Animated Solution for Chemistry - Chemical Kinetics: At $518^{\circ }\text{C}$, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of $363\text{ Torr}$, was $1.00{\text{ Torr s}}^{-1}$ when $5\%$ had reacted and $0.5{\text{ Torr s}}^{-1}$ when $33\%$ had reacted. The order of the reaction is :

The problem of finding the order of a chemical reaction can sometimes feel like a puzzle with missing pieces. You might look at the initial pressure of $363\text{ Torr}$ and wonder, "How does this fit into the equation?" But as we'll see, the beauty of ratios is that they often make the "missing" or "extra" pieces completely irrelevant!

We are given the decomposition of gaseous acetaldehyde:

The problem provides the rate of reaction at two different stages: 1. When $5\%$ of the reactant has decomposed, the rate is $1.00{\text{ Torr s}}^{-1}$. 2. When $33\%$ of the reactant has decomposed, the rate is $0.5{\text{ Torr s}}^{-1}$.

The crucial insight here is to focus on what remains, not what has reacted. The rate of a reaction depends on the concentration (or partial pressure) of the reactants currently present in the system.

If the initial amount is $a$, then: - In the first case, the unreacted amount is $(100\%-5\%)=95\%$ of $a$, which is $0.95a$. - In the second case, the unreacted amount is $(100\%-33\%)=67\%$ of $a$, which is $0.67a$.

For an $m$-th order reaction, the rate law is expressed as: where $r$ is the rate, $k$ is the rate constant, and $(a-x)$ is the unreacted concentration.

Let's plug our two cases into this master equation:

We have two equations and three unknowns ($k$, $a$, and $m$). This is where the magic of ratios comes in. By dividing the first equation by the second, we can eliminate both $k$ and $a$ in one swift move!

Notice how $k$ and $a^m$ cancel out perfectly:

Now, we just need to simplify the numbers:

At first glance, solving for $m$ might seem to require logarithms. But let's look closely at the number $1.417$. If you've practiced your square roots, you'll immediately recognize that $\sqrt{2}\approx 1.414$.

This means that $1.417$ is approximately $\sqrt{2}$, or $2^{1/2}$.

For this equation to hold true, $m$ must be exactly $2$.

The reaction is second order! We solved the entire problem without ever needing the initial pressure of $363\text{ Torr}$. It was a classic distractor, placed there to test your confidence in the fundamental rate laws.