### Video Transcript

Aqueous hydrogen peroxide, H2O2, decomposes into oxygen and water. The rate of decomposition of hydrogen peroxide is found to be 3.20 times 10 to the minus two molars per hour. Write a balanced chemical equation for this reaction.

We’ve already been given the symbol for our reactant, hydrogen peroxide. The symbol for our first product, oxygen, is O2. And the symbol for water is H2O. We should include state symbol. So, for hydrogen peroxide, that’s aqueous, aq, gas for oxygen. And we can assume that water is produced in liquid form, since there is no mention of high temperature in the question. The first part of our chemical equation is the decomposition of hydrogen peroxide into water and oxygen.

It doesn’t matter which way around you put the water and the oxygen. The question asks for a balanced chemical equation. So, we should check that this is balanced, and if it isn’t, balance it. An initial count shows that we have balanced hydrogens but imbalanced oxygens. We have two oxygen atoms on the left and three on the right. The easiest thing to try at this stage is simply to double up our reactants. This brings our hydrogen out of balance once more.

However, doubling up the amount of water we produce bounces both hydrogen and oxygen, giving us a balanced chemical equation: 2H2O2 aqueous reacts to form 2H2O liquid plus O2 gas. It’s, of course, perfectly valid if you balance the equation in a different way, for instance, using half an oxygen molecule rather than the full oxygen molecule. The way I’ve shown just gives nice round coefficients for each chemical. I’m gonna store away that chemical equation so that we can reference it later.

What is the rate of formation of water?

To solve this part of the question, we’re going to need to combine what we learned in part (a) with the other half of the question, which tells us the rate of decomposition of hydrogen peroxide is 3.20 times 10 to the minus two molars per hour. The rate of decomposition of hydrogen peroxide will be for that particular instant when it was measured. The rate of decomposition would’ve been obtained from the gradient of the concentration–time graph. Given that we are consuming hydrogen peroxide, it’s decomposing, the rate of change of hydrogen peroxide concentration would be equal to negative 3.20 times 10 to minus two molars per hour.

So, how do we use this information to work out the rate of production of water? Well, we know from the balanced chemical equation that if we consume two molecules of hydrogen peroxide, we’ll produce two molecules of water. Therefore, for each molecule of hydrogen peroxide consumed, we produce one molecule of water. Therefore, the rate of production of water is equal to the negative rate of change of hydrogen peroxide.

So, the rate of production of water is equal to the negative of negative 3.20 times 10 to the minus two molars per hour, which is equal to positive 3.20 times 10 to the minus two molars per hour. It makes sense from the equation that we would produce water as quickly as we consume hydrogen peroxide.

What is the rate of formation of oxygen?

We can determine the rate of formation of oxygen in exactly the same way as we did for water. As we know from part (a), when we consume two molecules of hydrogen peroxide, we generate one molecule of water. Therefore, for each molecule of hydrogen peroxide we consume, we’re going to generate half a molecule of oxygen.

So, the rate at which we form oxygen molecules is equal to the negative half times the rate of change of hydrogen peroxide. So, the rate of formation of oxygen molecules is equal to the negative half times negative 3.20 times 10 to the minus two molars per hour, which evaluates to 1.60 times 10 to the minus two molars per hour.

It’s perfectly reasonable to have done these calculations using positive numbers only. The reason I worked with negative numbers here was to show clearly that certain things were being consumed and certain things were being produced. Either way, the final answer for the rate of formation of oxygen should be 1.60 times 10 to minus two molars per hour.