Nitrogen oxides and carbon monoxide produced in the internal combustion engine of a
car are passed over precious metal catalysts as they pass through the catalytic
converter. Carbon monoxide is converted to carbon dioxide by reaction with oxygen. Balance the equation for this reaction.
So to balance the equation, we’re going to look at the number of atoms of each
element on the reactant side and the product side and adjust the coefficients of the
reactants and products until these match. Let’s start with carbon.
Assuming we have one equivalent of everything to start with, we have one carbon atom
on the left and one on the right. However, with oxygen, we have three on the left and two on the right. Now, one way of balancing this is to add only half an equivalent of oxygen rather
than a full equivalent. This gives us two atoms of oxygen on the left and two on the right.
However, I’m going to multiply the whole reaction equation through by two and remove
that fraction. This gives us two carbon atoms on the left, two carbon atoms on the right, four
oxygen atoms on the left, and four on the right. So both elements are balanced, and the overall reaction equation is balanced.
If you prefer, you can remove the one before O₂ because it is assumed that if there
is no number there, then there is one equivalent. So, the balanced equation for the conversion of carbon monoxide to carbon dioxide by
reaction with oxygen is 2CO plus O₂ react to form 2CO₂.
The precious metal catalysts in a catalytic converter are plated onto a fine
honeycomb structure. What is the purpose of this structure? Tick one box. To increase the catalyst surface area, to decrease the catalyst concentration, to
increase the strength of the converter, or to decrease the gas flow rate.
For this question, the first thing I’m going to do is review the introduction. Waste gases from an internal combustion engine are passed over precious metal
catalysts. Another word for precious is “expensive.” As with anything expensive, the designers of a catalytic converter will want to get
the most out for the smallest amount spent. This means they’ll want to maximise the efficiency of the catalysts and reduce the
cost. Therefore, they’ll want to use as little of the precious metals as possible. This is how the honeycomb structure looks.
However, in a catalytic converter, this structure would be much much smaller. Nitrogen oxides and carbon monoxide pass over this fine structure plated with the
precious metals and react.
Now let’s have a look at the options and see which one is most realistic. Increasing the catalyst surface area is a good thing to do. It ties in with the desire to maximise efficiency and reduce cost. The honeycomb structure does provide a very high surface area. Therefore, this answer is a good answer. But it may not be the best, so let’s look at the other options.
What about decreasing the catalyst concentration? Well, the honeycomb structure does spread out the catalyst over space. So the concentration does decrease. But decreasing the catalyst concentration isn’t fundamentally something that’s going
to help maximise efficiency. In fact, decreasing the catalyst concentration is only going to make the catalytic
converter less effective.
Now what about increasing the strength of the converter? Now while the honeycomb structure is relatively strong, in the question, we see that
it’s a very fine honeycomb structure. Given how little catalyst we want to be using, it’s unlikely that this honeycomb
structure is going to be particularly strong. And the outside housing of a catalytic converter likely made of aluminium or steel is
going to be much stronger than this honeycomb structure in the first place and
strong enough to handle anything that’s going to happen to it.
What about decreasing the gas flow rate? If you put a fine mesh in front of a gas, it will slow down a little. That’s true. However, the effect on the gas flow rate is going to be negligible. Therefore, of these four options, increasing the catalyst surface area is the one
that makes most sense. Therefore, this is the correct answer.
A freshly started car will gradually warm up. As the temperature changes, the efficiency of the precious metal catalysts changes.
Figure 1 shows the relative efficiency of a palladium catalyst converting carbon
monoxide to carbon dioxide. Describe the overall trend shown in Figure 1.
As we can see from Figure 1, the relationship between the temperature and the
relative efficiency of the catalyst isn’t a simple one. Up until about 115 degrees Celsius, there is a slow increase in the relative
efficiency. And between about 115 degrees C and 165 degrees C, there is a much more rapid
increase in the relative efficiency. And finally, between 165 degrees Celsius and 250 degrees Celsius, there is again a
more gradual increase in the relative efficiency.
Now the question doesn’t ask for a detailed breakdown of the behaviour of the
data. Instead, we’ve been asked to describe the overall trend. So what each of these regions have in common? In all these regions, as the temperature increases, so does the relative
efficiency. So the overall trend shown in Figure 1 is that as the temperature rises, the relative
efficiency of the catalyst increases.
Use Figure 1 to determine the difference in the relative efficiency of the palladium
catalyst at 125 degrees Celsius and 150 degrees Celsius.
Let’s start by having another look at Figure 1. What we need to do is find the values 125 and 150 on the 𝑥-axis; trace upwards,
preferably using a ruler, until we hit the line; and then trace sideways to the
𝑦-axis and read off the values. The difference in relative efficiency between these two temperatures is the
percentage at the top minus the percentage at the bottom, which is 70 percent minus
18 percent. This is equal to 52 percent. Therefore, the difference in the relative efficiency of the palladium catalyst at 125
degrees Celsius and 150 degrees Celsius is 52 percent.
There’s a little bit of margin for error in this question. You could have got this value plus or minus about four percent. Just remember, when you’re calculating a difference, you take the smaller value away
from the bigger one.