn}}$$

The decomposition of $\ce{H2O2(aq)}$ is represented by the equation above. Assume that the bond enthalpies of the oxygen-hydrogen bonds in $\ce{H2O}$ are not significantly different from those in $\ce{H2O2}$. Based on the value of $\Delta H^\circ$ of the reaction, which of the following could be the bond enthalpies (in kJ/mol) for the bonds broken and formed in the reaction?

$$\begin{array}{l c c c} & \ce{O-O} & \ce{O=O} & \ce{O-H} \\ & \text{in } \ce{H2O2} & \text{in } \ce{O2} & \\ \hline A & 300 & 500 & 500 \\ \hline B & 150 & 500 & 500 \\ \hline C & 500 & 300 & 150 \\ \hline D & 250 & 300 & 150 \\ \hline \end{array}$$

Summary Submit Skip Question

The enthalpy change can be estimated by taking the sum of the energy required to break bonds and the sum of the energy required to form bonds:

$$\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$$

Energy is released when bonds are formed, hence the negative sign.

Write the lewis structures for the reactants and products. We can also infer the types of bonds from the table.

Our equation should look something like this:

$$\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$$ $$\Delta H = \underbrace{2[(\ce{O-O})+2(\ce{O-H})]}_{\text{hydrogen peroxide}} - \underbrace{2[2(\ce{O-H})]}_{\text{water}} - \underbrace{(\ce{O=O})}_{\text{oxygen}}$$ $$= 2(\ce{O-O})+4(\ce{O-H}) -4(\ce{O-H}) - (\ce{O=O})$$ $$= 2(\ce{O-O}) - (\ce{O=O})$$

Using the values from the table:

$$\begin{array}{l c c c | c} & \ce{O-O} & \ce{O=O} & \ce{O-H} & 2(\ce{O-O}) - (\ce{O=O}) \\ & \text{in } \ce{H2O2} & \text{in } \ce{O2} & \\ \hline A & 300 & 500 & 500 & 2(300)-500 =100 \\ \hline B & 150 & 500 & 500 & 2(150)-500=\boxed{-200} \\ \hline C & 500 & 300 & 150 & 2(500)-150=850\\ \hline D & 250 & 300 & 150 & 2(250)-150=350 \\ \hline \end{array}$$