# Video: Bond Energy Changes in the Andrussov Oxidation

In the Andrussow oxidation, methane, ammonia, and oxygen react to produce hydrogen cyanide (HCN) and water. The energies of selected bonds are listed in the table. (a) Give a balanced chemical equation for this reaction. (b) Calculate the total energy change in bond energy for this reaction per mole of hydrogen cyanide produced.

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### Video Transcript

In the Andrussow oxidation methane, ammonia, and oxygen react to produce hydrogen cyanide, HCN, and water. The energies of selected bonds are listed in the table. Give a balanced chemical equation for this reaction.

For this question, you do not need to be familiar with the Andrussow oxidation. You just need to identify that the reactants are methane, ammonia, and oxygen and that the products are hydrogen cyanide and water. As we’ve been asked to make the balanced chemical equation, we’re going to need the chemical formulas of all the components.

Methane is the simplest hydrocarbon and has the formula CH4. Ammonia has the symbol NH3. And oxygen here refers to the oxygen molecule O2. The symbol for hydrogen cyanide has already been given, HCN. And water has the formula H2O. Now, we can start writing out the chemical equation, starting with methane, ammonia, and oxygen as the reactants and hydrogen cyanide and water as the products. Now, we can move on to balancing the chemical equation.

We have one equivalent of carbon on both sides, a total of seven hydrogens on the left-hand side and three hydrogens on the right-hand side, one nitrogen on both sides, and two equivalents of oxygen on the left-hand side and one on the right-hand side. Carbon and nitrogen are balanced, while oxygen and hydrogen are imbalanced. Balancing hydrogen is going to be quite complicated because it appears in two reactants and two products. Oxygen, on the other hand, appears in one reactant and one product. So, let’s start with balancing the oxygen.

If we add one more hydrogen molecule to the right-hand side, we double the amount of oxygen, meaning that we have two equivalents on both sides. So, the oxygen is balanced. The hydrogen, however, is not balanced. We have seven equivalents on the left-hand side but only five on the right. Now, this is quite a difficult equation to balance. Because hydrogen is in so many components, we could try many different combinations and get nowhere.

But we do know that we currently can’t get the system to balance with only one equivalent of methane, ammonia, oxygen, and hydrogen cyanide. The easiest way to approach this is simply to try slightly more methane. This brings the carbon out of balance, so we need to add one more hydrogen cyanide molecule to balance the carbons. But this brings the nitrogen out of balance. This can be corrected by doubling the amount of ammonia, bringing the number of equivalents of hydrogen on both sides to two.

Now, we have a new equation that we can try to balance. Since we’re introducing oxygen in its pure form, we can modify the amount of oxygen to balance out any amount of oxygen on the product side. So, what we can do is change the amount of water we have to balance the hydrogen and then change the amount of oxygen to balance the oxygens again. We have 14 hydrogens on the reactant side, but only six on the product side. If we bring the total of water molecules up to six, we have 12 hydrogens from water, two from hydrogen cyanide, giving the 14 we need on the product side to balance it out.

But this has now unbalanced the oxygen. This can be fixed by adding two more oxygen molecules to the reactant side, giving us six equivalents of oxygen on both sides, meaning the equation is now balanced. This gives us our final equation 2CH4 plus 2NH3 plus 3O2 react to form 2HCN plus 6H2O. Now, it would’ve been possible to balance without doubling up the amount of methane, but it would’ve meant introducing a fractional term. With this one and a half oxygen molecules, the equation is balanced, and it’s a perfectly valid answer. However, I prefer to present it with full round numbers. As long as you have a balanced chemical equation, then your answer is correct.

Calculate the total energy change in bond energy for this reaction per mole of hydrogen cyanide produced.

To answer this part of the question, we’re going to need to fall back on our balanced chemical equation from part (a). The bond energies in the table indicate the energy required to break bonds. For instance, it would take 142 kilojoules to break one mole of oxygen oxygen single bonds. So, what we have to do for this question is work out the bonds in the reactants and the products, work out their energies, and then work out the difference.

The first step is to work out the energy of the reactants. So, we can start with methane, which has this structure. One molecule of methane has four CH single bonds. Since we have two molecules of methane, we have two times four CH bonds. This is the structure of ammonia. So, one molecule of ammonia has three NH bonds. So, our two molecules of ammonia have two times three NH bonds.

And finally, for the reactant side, we have oxygen with the structure O double bond O. Since we have three equivalents of oxygen, we have three equivalents of the oxygen oxygen double bond. So, this is the combined bonds of the reactants. From this, we want to take away the energy of the bonds of the products. Hydrogen cyanide has this structure. So, one molecule of hydrogen cyanide has one HC single bond and one CN triple bonds. So, in our two molecules of hydrogen cyanide, we have two HC bonds and two CN triple bonds.

And lastly, we come to water. One molecule of water contains two OH bonds. So, in six molecules of water, we have six times two OH bonds. Now, we have all the bonds for the products. The next step is to substitute in the values from the table: 411 for the CH bonds, 386 for the NH bonds, 494 for the oxygen oxygen double bonds, 411 for the HC bonds, 887 for the CN triple bonds, and lastly 459 for the oxygen hydrogen bonds.

The next step is to evaluate this and get our total energy change. Evaluating each term gives us this, which sums to 7086 kilojoules per mole for the reactants and 8104 kilojoules per mole for the products. This gives us a total energy change of 1018 kilojoules per mole.

However, this value is for the equation where we get two equivalents of hydrogen cyanide. The question asks us for the total energy change per mole of hydrogen cyanide produced. So, to get our total energy change per mole of hydrogen cyanide, we need to divide minus 1018 by two. This gives us a final figure of minus 509 kilojoules per mole of hydrogen cyanide.

The negative sign indicates that the total bond energy of the products is greater than the total bond energy of the reactants and that there is, overall, a release of energy. This means that the reaction of methane, ammonia, and oxygen to produce hydrogen cyanide and water is exothermic.