AS-Level Chemistry paper 1 2022 AQA Walk through

[@abdurkhan0621]

This was really useful, your explanation was clear and it was easy to understand so thank you

[@Ilovecloudsthatrain]

Thank you so much I was struggling with paper 1. And this has helped me. You also explained it quite well, especially the questions that involved equations. Thank you. :)

[@trxnqui]

for 7.2 how did you know that the forward reaction results in an increase in the amount of gas molecules (59:50)

[@letsgettothemarks]

When you have a LOW pressure for both lines. You can see yield is high. Yield means product which is always on the right or forward direction. So the graph shows us a LOW pressure results in the reactions equilibrium shifting to the right hand side or forwards. Le chatliers principle tells us low pressure forces the reaction to the side with most gas molecules. As low pressure increases the forward reaction (yield) it must be the side with the highest number of gas molecules

[@Vanessa-cv5yy]

on q7.5, why do you do one over and not just calculate Kc using the values for concentration?

[@letsgettothemarks]

For 7.5 we havent been given the concentrations. So how do we get Kc? Without concs. We learn in the stem at too of page that the Kc for the reverse of this reaction which is N2 + 3H2 goes to 2NH3 is 0.118 at 745k. In 745 the reaction is flipped around but still at 745k so we do 1 over kc to get the kc

[@letsgettothemarks]

You could get close or even the same answer if you put the concentrations in from the earlier question 7.4 as the temp is the same and you should get the same answer for Kc roughly 8.47 it will take just a bit longer. Just remember to reverse the reactants and products

[@Vanessa-cv5yy]

@@letsgettothemarks thank you

[@riemashanaachandragantha4655]

thank you very much for this video!!!! Please can you make more videos like this as it was really helpful when someone explains why the answer is the answer???

[@letsgettothemarks]

More to come!

[@mrwahssan]

hi i am stuck on a chem question - the question asked me to write the third ionisation energy of magnesium which i can do but then it asked me 'which sub-shell loses this electron' and i do not understand why it is the 2p shell

[@letsgettothemarks]

Mg has electronic configuration of 1S2 2S2 2P6 3S2 so the first electron is removed from 3S2 making mg+ 1s2 2s2 2p6 3s1 the 2nd electron removed makes mg2+ with a config of 1s2 2s2 2p6 then when you remove the 3rd electron you have to break into a new shell removing a 2p6 electron and leaving the config as 1s2 2s2 2p5 breaking into the new shell requires a lot more energy as the 2nd energy level is closer to the nucleus and so feels more attraction also there is less shielding and repulsion.

[@danielwalker9105]

if you dont show the chunky and dash lines for AsF5 in the right position do you loose the mark, also for question 7.5 why is it 1/kc for the reverse reaction is it because its basically the inverse of the quation the other way around

[@letsgettothemarks]

Aslong as you have the same number of dashed and chunky it doesnt matter the order if they are just reversed. Yes 1 over kc because equation reversed. But you could also use the concentrations from the first part and just reverse them in kc equation

[@danielwalker9105]

Thanks 🙏

[@sleepdeprivedalevelstudent]

hello! for question 01.1 would it then be right to say effective nuclear charge decreases down the group as no. shells increases/more shielding

[@letsgettothemarks]

@@sleepdeprivedalevelstudent Thats right. However easy mistake to make if you miss out the word effective. You used to gain marks in the old syllabus for talking about nuclear charge. If your worried about making mistakes you can just talk about force of attraction to the nucleus. More shielding less force of attraction to rhe nucleus.

[@ayeshaahmed8396]

Hello on q17 why is suphur ion larger than cl ion when they both have the same configuration?

[@letsgettothemarks]

Good question. Cl- has an extra proton in nucleus so greater attraction to outer shells making it slightly smaller. Same shielding same e- but one has more protons

[@danialmalik2513]

On 7.2: Yield is high when pressure is low and high yield always means moving to the right. Why can’t we say when pressure is high, yield is low and when pressure is low it means moving to the left. Why must we talk about the yield being high when pressure is low and not the other way around?

[@letsgettothemarks]

All depends on the reaction. If there are less gas moles on right then high pressure will move the eq position to the right increasing yield (product we are wanting normally written on right). If there are more gas molecules on the right then high pressure forces eq to left or reactant side so decreases the yield

[@vsiva2005]

@@letsgettothemarks For 7.2, How do you know it is increasing?

[@letsgettothemarks]

@@vsiva2005 when you look at graph an increase in pressure along x axis causes percentage yield to decrease for both temps . Think of high yield as equilibrium shift to right. So as increasing pressure causes low yield this must mean an eq shift to the left hand side (reactant side). Why would it shift to left hand side with high pressure? Thats because left hand side must have less gas molecules and a shift to that side will counter the change (increase in pressure). So we know the forward or right hand side must contain more gas molecules. If the right hand side or forward direction had LESS gas molecules then increasing pressure would increase YIELD. Also we know the sides do not have same ampunt of gas moles on each side as on the graph the lines are curved which shows pressure having effect.

[@letsgettothemarks]

Hi Danial your first sentence is correct but not your second one. When pressure is low on the graph it shows high yield which means it is moving to the right the forward side (product side)

[@zara-fe5sk]

hello sir, for question 6.3, how did you work out the equation? did you just use the information given to make an equation off the top of your head? i hope this makes sense i'm not fully sure how it works

[@letsgettothemarks]

So you have to know this one off by heart. Its in the section on halide ions as reducing agents. Normally in cgp and text book it is in the context of NaI and H2S04. For this equation it tells us Iodide reacts with sulfuric acid. As Iodide means I- this reacts with H2SO4. So we have I- + H2SO4 the reactants. Now we are told to show the equation to form the yellow solid. We have to know off by heart that sulfur is the yellow solid. So we write I- + H2S04 --> S Then we see we have H and O with sulfur so to balance this we need water on right so we have I- + H2SO4 --> S +.H20 NOW we must balance water as we have 4 x O in H2sO4 we get 4H2O on right I- + H2S04 --> S + 4H20 so we need 6 extra H+ iins on left to balance the water so I- + H2SO4 + 6H+ --> S + 4H20 We need I2 on right 2I- + H2so4 + 6H+ --> S +.4H30 +I2 However finally on left it is electrically charged 6H+.i need to cancel.this by having 6I- So.finally 6I- + 6H+ +H2SO4 --> S + 4H2O + 3I2

[@Vanessa-cv5yy]

on Q6.2 when writing the RHS, why don't you include the 3 before the Iodine?

[@letsgettothemarks]

Aha so in the overall equation in the stem there are 3I2 on the right hand side. However the question asks for the oxidation of Iodide ions to iodine so we literally just need to show how iodide ions lose electrons and form Iodine the simplest ratio to do this is the one shown in my answer but you could have any multiple you want aslong as it balances so 2I- -> I2 + 2e- could also be written as 6I- -> 3I2 + 6e-

[@Vanessa-cv5yy]

@@letsgettothemarks ohh okay thank you

[@Iq-.-]

Thank you 😊

[@vsiva2005]

Hi, For Question 2.2 Is it okay to write it as 8.69 x 10^-3 as in the mark scheme it doesn't say anything about that and left it as a 0.00869. For Question 5 can you write 2.01 x 10^-15 and again in the mark scheme doesn’t say anything else apart from 2.009 x 10^-15.

[@letsgettothemarks]

Se answers should be fine

[@letsgettothemarks]

Those answers are fine