A LEVEL CHEMISTRY EXAM QUESTION WALKTHROUGH - RATES 21

Hinzugefügt: 2026-05-17

[00:00:00]got another question for the rates of reaction topic. So, we're up to number 21 now. If you want to check out the other videos in the playlist, I'll put the link to that at the top of the screen. Now, hope you like the video.

[00:00:11]Hope you find it helpful. And if you haven't already subscribed to the channel, obviously, I'd love you to do so. But as always, the link to the questions in the description of the video if you want to try it first. Okay, so make a start. So, first thing I'm going to do is complete the table. So, we've got to do 1 / t. So 1 over 573 and the lin of k. So that's the lin of 15.6 * 10 -4.

[00:00:36]So there's your two answers there. 1.75 * 10us 3 and -6.46.

[00:00:43]Moving on to part two. So we've got to identify the piece of information that tells us the overall order of the reaction. Well, the piece of information are the units of the rate constant K.

[00:00:58]So, because we've got seconds to the minus one, I'll explain this in a second. It's first order overall. So, why is that? Well, if you think about a generic rate equation for a first order reaction. So, we've got rate equals K concentration of the species to the power one. So if you then rearrange for K and then put the units in, you can see moles per decime cubed are going to cancel top and bottom. All you're left with are seconds to minus one. So moving on to the next part, we've got to draw the graph of the lin of K against 1 / t.

[00:01:31]So we get a straight line graph. Mine looks like that. And then we've got to use the graph to calculate the activation energy. So the first thing I want to do while the graph's still on the screen is calculate the gradient of the line. So obviously to do that I need my change in y. So you can see my little scribbles there. So my change in y is the difference between those two numbers. So 2.8 in my case and my change in x is the difference between these two numbers here. So I've got 1.4 * 10us4.

[00:02:06]So you can see my gradient's coming out at a lovely round number of minus 20,000. The range allowed was between - 19,400 to - 20,400.

[00:02:18]So from the arinius equation, the gradient of the line is equal to minus EA / R. So the first thing I'm going to do is get rid of those minus signs, cancel them out. So therefore to find the activation energy we need to multiply the gradient by the gas constant. So my activation energy is coming out at 166,200 jewels per mole. Don't forget that the activation energy calculated from an urinous graph is in jewels per mole.

[00:02:51]That's because you've got jewels in your gas constant units. So all we need to do now is turn it into kilogjles per mole by dividing by a th00and. So my answer would be 166 kJ per mole and the range was between 161 and 170. And finally part five. So we're told that this substance here decomposes in a similar way to but 3N1. And I've just copied up the equation from earlier in the question. So the way I processed this was I looked at where the break in the molecule was in the example given and I've literally just applied it to the new molecule in the question which means that the alken produces the same one as before. So propane again whereas the carbonile compound this time is propon.