Question

using the appropriate bond energies, calculate the heat of reaction \\( \delta h \\) for the following reaction:
\\( \begin{array} { r } { 2 mathrm { h } - mathrm { n } - mathrm { h } } \\\\ { 2 mathrm { h } - mathrm { n } - mathrm { h } } end{array} \
ightarrow 3 mathrm { h } - mathrm { h } + mathrm { n } \equiv mathrm { n } \\)
you can find a table of bond energies by using the data button on the aleks to
round your answer to the nearest \\( \mathrm { kj } / \mathrm { mol } \\).
note: for clarity, all lone pairs have been omitted from the molecular structure!

Explanation:

Step1: Calculate the energy required to break bonds

The reaction is \(2\ce{H - N - H}
ightarrow3\ce{H - H}+\ce{N\equiv N}\).
For the reactant \(2\ce{H - N - H}\), there are \(4\) \(N - H\) bonds. The bond - energy of \(N - H\) is \(391\ \text{kJ/mol}\).
The energy required to break bonds in reactants \(E_{break}=4\times391\ \text{kJ/mol}\).
$$E_{break}=1564\ \text{kJ/mol}$$

Step2: Calculate the energy released when bonds are formed

For the products, there are \(3\) \(H - H\) bonds (bond - energy of \(H - H\) is \(436\ \text{kJ/mol}\)) and \(1\) \(N\equiv N\) bond (bond - energy of \(N\equiv N\) is \(945\ \text{kJ/mol}\)).
The energy released when bonds are formed \(E_{form}=(3\times436 + 945)\ \text{kJ/mol}\).
$$E_{form}=(1308+945)\ \text{kJ/mol}=2253\ \text{kJ/mol}$$

Step3: Calculate the heat of reaction \(\Delta H\)

Using the formula \(\Delta H=E_{break}-E_{form}\).
$$\Delta H = 1564\ \text{kJ/mol}-2253\ \text{kJ/mol}=-689\ \text{kJ/mol}$$

Answer:

\(-689\ \text{kJ/mol}\)