### Video Transcript

A student designs an experiment to measure the rate of reaction of a 0.25-gram zinc
plate with dilute hydrochloric acid, HCl. The apparatus used in the experiment is shown in figure one. Complete and balance the equation for the reaction between zinc and hydrochloric
acid. Blank plus blank reactive form ZnCl₂ plus blank.

For this question, almost everything you need to know is in figure one. Zinc and hydrochloric acid are the reactants and hydrogen gas is one of the
products. We already know from the equation that one of the products is zinc chloride. Therefore, we have all the information we need to fill it in.

The symbol for zinc is Zn. You can find this on your periodic table. The symbol for hydrochloric acid is HCl as given in the question. And the symbol for the second product hydrogen gas is H₂ since hydrogen is a
diatom.

Now, we can move on to the balancing portion of this question. We have one zinc on each side, one hydrogen on the left, and two hydrogens on the
right and one chlorine on the left and two chlorines on the right. We can balance the hydrogens and the chlorines by adding one more equivalent of
hydrochloric acid.

You could have approached this question from another direction, remembering that a
metal plus an acid react to form a metal salt plus hydrogen. So the complete and balanced the equation for the reaction between zinc and
hydrochloric acid is Zn plus 2HCl react to form ZnCl₂ plus H₂.

The results of this experiment are shown in table one. Plot these results on the grid in figure two and draw a line of best fit.

The first thing you’ll notice about figure two is that there are no numbers on the
axes. This is the first thing that we need to make a decision on. Once we’ve added values to the axes, we can then plot the points and then draw the
line of best fit.

The first step of adding values to the graph is easy. We can see from the table that the lowest values for the two columns are zero. Therefore, we can put zero, zero at the origin.

The next step is to find the largest value for the 𝑥-axis, which is 540 seconds. If we put 540 at the far right, we would be able to fit all the values on the
graph. We have seven major graduations along the 𝑥-axis. But 540 doesn’t divide evenly by seven. We would have to do some very interesting multiplication and division in order to
place the other points.

It’s much easier if we pick a sensible value to assign to each increment, in this
case 100. This way, we can easily fit 540 seconds on the graph and be able to place the other
points with simple calculations. Now, onto the 𝑦-axis, the largest value we need to place on the 𝑦-axis is 85
centimetres cubed. We have five graduations separated by 10 small squares. If we assign the value of 20 to this increment, we can easily accommodate 85
centimetres cubed. And this allows us to place the other points easily.

Now that we’ve set up our axes, we can start plotting points. Zero, zero is straightforward to place at the origin. For the second point, we go six small squares to the right, each with a value of 10
making 60 seconds in total, and then we move upwards and stop just short of 20 at
19.8 centimetres cubed.

It may help you to understand that each small square is worth 10 along the 𝑥-axis
and two along the 𝑦-axis. Now, onto the third point, we place this at 120 seconds and 38.1 centimetres cubed,
and so on with the fourth point, and so on and so forth.

Now that we’ve plotted the points, we can move on to draw the line of best fit. Before we draw the line of best fit, we need to identify that there’s an anomaly in
the data. The value at 300 seconds is lower than what the rest of the data would indicate. We should ignore this value when drawing the line of best fit. You can see that this line is actually a curve. This is perfectly okay.

The student repeats the experiment using 0.25 grams of thin zinc wire in place of the
zinc plate. Draw a line on figure two to show the predicted results of this experiment. Label the line A.

So the only change the student made in this experiment compared to the old one was to
substitute a zinc plate with a zinc wire. The two masses are exactly the same. All we need to do is make a judgement call on how this change is going to affect the
rate of production of hydrogen gas throughout the experiment and of course label the
line A.

The question we need to answer here is, “Plate versus wire, which is going to produce
the faster reaction?” One thing we can be certain of is that the wire will have a much greater surface area
than the plate. Therefore, with the rate of the reaction, at least to begin with is going to be much
faster for the zinc wire. As a consequence of this, we would expect a steeper line for the volume versus time
graph.

However, the mass of the wire and the mass of the plate are the same. Therefore, we would expect the two lines to have exactly the same endpoint: 85
centimetres cubed of gas. This is what a good answer might look like. Yours can be steeper or shallower as long as it’s steeper than the original line. And of course, it needs to finish at 85 centimetres cubed. To finish the answer, we add the label A.

Use collision theory to explain the changes in rate observed during the reaction of
the zinc plate with hydrochloric acid.

In our conical flask, we have our zinc, our acid. And the reaction between them depends on collisions between the acid particles and
the zinc. Over time, the concentration of these acid particles will decrease. Therefore, as the concentration decreases, the collision frequency also
decreases. Therefore, as the reaction progresses, the rate of the reaction decreases.

Now, we have to put all this into full sentences. As the reaction proceeds, the acid is consumed and the acid concentration
decreases. Protons from the acid collide less frequently with the surface of the zinc. So there are fewer reactions per unit time. Therefore, the rate of reaction decreases.

Now, alternatively, you could have talked about the surface area of the zinc, which
gradually reduces throughout the reaction. This would have been perfectly valid.

The student repeats the experiment that monitors the reaction by using a pH metre to
determine the HCl concentration. The student plots the results of this experiment and draws a line of best fit. This line is shown in figure three. Use Figure three to calculate the mean reaction rate between the start and the end of
the reaction. Give your answer to three significant figures.

The first thing the question asks us to calculate is the mean reaction rate between
the start and the end. That means the average of the whole reaction. So we can ignore the profile of the reaction and focus on the beginning and the
end.

Let’s begin with the start point. The reaction begins at time zero with an HCl concentration of 1.032 molar. If you read this value slightly differently, that’s okay. Now, the endpoint is slightly trickier. Is it here? Well, even though the data extends that far, the graph itself starts to level out
before then. This means that the reaction ended sooner than 720 seconds.

So the endpoint we’re actually looking at is 600 seconds, the point at which the
graph started to level out. We go up from 600 and then trace along to the 𝑦-axis to a concentration of 0.879
molar. Again, if you got a slightly different value, that’s okay.

Now that we have the start and the end points, we can start to think about how to
calculate the mean reaction rate. The rate of reaction is equal to the change in concentration divided by the reaction
time. The change in concentration is equal to the concentration at the start 1.032 minus
the concentration at the end 0.879 which is equal to 0.153 molar. The reaction time is equal to 600 minus zero which is equal to 600 seconds. The rate of the reaction is equal to 0.153 divided by 600 which is equal to 2.55
times 10 to the minus four molars per second.

The question asks that we give the answer to three significant figures. So the mean reaction rate for the reaction between zinc and hydrochloric acid is 2.55
times 10 to the minus four molars per second.

Showing your working on figure three, estimate the instantaneous rate of reaction
after 200 seconds. Give your answer in standard form.

The first thing we need to answer is what exactly is an instantaneous rate of
reaction. An instantaneous rate of reaction is the rate of the reaction at a specific instant
in time. Now, a gradient is a measure of the rate of change of one variable relative to
another.

So the rate at 200 seconds is equal to the gradient of the tangent at 200
seconds. This can be expressed as a change in 𝑦 divided by a change in 𝑥. Now, we can start by drawing the tangent. Now, a properly drawn tangent should exactly touch the point at 200 seconds.

We could measure the gradient of this tangent at any point, taking any change in 𝑦
and dividing it by the equivalent change in 𝑥. It’s easier to extend the tangent to the axes and read directly from there. This gives us values of 1.002 on the 𝑦-axis at the top and 0.86 on the bottom and
433 on the 𝑥-axis and zero at the bottom. Δ𝑦 is equal to 1.002 minus 0.860 which is equal to 0.142 molar. Δ𝑥 is equal to 433 minus zero which is equal to 433 seconds.

To calculate the rate at 200 seconds, we take 0.142 and divide it by 433. This is equal to 0.00032794 molars per second which is equal to 3.28 times 10 to the
minus four molars per second in standard form and to three significant figures.