Video: Determining the Rate Law for the Formation of Hypoiodite

Hypochlorite (ClO⁻) and iodide ions react to form hypoiodite (IO⁻) and chloride ions: ClO⁻(aq) + l⁻(aq) ⟶ IO⁻(aq) + Cl⁻(aq). The initial rate of reaction was measured at three different reactant concentrations, as shown in the table. a) Determine the rate law for this reaction. b) Determine the appropriate units for the rate constant of this reaction. c) Expressing concentrations in molars and time in seconds, estimate, to 3 significant figures, the numerical value of the rate constant, 𝑘.

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Video Transcript

Hypochlorite ClO⁻ and iodide ions react to form hypoiodite IO⁻ and chloride ions. ClO⁻ aqueous plus I⁻ aqueous react to form IO⁻ aqueous plus Cl⁻ aqueous. The initial rate of reaction was measured at three different reacted concentrations, as shown in the table. Determine the rate law for this reaction.

A rate law is an equation that relates the rate of a reaction to the rate constant multiplied by the concentration of various reactants to various exponents. The rate constant is empirically determined and can be dependent on temperature and a few other factors. The power to which any given concentration is raised is referred to as the order with respect to that component.

If the order with respect to a component is zero, then the rate is independent of that component’s concentration. If the order is one, then the rate is proportional to that component’s concentration. So, if we double the concentration of that component, we’ll double the rate. If the order with respect to 𝑥 is two, then the rate is proportional to the square of 𝑥’s concentration. Under this scenario, if we double the concentration of 𝑥, we quadruple the rate, or multiply it by four. We can use these relationships to figure out the order in reverse.

The standard form for our rate equation will be 𝑘 multiplied by the concentrations of all the reactants raised to some unknown power, in this case 𝑥 and 𝑦. In the table, we have some initial concentrations for our hypochlorite ion and iodide ion. The initial rate is determined from the slope of the line right at the beginning of the concentration time graph. I’m going to label these initial conditions one, two, and three. By looking at the relationships between these initial conditions, we can figure out the rate law.

Let’s start by figuring out the order with respect to the hypochlorite ion. To do this, we use what’s called a pseudo rate law. If we don’t vary the concentration of I⁻, we can pretend that our rate law only has a term for ClO⁻. In this case, our rate constant 𝑘 prime simply wraps up all the terms from the I⁻ concentration. Using this pseudo rate law, we can compare any situations where the concentration of iodide ions does not change, in this case that’s scenarios one and three.

We can’t look at scenario two in this case because the concentration of I⁻ doesn’t match the concentration in the other two scenarios. If we look at scenario three and scenario one, we can see that the concentration of hypochlorite in scenario one is twice that for scenario three. But the rate for scenario one is four times that for scenario three. Four is two to the power of two. So, there we have our order.

We discussed before how doubling the concentration of an order-two component will quadruple the rate. And that’s what we’re seeing here. Now, we’re well on our way to figuring out the whole of the rate equation. All we have left to figure out is the order with respect to iodide. This time, we’re going to make a pseudo rate law by fixing the value of the concentration of hypochlorite, giving us a rate law in terms of the concentration of I⁻. This rate law is relevant to scenarios two and three, where the concentration of ClO⁻ doesn’t change.

We can’t use scenario one because the concentration of hypochlorite differs from the values in the other two scenarios. We can see that the concentration of iodide in scenario two is exactly double that for scenario three. Now, the rate has only increased by a factor of two, not four. This is what we’d expect if the reaction had an order with respect to iodide of one. So, our pseudo rate law is that the rate is proportional to the concentration of iodide. This gives us a final rate law of rate equal to the rate constant multiplied by the concentration of hypochlorite squared multiplied by the concentration of iodide. I’ll store away that expression because we’ll need it in a moment.

Determine the appropriate units for the rate constant of this reaction.

Here’s our rate constant in the rate expression. In circumstances like this, the rate is always given in molars per second. And as it’s standard, all our concentrations are in molars. Which means the units for ClO⁻ to the power of two is molars squared. In order to work out the units of 𝑘, we’ll need to rearrange all of these units. We can combine the molar units on the right-hand side into molars cubed. We can then divide through by molars cubed to give us the units of 𝑘 equal to one over molar squared seconds. But another way of writing molars is moles per liter. So, we could also write the units of 𝑘 as liters squared per mole squared per second.

Expressing concentrations in molars and time in seconds, estimate to three significant figures, the numerical value of the rate constant, 𝑘.

All the question means by numerical value is that it wants the estimate to be written without any units. Our first job is simply to rearrange the rate equation in terms of the rate constant 𝑘, which we can do by just dividing through by the concentration terms. Our next job is to pick a scenario and simply plug in the values from it. I’m going to pick the scenario with the biggest numbers so that it’s as accurate as possible. But you could pick the values from any scenario.

I’ve picked scenario one, so the rate I’m plugging in is 1.84 times 10 to the minus three molars per second. The question does say that we don’t need units at the end, so we could either remove them now or at the end. The concentration of iodide in scenario one is 2.00 times 10 to the minus three molars. And the concentration of hypochlorite is 4.00 times 10 to the minus three molars. Just be careful to make sure you square the concentration.

Evaluating the denominator of this fraction gives us 3.2 times 10 to the minus eight molars cubed. And finishing the evaluation of 𝑘 gives us 5.75 times 10 to the power four per molars squared per second. The question asks for our answer to three significant figures. And all it wants is the numerical value of 5.75 times 10 to the power of four.

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