Question

Use bond energies from the table below to calculate the heat of reaction.

Use bond energies from the table below to calculate the heat of reaction. KJ

kJ

Answer 1

Concepts and reason

This problem can be solved by using the general equation for the heat of a reaction.

The heat of a reaction or change in enthalpy of a reaction can be estimated from bond energy. Heat energy is estimated after balancing the given equation.

Bond energy is defined as the amount of energy required to break the bond of 1 mole of a gaseous compound to produce its constituents.

Fundamentals

The change in enthalpy of a reaction is the difference of summation of bond energies of bond broken and the summation of the bond energies of the bonds formed. It is written as follow.

${\rm{\Delta }}{H^o} = \sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} - \sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}}$ …… (1)

First calculate the bond energy of the reactants or the bonds broken as follow.

$\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} = 2{\rm{mol}} \times {\rm{\Delta }}H\left( {{\rm{CO}}} \right) + 6{\rm{ mol}} \times {\rm{\Delta }}H\left( {{\rm{NH}}} \right)$

Substitute 799 kJ/mol for ${\rm{\Delta }}H\left( {{\rm{CO}}} \right)$ and 391 kJ/mol for ${\rm{\Delta }}H\left( {{\rm{NH}}} \right)$ .

$\begin{array}{c}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} = 2{\rm{mol}} \times \left( {799{\rm{ kJ/mol}}} \right) + 6{\rm{mol}} \times \left( {391{\rm{ kJ/mol}}} \right)\\\\ = \left( {1598 + 2346} \right){\rm{kJ}}\\\\ = 3944{\rm{ kJ}}\\\end{array}$

Now, calculate bond energy of product bonds formed.

$\begin{array}{l}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} = \left[ {2{\rm{ mol}} \times {\rm{\Delta }}H\left( {{\rm{CN}}} \right) + 4{\rm{mol}} \times {\rm{\Delta }}H\left( {{\rm{NH}}} \right) + 1{\rm{mol}} \times {\rm{\Delta }}H\left( {{\rm{CO}}} \right)} \right] + \\\\{\rm{ }}\left[ {2{\rm{mol}} \times {\rm{\Delta }}H\left( {{\rm{OH}}} \right)} \right]\\\end{array}$

Substitute 745 kJ/mol for ${\rm{\Delta }}H\left( {{\rm{CO}}} \right)$ , 391 kJ/mol for ${\rm{\Delta }}H\left( {{\rm{NH}}} \right)$ , 467 kJ/mol for ${\rm{\Delta }}H\left( {{\rm{OH}}} \right)$ and 305 kJ/mol for ${\rm{\Delta }}H\left( {{\rm{CN}}} \right)$ .

$\begin{array}{c}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} = 4{\rm{mol}}\left( {391{\rm{ kJ/mol}}} \right) + 2{\rm{mol}}\left( {305{\rm{ kJ/mol}}} \right) + 1{\rm{mol}}\left( {745{\rm{ kJ/mol}}} \right) + \\\\{\rm{ }}2{\rm{mol}}\left( {467{\rm{ kJ/mol}}} \right)\\\\ = \left( {1564 + 610 + 745 + 934} \right){\rm{ kJ}}\\\\ = 3853{\rm{ kJ}}\\\end{array}$

Now, substitute 3944 kJ for $\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}}$ and 3853 kJ for $\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}}$ in equation (1).

$\begin{array}{c}\\{\rm{\Delta }}{H^o} = \sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} - \sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} \\\\ = \left( {3944 - 3853} \right){\rm{ kJ}}\\\\ = 91{\rm{ kJ}}\\\end{array}$

Ans:

The heat of the reaction is 91 kJ.