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.
