Video: Calculating the Rate of Formation of Nitrogen and Hydrogen Using Decomposition Rate of Ammonia from a Balanced Chemical Equation

Ammonia decomposes into hydrogen and nitrogen at high temperature. The rate of decomposition of ammonia is found to be 2.10 × 10⁻⁶ M/s. a) Write a balanced chemical equation for this reaction. b) What is the rate of formation of nitrogen? c) What is the rate of formation of hydrogen?

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Video Transcript

Ammonia decomposes into hydrogen and nitrogen at high temperature. The rate of decomposition of ammonia is found to be 2.10 times 10 to the minus six molars per second. Write a balanced chemical equation for this reaction.

Ammonia has the symbol NH₃. Hydrogen in its natural state goes around as a diatom, H₂. And, so too does nitrogen. Our job is to write the balanced chemical equation for this reaction at high temperature. So, our first assumption is that all the components of the reaction are gases. We start off with the direct decomposition of an ammonia molecule, NH₃, into nitrogen and hydrogen gases. If we count up the number for each element, we can see that neither nitrogen nor hydrogen are balanced.

Since we have more nitrogen on the right-hand side, let’s balance that first. Doubling up the number of ammonia molecules on the left-hand side balances the nitrogens, giving us two on each side. But, we now have six hydrogens on the left and only two on the right. Increasing the number of hydrogen molecules on the right to three means that we have balanced the hydrogens, six on both sides. We now have a balanced chemical equation because all the elements are in balance. So, the final balanced chemical equation is 2NH₃ gas react to form N₂ gas plus 3H₂ gas. If you put the nitrogen and the hydrogen the other way around, that’s fine. I’m gonna store that chemical equation so that we could reference it later.

What is the rate of formation of nitrogen?

To answer this question, we’re going to need to combine the results from part a with the other bit of the question, which tells us the rate of decomposition of ammonia. The rate of decomposition of ammonia here is equal to 2.10 times 10 to the minus six molars per second. What this means is if the rate of decomposition was constant and we chose a random concentration for the ammonia of 10.5 times 10 to minus six molar, after a second we’d have only 8.4 times 10 to the minus six molar. This means that we’re actually using up the ammonia.

So, in the traditional way of writing it, we would say the rate of change of the ammonia concentration is equal to a negative 2.10 times 10 to the minus six molars per second. So, how does this help us find the rate of formation of nitrogen? Well, from the equation in part a, we can see that consuming two molecules of ammonia gives us one molecule of nitrogen. And therefore, for each molecule of ammonia we consume, we generate half a molecule of nitrogen.

So, our rate of formation of our product, nitrogen, is equal to a negative half times the rate of change for the ammonia. So, the rate of formation of nitrogen is equal to a negative half times negative 2.10 times 10 to the minus six molars per second. Giving us our final answer for the rate of formation of nitrogen equal to 1.05 times 10 to the minus six molars per second. With a ratio of two to one, it makes sense that we consume ammonia twice as fast as we generate nitrogen.

What is the rate of formation of hydrogen?

Calculating the rate of formation of hydrogen is a little bit more difficult than doing it for nitrogen. Ammonia and hydrogen are not in a simple ratio. It takes two molecules of ammonia to produce three molecules of hydrogen. Two ammonia molecules in and three hydrogen molecules out. Therefore, for each ammonia molecule consumed, we produce one and a half or three over two hydrogen molecules.

So, the rate of formation of hydrogen is equal to a negative three over two multiplied by the rate of change of ammonia. Which we demonstrated in part b was equal to negative 2.10 times 10 to the minus six molars per second. So, the rate of formation of hydrogen is equal to negative three over two multiplied by negative 2.10 times 10 to the minus six molars per second. Which equates to 3.15 times 10 to the minus six molars per second.

Now, you may have noticed that I chose to work with negative numbers here. I could’ve cancelled the negatives earlier and just done a simpler calculation. However, what’s shown here is a clear difference between consuming one chemical and generating another, that isn’t necessarily clear if you only work with positive numbers. If you worked only with positive numbers and got the same result, that’s fine. Either way, your answer to what is the rate of formation of hydrogen should be 3.15 times 10 to the minus six molars per second.

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